7.6 Exercise Problems and Solutions

Exercises

Q1. Assign oxidation number to the underlined elements in each of the following species:
(a) $\mathrm{NaH}_2 \mathrm{\underline{P}O}_4$
(b) $\mathrm{NaH}\mathrm{\underline{S}} \mathrm{O}_4$
(c) $\mathrm{H}_4 \mathrm{\underline{P}}_2 \mathrm{O}_7$
(d) $\mathrm{K}_2 \mathrm{\underline{Mn}O}_4$
(e) $\mathrm{Ca\underline{O}}_2$
(f) $\mathrm{Na\underline{B}H}_4$
(g) $\mathrm{H}_2 \mathrm{\underline{S}}_2 \mathrm{O}_7$
(h) $\mathrm{KAl}\left(\mathrm{\underline{S}O}_4\right)_2 \cdot 12 \mathrm{H}_2 \mathrm{O}$

Answer: Let the oxidation no. of underlined element in all the given compounds $=x$
(a) We know that, Oxidation number of $\mathrm{Na}=+1$
Oxidation number of $\mathrm{H}=+1$
Oxidation number of $\mathrm{O}=-2$
$+1 \quad +1 \quad x-2$
$\mathrm{Na} \quad \mathrm{~H}_2 \quad \mathrm{~PO}_4$
Since the sum of oxidation number of various atoms in $\mathrm{NaH}_2 \mathrm{PO}_4$ (neutral) is zero.
Thus, the oxidation number of $\mathrm{P}$ in $\mathrm{NaH}_2 \mathrm{PO}_4$ $=+5$.
(b) $+1 \quad +1 \quad x-2$
$\text { In } \mathrm{Na} \quad \mathrm{H} \quad \quad \mathrm{SO}_4: 1(+1)+1(+1)+x+4(-2)=0$
Therefore, $x=+6$
Thus, the oxidation number of $\mathrm{S}$ in $\mathrm{NaHSO}_4=+6$.
(c) $+1 \quad x \quad -2$
$\text { In } \mathrm{H}_4 \quad \mathrm{P}_2 \quad \mathrm{O}_7: 4(+1)+2(x)+7(-2)=0$
Therefore, $x=+5$
Thus, the oxidation number of $\mathrm{P}$ in $\mathrm{H}_4 \mathrm{P}_2 \mathrm{O}_7=+5$.
(d) $+1 \quad x \quad -2$
$\text { In } \mathrm{K}_2 \quad \mathrm{Mn} \quad \mathrm{O}_4: 2(+1)+1(x)+4(-2)=0$
Therefore, $x=+6$
Thus, the oxidation number of $\mathrm{Mn}$ in $\mathrm{K}_{2} \mathrm{MnO}_4$ $=+6$.
(e) $+2 \quad x$
$\text { In } \mathrm{Ca} \quad \mathrm{O}_2: 2+2 x=0 \text { or } x=-1$
Thus, the oxidation number of oxygen in $\mathrm{CaO}_2=-1$
(f) In $\mathrm{NaBH}_4$, hydrogen is present as hydride ion. Therefore, its oxidation number is -1 . Thus,
$+1 \quad \quad x \quad -1$
$\text { In } \mathrm{Na} \quad \mathrm{B} \quad \mathrm{H}_4: 1(+1)+x+4(-1)=0 \text { or } x=+3$
Thus, the oxidation number of $\mathrm{B}$ in $\mathrm{NaBH}_4=+3$
(g) $+1 \quad x \quad -2$
$\ln \mathrm{H}_2 \quad \mathrm{~S}_2 \quad \mathrm{O}_7: 2(+1)+2(x)+7(-2)=0$
Therefore, $x=+6$
Thus, the oxidation number of $\mathrm{S}$ in $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7=+6$.
(h) $+1 \quad +3 \quad x-2$
$\text { In } \mathrm{K} \quad \mathrm{Al} \quad \quad \left(\mathrm{SO}_4\right)_2 \cdot 12 \mathrm{H}_2 \mathrm{O} \text { : }$
$+1+3+2 x+8(-2)+12 \times 0=0 \text { or } x=+6$
Thus, the oxidation number of $\mathrm{S}$ in
$\mathrm{KAl}\left(\mathrm{SO}_4\right)_2 \cdot 12 \mathrm{H}_2 \mathrm{O}=+6$

Q2. What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) $\mathrm{K\underline{I}}_3$
(b) $\mathrm{H}_2 \mathrm{\underline{S}}_4 \mathrm{O}_6$
(c) $\mathrm{\underline{Fe}}_3 \mathrm{O}_4$
(d) $\mathrm{\underline{C}H}_3 \mathrm{\underline{C}H}_2 \mathrm{OH}$
(e) $\mathrm{\underline{C}H}_3 \mathrm{\underline{C}OOH}$

Answer: (a) In $\mathrm{KI}_3$, since the oxidation number of $\mathrm{K}$ is +1 , therefore, the average oxidation number of iodine $=-1 / 3$. In the structure, $\mathrm{K}^{+}[\mathrm{I}-\mathrm{I}<\mathrm{-I}]^{-}$, a coordinate bond is formed between I2 molecule and I ion. The oxidation number of two iodine atoms forming the 12 molecule is zero while that of iodine ion forming the coordinate bond is -1 . Thus, the $\mathrm{O} . \mathrm{N}$. of three iodine atoms in $\mathrm{Kl}_3$ are 0,0 and -1 respectively.
(b) The structure of $\mathrm{H}_2 \mathrm{~S}_4 \mathrm{O}_6$ is shown below:
The O.N. of each of the S atoms linked with each other in the middle is zero while that of each of the remaining two S-atoms is +5.

(c) $x \quad-2$
$\mathrm{Fe}_3 \mathrm{O}_4$
Let O.N. of $\mathrm{Fe}=x$, then $3 x+4(-2)=0$ or $\quad x=+\frac{8}{3}$ (average)
By stoichiometry $\mathrm{Fe}_3 \mathrm{O}_4$ is
$+2 \quad-2 \quad+3 \quad-2$
$\mathrm{Fe} \quad \mathrm{~O} \cdot \quad \quad \mathrm{Fe}_2 \quad \mathrm{O}_3$.
Thus, Fe has O.N. of +2 and +3.
(d)

In this molecule, $\mathrm{C}-2$ is attached to three $\mathrm{H}$-atoms (less electronegative than carbon) and one $\mathrm{CH}_2 \mathrm{OH}$ group (more electronegativity than carbon).
Therefore O.N. of C-2 $=3(+1)+x+1(-1)=0$ or $x=-2$.
$\mathrm{C}-1$ is however, attached to one $\mathrm{OH}$ (charge $=-1$ ) and one $\mathrm{CH}_3$ (charge $=+1$ ), and two $\mathrm{H}$-atoms, O.N. of +1 .
Therefore, O.N. of C-1 $=1(+1)+2(+1)+x+1(-1)=0$ or $x=-2$

(e)

In this molecule, $\mathrm{C}-2$ is attached to three $\mathrm{H}$-atoms (less electronegative than carbon) and one-COOHgroup(more electronegativity than carbon).
Therefore, O.N. of C-2 $=3(+1)+x+1(-1)=0$ or $x=-2$.
C-1 is, however, attached to one oxygen atom by a double bond, one $\mathrm{OH}$ (charge $=-1$ ) and one $\mathrm{CH}_3$ (charge $=+1$ ) group, therefore, O.N. of C-1 = $1(+1)+x+1(-2)+1(-1)=0$ or $x=+2$.

Q3. Justify that the following reactions are redox reactions:
(a) $\mathrm{CuO}(\mathrm{s})+\mathrm{H}_2(\mathrm{~g}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$
(b) $\mathrm{Fe}_2 \mathrm{O}_3(\mathrm{~s})+3 \mathrm{CO}(\mathrm{g}) \rightarrow 2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{CO}_2(\mathrm{~g})$
(c) $4 \mathrm{BCl}_3(\mathrm{~g})+3 \mathrm{LiAlH}_4(\mathrm{~s}) \rightarrow 2 \mathrm{~B}_2 \mathrm{H}_6(\mathrm{~g})+3 \mathrm{LiCl}(\mathrm{s})+3 \mathrm{AlCl}_3(\mathrm{~s})$
(d) $2 \mathrm{~K}(\mathrm{~s})+\mathrm{F}_2(\mathrm{~g}) \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}(\mathrm{s})$
(e) $4 \mathrm{NH}_3(\mathrm{~g})+5 \mathrm{O}_2(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$

Answer: (a)

$+2 -2 \quad 0 \quad \quad \quad \quad 0 \quad +2 -2$
$\mathrm{CuO}_{(s)}+\mathrm{H}_{2(g)} \rightarrow \mathrm{Cu}_{(s)}+\mathrm{H}_2 \mathrm{O}_{(g)}$
O.N. of $\mathrm{Cu}$ decreases from +2 in $\mathrm{CuO}$ to 0 in $\mathrm{Cu}$ but that of $\mathrm{H}$ increases from 0 in $\mathrm{H}_2$ to +1 in $\mathrm{H}_2 \mathrm{O}$. Therefore, $\mathrm{CuO}$ is reduced to $\mathrm{Cu}$ but $\mathrm{H}_2$ is oxidised to $\mathrm{H}_2 \mathrm{O}$. Thus, this is a redox reaction.
(b)

$+3 -2 \quad +2 \quad \quad \quad \quad 0 \quad \quad +4$
$\mathrm{Fe}_2 \mathrm{O}_{3(s)}+3 \mathrm{CO}_{(g)} \rightarrow 2 \mathrm{Fe}_{(s)}+3 \mathrm{CO}_{2(g)}$
$\mathrm{O} . \mathrm{N}$. of $\mathrm{Fe}$ decreases from +3 in $\mathrm{Fe}_2 \mathrm{O}_3$ to 0 in Fe while that of $\mathrm{C}$ increases from +2 in $\mathrm{CO}$ to +4 in $\mathrm{CO}_2$. Thus, $\mathrm{Fe}_2 \mathrm{O}_3$ is reduced and $\mathrm{CO}$ is oxidised. Thus, this is a redox reaction.
(c)

$+3 -1 \quad +1 +3 -1 \quad -3 +1 \quad +1 -1 \quad +3 -1$
$4 \mathrm{BCl}_{3(g)}+3 \mathrm{LiAlH}_{4(s)} \rightarrow 2 \mathrm{~B}_2 \mathrm{H}_{6(g)}+3 \mathrm{LiCl}_{(s)}+3 \mathrm{AlCl}_{3(s)}$
O.N. of $\mathrm{B}$ decreases from +3 in $\mathrm{BCl}_3$ to -3 in $\mathrm{B}_2 \mathrm{H}_6$ while that of $\mathrm{H}$ increases from -1 in $\mathrm{LiAlH}_4$ to +1 in $\mathrm{B}_2 \mathrm{H}_6$. Therefore, $\mathrm{BCl}_3$ is reduced while $\mathrm{LiAlH}_4$ is oxidised. Thus, this is a redox reaction.
(d)

$2 \mathrm{~K}_{(s)}+\mathrm{F}_{2(g)} \rightarrow 2 \mathrm{~K}^{+} \mathrm{F}^{-}_{(s)}$
Each $\mathrm{K}$ atom has lost one electron to form $\mathrm{K}^{+}$ while $\mathrm{F}_2$ has gained two electrons to form two $\mathrm{F}^{-}$ions. Therefore, $\mathrm{K}$ is oxidised while $\mathrm{F}_2$ is reduced. Thus, it is a redox reaction.
(e)

$-3 +1 \quad \quad 0 \quad \quad +2 -2 \quad +1 -2$
$4 \mathrm{NH}_{3(g)}+5 \mathrm{O}_{2(g)} \rightarrow 4 \mathrm{NO}_{(g)}+6 \mathrm{H}_2 \mathrm{O}_{(g)}$
$\mathrm{O} . \mathrm{N}$. of $\mathrm{N}$ increases from -3 in $\mathrm{NH}_3$ to +2 in $\mathrm{NO}$ while that of $\mathrm{O}$ decreases from 0 in $\mathrm{O}_2$ to -2 in $\mathrm{NO}$ or $\mathrm{H}_2 \mathrm{O}$. Therefore, $\mathrm{NH}_3$ is oxidised while $\mathrm{O}_2$ is reduced. Thus, it is a redox reaction.

Q4. Fluorine reacts with ice and results in the change:
$\mathrm{H}_2 \mathrm{O}(\mathrm{s})+\mathrm{F}_2(\mathrm{~g}) \rightarrow \mathrm{HF}(\mathrm{g})+\mathrm{HOF}(\mathrm{g})$
Justify that this reaction is a redox reaction.

Answer:

$+1-2 \quad 0 \quad \quad+1-1 \quad+1-2+1$
$\mathrm{H}_2 \mathrm{O}_{(\mathrm{s})}+\mathrm{F}_{2(\mathrm{~g})} \rightarrow \mathrm{HF}_{(\mathrm{g})}+\mathrm{HOF}_{(\mathrm{g})}$
Since fluorine can undergo oxidation as well as reduction hence, it is an example of redox reaction.

Q5. Calculate the oxidation number of sulphur, chromium and nitrogen in $\mathrm{H}_2 \mathrm{SO}_5$. $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ and $\mathrm{NO}_3^{-}$. Suggest structure of these compounds. Count for the fallacy.

Answer: Oxidation number of sulphur in $\mathrm{H}_2 \mathrm{SO}_5$ :
Let the oxidation number of $\mathrm{S}=x$ then $(+1) \times 2+x+(-2) \times 5=0$ or $2+x-10=0$
$\begin{aligned} & \Rightarrow \quad x-8=0 \\ & \therefore \quad x=+8 \end{aligned}$
The maximum O.N. of S cannot be more than 6 since it has only 6 electrons in the valence shell. This fallacy is overcome if we calculate the O.N. of sulphur by chemical bonding method. The structure of $\mathrm{H}_2 \mathrm{SO}_5$ is

It has two peroxide oxygen with O.N. $=-1$ and three oxygens with O.N. $=-2$
Thus, $2 \times(+1)+x+2(-1)+3 \times(-2)=0$ $+2+x-2-6=0 \Rightarrow x-6=0 \Rightarrow x=+6$ Thus, O.N. of sulphur in $\mathrm{H}_2 \mathrm{SO}_5=+6$
Oxidtion number of chromium in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ :
Let the oxidation number of chromium $=x$
$\begin{aligned} & \therefore 2 x+7(-2)=-2 \Rightarrow 2 x-14=-2 \\ & \Rightarrow 2 x=-2+14 \\ & \Rightarrow 2 x=+12 \Rightarrow x=+6 \end{aligned}$
Thus, the oxidation number of chromium $=+6$
Oxidation number of nitrogen in $\mathrm{NO}_3^{-}$:
Let the oxidation number of nitrogen $=x$ then $x+(-2) \times 3=-1 \Rightarrow x-6=-1$ $\Rightarrow x=-1+6 \Rightarrow x=+5$ Thus, the oxidation number of nitrogen $=+5$

Q6. Write formulas for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin{IV}) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate
(f) Chromium(III) oxide

Answer: (a) $\mathrm{HgCl}_2$
(b) $\mathrm{NiSO}_4$
(c) $\mathrm{SnO}_2$
(d) $\mathrm{Tl}_2 \mathrm{SO}_4$
(e) $\mathrm{Fe}_2\left(\mathrm{SO}_4\right)_3$
(f) $\mathrm{Cr}_2 \mathrm{O}_3$

Q7. Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5 .

Answer: 

$\begin{array}{|c|c|c|c|} \hline \text { Compound } & \begin{array}{l} \text { O.N. of } \\ \text { Carbon } \end{array} & \text { Compound } & \begin{array}{l} \text { O.N. of } \\ \text { Nitrogen } \end{array} \\ \hline \mathrm{CH}_4 & -4 & \begin{array}{l} \mathrm{NH}_3 \\ \mathrm{NH}_4 \mathrm{Cl} \end{array} & -3 \\ \hline \mathrm{CH}_3-\mathrm{CH}_3 & -3 & \mathrm{NH}_2-\mathrm{NH}_2 & -2 \\ \hline \begin{array}{l} \mathrm{CH}_2=\mathrm{CH}_2 \\ \text { or } \mathrm{CH}_3 \mathrm{Cl} \end{array} & -2 & \begin{array}{l} \mathrm{NH}=\mathrm{NH}, \\ \mathrm{NH}_2 \mathrm{OH} \end{array} & -1 \\ \hline \mathrm{CH} \equiv \mathrm{CH} & -1 & \mathrm{~N} \equiv \mathrm{N} & 0 \\ \hline \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6 & 0 & \mathrm{~N}_2 \mathrm{O} & +1 \\ \hline \mathrm{C}_6 \mathrm{Cl}_6 & +1 & \text { NO } & +2 \\ \hline \mathrm{CHCl}_3 & +2 & \begin{array}{l} \mathrm{N}_2 \mathrm{O}_3 \\ \mathrm{HNO}_2 \end{array} & +3 \\ \hline(\mathrm{COOH})_2 & +3 & \mathrm{~N}_2 \mathrm{O}_4 & +4 \\ \hline \mathrm{CCl}_4, \mathrm{CO}_2 & +4 & \begin{array}{l} \mathrm{N}_2 \mathrm{O}_5 \\ \mathrm{HNO}_3 \end{array} & +5 \\ \hline \end{array}$

Q8. While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?

Answer: The oxidation state of sulphur in sulphur dioxide is +4 . It can be oxidised to +6 oxidation state or reduced to +2. Therefore, sulphur dioxide acts as a reducing agent as well as oxidising agent. Similarly, the oxidation state of oxygen in hydrogen peroxide is – 1 . It can be oxidised to $\mathrm{O}_2$ (zero oxidation state) or reduced to $\mathrm{H}_2 \mathrm{O}$ or $\mathrm{OH}^{-}(-2$ oxidation state) and therefore, acts as reducing as well as oxidising agents. However, both ozone and nitric acid can only decrease their oxidation number and therefore, act only as oxidising agents.

$$\begin{array}{l}\mathrm{N}\mathrm{H}=\mathrm{N}\mathrm{H},\\ {\mathrm{N}\mathrm{H}}_2\mathrm{}\mathrm{}\end{array}$$

Q9. Consider the reactions:
(a) $6 \mathrm{CO}_2$ (g) $+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{aq})+6 \mathrm{O}_2(\mathrm{~g})$
(b) $\mathrm{O}_3$ (g) $+\mathrm{H}_2 \mathrm{O}_2(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})+2 \mathrm{O}_2$ (g)
Why it is more approprlate to write these reactions as :
(a) $6 \mathrm{CO}_2(\mathrm{~g})+12 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6(\mathrm{aq})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{l})+6 \mathrm{O}_2(\mathrm{~g})$
(b) $\mathrm{O}_3(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})$
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Answer: (a) Most accepted mechanism of photosynthesis is supposed to occur in two steps. In the first step, $\mathrm{H}_2 \mathrm{O}$ decomposes to give $\mathrm{H}_2$ and $\mathrm{O}_2$ in presence of chlorophyll and the $\mathrm{H}_2$ thus produced reduces $\mathrm{CO}_2$ to $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{C}_6$
and $\mathrm{H}_2 \mathrm{O}$ molecules are also produced as shown below:
$12 \mathrm{H}_2 \mathrm{O}_{(l)} \rightarrow 12 \mathrm{H}_{2(g)}+6 \mathrm{O}_{2(g)} \dots(i)$
$6 \mathrm{CO}_{2(g)}+12 \mathrm{H}_{2(g)} \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(s)}+6 \mathrm{H}_2 \mathrm{O}_{(l)} \dots(ii)$
$\overline{6 \mathrm{CO}_{2(g)}+12 \mathrm{H}_2 \mathrm{O}_{(l)} \rightarrow \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_{6(s)}+6 \mathrm{H}_2 \mathrm{O}_{(l)}+6 \mathrm{O}_{2(g)}} \dots(iii)$
Therefore, it is more appropriate to write the equation for photosynthesis as (iii) because it emphasises that 12 molecules of $\mathrm{H}_2 \mathrm{O}$ are used per molecule of carbohydrate formed and 6 molecules of $\mathrm{H}_2 \mathrm{O}$ are produced during the process.
(b) The purpose of writing $\mathrm{O}_2$ two times suggests that $\mathrm{O}_2$ is being obtained from each of the two reactants.
$\begin{gathered} \mathrm{O}_{3(g)} \longrightarrow \mathrm{O}_{2(g)}+\mathrm{O}_{(g)} \\ \frac{\mathrm{H}_2 \mathrm{O}_{2(g)}+\mathrm{O}_{(g)} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(l)}+\mathrm{O}_{2(g)}}{\mathrm{O}_{3(g)}+\mathrm{H}_2 \mathrm{O}_{2(l)} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(l)}+\mathrm{O}_{2(g)}+\mathrm{O}_{2(g)}} \end{gathered}$
The path of reactions (a) and (b) can be determined by tracer technique method using labelled $\mathrm{H}$ or $\mathrm{O}, \mathrm{H}_2 \mathrm{O}^{18}$ or $\mathrm{H}_2 \mathrm{O}$ in reaction (a) or by using $\mathrm{H}_2 \mathrm{O}_2^{18}$ or $\mathrm{O}_3^{18}$ in reaction (b).

Q10. The compound $\mathrm{AgF}_2$ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?

Answer: The oxidation state of $\mathrm{Ag}$ in $\mathrm{AgF}_2$ is +2 . But, +2 is an unstable oxidation state of $\mathrm{Ag}$. Therefore, whenever $\mathrm{AgF}_2$ is formed, silver readily accepts an electron to form $\mathrm{Ag}^{+}$.
$\mathrm{Ag}^{2+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}^{+}$
This helps to bring the oxidation state of $\mathrm{Ag}$ down from +2 to a more stable state of +1 . As a result, $\mathrm{AgF}_2$ acts as a very strong oxidizing agent.

Q11. Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxddation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Answer: (i) $\mathrm{C}$ is a reducing agent while $\mathrm{O}_2$ is an oxidising agent. If excess of carbon is burnt in a limited supply of $\mathrm{O}_2$, $\mathrm{CO}$ is formed in which oxidation state of $\mathrm{C}$ is +2 but when $\mathrm{O}_2$ is in excess $\mathrm{CO}$ formed gets oxidised to $\mathrm{CO}_2$ in which oxidation state of $\mathrm{C}$ is +4.

(ii) $\mathrm{P}_4$ is a reducing agent while $\mathrm{Cl}_2$ is an oxidising agent. When excess of $\mathrm{P}_4$ is used, $\mathrm{PCl}_3$ is formed in which the oxidation state of $\mathrm{P}$ is +3 . When excess of $\mathrm{Cl}_2$ is used, the initially formed $\mathrm{PCl}_3$ reacts further to form $\mathrm{PCl}_5$ in which the oxidation state of $\mathrm{P}$ is +5.

(iii) $\mathrm{Na}$ is a reducing agent and $\mathrm{O}_2$ is an oxidising agent. When excess of $\mathrm{Na}$ is used, sodium oxide is formed in which the oxidation state of $\mathrm{O}$ is -2 . If, excess of $\mathrm{O}_2$ is used, $\mathrm{Na}_2 \mathrm{O}_2$ is formed in which the oxidation state of $\mathrm{O}$ is -1.

Q12. How do you count for the following observations?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas $\mathrm{HCl}$, but if the mixture contains bromide then we get red vapour of bromine. Why ?

Answer:(a) In neutral medium, $\mathrm{KMnO}_4$ acts as an oxidant as follows:
$\mathrm{MnO}_4{ }^{-}+2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2+40 \mathrm{H}^{-}$
In laboratory, alkaline $\mathrm{KMnO}_4$ is used to oxidise toluene to benzoic acid.

In industry alcoholic $\mathrm{KMnO}_4$ is preferred due to following reasons :
1. Cost of adding a base or an acid is avoided as a base ( $\mathrm{OH}^{-}$ions) is produced during the reaction itself.
2. An organic polar solvent, ethyl alcohol, helps in mixing of two reactants $\mathrm{KMnO}_4$ (due to its polar nature) and toluene (because, an organic compound).
(b) When a chloride such as $\mathrm{NaCl}$ is heated with conc. $\mathrm{H}_2 \mathrm{SO}_4, \mathrm{HCl}$ is evolved which is not a good reducing agent and is not oxidised with conc. $\mathrm{H}_2 \mathrm{SO}_4$
$\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \stackrel{\Delta}{\longrightarrow} \mathrm{NaHSO}_4+\mathrm{HCl}$
But when $\mathrm{NaBr}$ is heated with conc. $\mathrm{H}_2 \mathrm{SO}_4$, $\mathrm{HBr}$ produced being a good reducing agent is oxidised by conc. $\mathrm{H}_2 \mathrm{SO}_4$ to give $\mathrm{Br}_2$ (red vapours).
$\mathrm{H}_2 \mathrm{SO}_{4(aq)} \xrightarrow{\text { heat }} \mathrm{SO}_{2(g)}+\mathrm{H}_2 \mathrm{O}_{(l)}+[\mathrm{O}]$
$\mathrm{NaBr}_{(\mathrm{s})}+\mathrm{H}_2 \mathrm{SO}_{4(\mathrm{aq})} \xrightarrow{\text { heat }} \mathrm{NaHSO}_{4(a q)}+\mathrm{HBr}_{(\mathrm{g})}$
$2 \mathrm{HBr}_{(g)}+[\mathrm{O}] \longrightarrow \mathrm{H}_2 \mathrm{O}_{(l)}+\mathrm{Br}_{2(g)}$

Q13. Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:
(a) $2 \mathrm{AgBr}$ (s) $+\mathrm{C}_6 \mathrm{H}_6 \mathrm{O}_2$ (aq) $\rightarrow 2 \mathrm{Ag}$ (s) $+2 \mathrm{HBr}$ (aq) $+\mathrm{C}_6 \mathrm{H}_4 \mathrm{O}_2$ (aq)
(b) $\mathrm{HCHO}(\mathrm{l})+2\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{HCOO}^{-}(\mathrm{aq})+4 \mathrm{NH}_3(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
(c) $\mathrm{HCHO}(\mathrm{l})+2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{Cu}_2 \mathrm{O}(\mathrm{s})+\mathrm{HCOO}^{-}(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
(d) $\mathrm{N}_2 \mathrm{H}_4(\mathrm{l})+2 \mathrm{H}_2 \mathrm{O}_2(\mathrm{l}) \rightarrow \mathrm{N}_2(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
(e) $\mathrm{Pb}$ (s) $+\mathrm{PbO}_2$ (s) $+2 \mathrm{H}_2 \mathrm{SO}_4$ (aq) $\rightarrow 2 \mathrm{PbSO}_4$ (s) $+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$

Answer:

$\begin{array}{|c|c|c|} \hline & \begin{array}{c} \text { Reducing agent } \\ \text { (Substance } \\ \text { oxidised) } \end{array} & \begin{array}{c} \text { Oxidising agent } \\ \text { (Substance } \\ \text { reduced) } \end{array} \\ \hline \text { (a) } & \mathrm{C}_6 \mathrm{H}_6 \mathrm{O}_2 \text { (aq) } & \mathrm{AgBr}_{(s)} \\ \hline \text { (b) } & \mathrm{HCHO}_{(l)} & {\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]_{(aq)}^{+}} \\ \hline \text { (c) } & \mathrm{HCHO}_{(l)} & \mathrm{Cu}^{2+}{ }_{(aq)} \\ \hline \text { (d) } & \mathrm{N}_2 \mathrm{H}_{4(l)} & \mathrm{H}_2 \mathrm{O}_{2(l)} \\ \hline \text { (e) } & \mathrm{Pb}_{(s)} & \mathrm{PbO}_{2(s)} \\ \hline \end{array}$

$$\begin{array}{c}\text{ Oxidising agent }\\ \text{ (Substance }\\ \end{array}$$

Q14. Consider the reactions :
$2 \mathrm{~S}_2 \mathrm{O}_3^{2-}(\mathrm{aq})+\mathrm{I}_2(\mathrm{~s}) \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})$
$\mathrm{S}_2 \mathrm{O}_3^{2-}(\mathrm{aq})+2 \mathrm{Br}_2(\mathrm{l})+5 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{SO}_4^{2-}(\mathrm{aq})+4 \mathrm{Br}^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq})$
Why does the same reductant, thiosulphate react differently with lodine and bromine?

Answer: The average O.N. of $\mathrm{S}$ in $\mathrm{S}_2 \mathrm{O}_3^{2-}$ is +2 while in $\mathrm{S}_4 \mathrm{O}_6^{2-}$ it is +2.5 . The O.N. of $\mathrm{S}$ in $\mathrm{SO}_4^{2-}$ is +6 . Since $\mathrm{Br}_2$ is a stronger oxidising agent, it oxidises $\mathrm{S}$ of $\mathrm{S}_2 \mathrm{O}_3^{2-}$ to a higher oxidation state of +6 and hence forms $\mathrm{SO}_4^{2-}$ ion. $\mathrm{I}_2$, however, being a weaker oxidising agent oxidises $\mathrm{S}$ of $\mathrm{S}_2 \mathrm{O}_3^{2-}$ ion to a lower oxidation state of +2.5 in $\mathrm{S}_4 \mathrm{O}_6^{2-}$ ion.

Q15. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Answer: The halogens $\left(X_2\right)$ have strong electron accepting tendency and have positive standard oxidation potential values. They are therefore, powerful oxidising agents. The decreasing order of oxidising powers of halogens is:
$\mathrm{X}_2: \mathrm{F}_2> \quad \quad \quad \quad \mathrm{Cl}_2> \quad \quad \quad \mathrm{Br}_2> \quad \quad \mathrm{I}_2$
$\mathrm{E}^{\ominus}:+2.87 \mathrm{~V} \quad \quad+1.36 \mathrm{~V} \quad+1.09 \mathrm{~V} \quad+0.54 \mathrm{~V}$
Fluorine is the strongest oxidising agent (oxidant) because it can liberate the other halogens from their respective compounds. For example,

$\begin{aligned} & 2 \mathrm{KCl}+\mathrm{F}_2 \longrightarrow 2 \mathrm{KF}+\mathrm{Cl}_2 \\ & 2 \mathrm{KBr}+\mathrm{F}_2 \longrightarrow 2 \mathrm{KF}+\mathrm{Br}_2 \\ & 2 \mathrm{KI}+\mathrm{F}_2 \longrightarrow 2 \mathrm{KF}+\mathrm{I}_2 \end{aligned}$
Among halogen acids $(\mathrm{HX})$, the $\mathrm{HI}$ is the strongest reducing agent or reductant because it has minimum bond dissociation energy:
Halogen acid: $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \mathrm{HF} \quad \mathrm{HCl} \quad \mathrm{HBr} \quad \mathrm{HI}$
Bond Dissociation: enthalpy ( $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ $566 \quad 431 \quad 366 \quad 299$
The iodination of methane is of reversible nature because $\mathrm{HI}$ produced in the reaction being a reducing agent converts iodomethane back to methane.
$\begin{aligned} & & & & \mathrm{CH}_4+\mathrm{I}_2 & \longrightarrow \mathrm{CH}_3 \mathrm{I}+\mathrm{HI} \\ & & & & \quad \quad \mathrm{CH}_3 \mathrm{I}+\mathrm{HI} & \longrightarrow \mathrm{CH}_4+\mathrm{I}_2 \end{aligned}$
$\text { Net reaction: } \overline{\mathrm{CH}_4+\mathrm{I}_2 \rightleftharpoons \mathrm{CH}_3 \mathrm{I}+\mathrm{HI}}$

Q16. Why does the following reaction occur?
$\mathrm{XeO}_6^{4-}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{XeO}_3(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
What conclusion about the compound $\mathrm{Na}_4 \mathrm{XeO}_6$ (of which $\mathrm{XeO}_6^{4-}$ is a part) can be drawn from the reaction.

Answer:

$\mathrm{XeO}_6^{4-}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq})+6 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow \mathrm{XeO}_3(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
In this reaction, O.N. of Xe decreases from +8 in $\mathrm{XeO}_6^{4-}$(perxenate ion) to +6 in $\mathrm{XeO}_3$ while that of $\mathrm{F}$ increases from -1 in $\mathrm{F}^{-}$to 0 in $\mathrm{F}_2$. Therefore, $\mathrm{XeO}_6^{4-}$ is reduced while $\mathrm{F}^{-}$is oxidised. From this reaction it is concluded that $\mathrm{Na}_4 \mathrm{XeO}_6$ is a stronger oxidising agent than $\mathrm{F}_2$.

Q17. Consider the reactions:
(a) $\mathrm{H}_3 \mathrm{PO}_2(\mathrm{aq})+4 \mathrm{AgNO}_3(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_3 \mathrm{PO}_4(\mathrm{aq})+4 \mathrm{Ag}(\mathrm{s})+4 \mathrm{HNO}_3(\mathrm{aq})$
(b) $\mathrm{H}_3 \mathrm{PO}_2(\mathrm{aq})+2 \mathrm{CuSO}_4(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_3 \mathrm{PO}_4(\mathrm{aq})+2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq})$
(c) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}(\mathrm{l})+2\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+4 \mathrm{NH}_3(\mathrm{aq})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})$
(d) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}(\mathrm{l})+2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \text { No change observed. }$
What inference do you draw about the behaviour of $\mathrm{Ag}^{+}$and $\mathrm{Cu}^{2+}$ from these reactions?

Answer: Reactions (a) and (b) indicate that both $\mathrm{AgNO}_3$ and $\mathrm{CuSO}_4$ oxidise $\mathrm{H}_3 \mathrm{PO}_2$ to $\mathrm{H}_3 \mathrm{PO}_4$. Hence, both are oxidising agents. Reactions (c) and (d) suggest that $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]^{+}$ oxidises $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$ (benzaldehyde) to $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}$ (benzoate ion) but $\mathrm{Cu}^{2+}$ ions cannot oxidise $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CHO}$ to $\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{-}$. Therefore, from the above reactions, we infer that $\mathrm{Ag}^{+}$ion is a stronger oxidising agent than $\mathrm{Cu}^{2+}$ ion.

Q18. Balance the following redox reactions by ion – electron method :
(a) $\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_2 \text { (s) }+\mathrm{I}_2(\mathrm{~s}) \text { (in basic medium) }$
(b) $\mathrm{MnO}_4^{-} \text {(aq) }+\mathrm{SO}_2 \text { (g) } \rightarrow \mathrm{Mn}^{2+} \text { (aq) }+\mathrm{HSO}_4^{-} \text {(aq) (in acidic solution) }$
(c) $\mathrm{H}_2 \mathrm{O}_2 \text { (aq) }+\mathrm{Fe}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \text { (l) (in acidic solution) }$
(d) $\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{SO}_2(\mathrm{~g}) \rightarrow \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{SO}_4^{2-} \text { (aq) (in acidic solution) }$

Answer: (a) The skeleton equation is:
$\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{I}^{-}(\mathrm{aq}) \rightarrow \mathrm{MnO}_2(\mathrm{~s})+\mathrm{I}_2(\mathrm{~s})$
(i) The O.N. of the atoms involved in the equation is:
$\stackrel{+7 -2}{\left(\mathrm{MnO}_4^{-}\right)}+\stackrel{-1}{\mathrm{I}^{-}} \longrightarrow \stackrel{+4 -2}{\mathrm{MnO}_2}+\stackrel{0}{\mathrm{I}_2}$
(ii) The species involved in the oxidation and reduction half reaction :

(iii) Oxidation half reaction : $\mathrm{I}^{-} \longrightarrow \mathrm{I}_2$
Reduction half reaction : $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{MnO}_2$
(iv) Balancing the oxidation half reaction
$2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 e^{-} \dots(i)$
(v) Balancing the reduction half reaction
(1) As the decrease in O.N. is 3, therefore, adding $3 e^{-}$on the reactant side
$\mathrm{MnO}_4^{-}+3 e^{-} \longrightarrow \mathrm{MnO}_2$
(2) To balance the oxygen atoms, add two $\mathrm{H}_2 \mathrm{O}$ molecules on the product side
$\mathrm{MnO}_4^{-}+3 e^{-} \longrightarrow \mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$
(3) To balance the charges, add $4 \mathrm{OH}^{-}$on the product side. Then to balance $\mathrm{H}$ atoms, add four $\mathrm{H}_2 \mathrm{O}$ molecules on the reactant side.

$\begin{aligned} & \mathrm{MnO}_4^{-}+3 e^{-}+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-}+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{MnO}_4^{-}+3 e^{-}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-} \dots(ii) \end{aligned}$
Thus, the reduction half reaction is balanced.
(vi) Adding the two half reactions In order to equate the electrons, multiply eqn. (i) by 3 and eqn. (ii) by 2 . Add the two equations.

(b) The skeleton equation is:
$\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{SO}_2(\mathrm{~g}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{HSO}_4^{-}(\mathrm{aq})$
(i) The O.N. of atoms involved in the equation is:
$+7-2 \quad+4-2 \quad+2 \quad -1+6-2$
$\mathrm{MnO}_4^{-}+\quad \mathrm{SO}_2 \longrightarrow \mathrm{Mn}^{2+}+\mathrm{HSO}_4^{-}$
(ii) The species involved in the oxidation and reduction half reactions:

(iii) Oxidation half reaction : $\mathrm{SO}_2 \longrightarrow \mathrm{HSO}_4^{-}$
Reduction half reaction : $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$
(iv) Balancing the oxidation half reaction (1) As the increase in O.N. is 2, therefore, add two electrons on the product side to balance change in O.N.
$\mathrm{SO}_2 \longrightarrow \mathrm{HSO}_4^{-}+2 e^{-}$
(2). In order to balance the number of oxygen atoms, add two $\mathrm{H}_2 \mathrm{O}$ molecules on the reactant side and then to balance $\mathrm{H}$ atoms add $3 \mathrm{H}^{+}$on the product side.
$\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HSO}_4^{-}+3 \mathrm{H}^{+}+2 e^{-} \dots(i)$
(v) Balancing the reduction half reaction
The reduction half reaction is :
$\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$
(1) As the decrease in O.N. is 5, therefore add $5 e^{-}$on the reactant side,
$\mathrm{MnO}_4^{-}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}$
(2) In order to balance the no. of oxygen atoms, add four $\mathrm{H}_2 \mathrm{O}$ molecules on the product side and then to balance $\mathrm{H}$ atoms add $8 \mathrm{H}^{+}$on the reactant side.
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \dots(ii)$
(vi) Adding the two half, reactions In order to equate the electrons, multiply eqn.
(i) by 5 and eqn. (ii) by 2 . Add the two eqns.

(c) Oxidation half equation :
$\mathrm{Fe}_{(a q)}^{2+} \longrightarrow \mathrm{Fe}_{(a q)}^{3+}+e^{-} \dots(i)$
Reduction half equation :
$\mathrm{H}_2 \mathrm{O}_{2(a q)}+2 \mathrm{H}_{(a q)}^{+}+2 e^{-} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}_{(l)} \dots(ii)$
Multiplying eqn. (i) by 2 and adding it to eqn. (ii), we get
$\mathrm{H}_2 \mathrm{O}_{2(a q)}+2 \mathrm{Fe}_{(a q)}^{2+}+2 \mathrm{H}_{(a q)}^{+} \longrightarrow 2 \mathrm{Fe}_{(a q)}^{3+}+2 \mathrm{H}_2 \mathrm{O}_{(l)}$
(d) Oxidation half equation :
$\mathrm{SO}_{2(g)}+2 \mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{SO}_4^{2-}{ }_{(a q)}+4 \mathrm{H}_{(aq)}^{+}+2 e^{-} \dots(i)$
Reduction half equation :
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}{ }_{(a q)}+14 \mathrm{H}_{(a q)}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}_{(a q)}^{3+}+7 \mathrm{H}_2 \mathrm{O}_{l} \dots(ii)$
Multiplying eqn. (i) by 3 and adding to eqn. (ii) we get,
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}{ }_{(a q)}+3 \mathrm{SO}_{2(\mathrm{g})}+2 \mathrm{H}_{(a q)}^{+} \longrightarrow 2 \mathrm{Cr}_{(a q)}^{3+}+3 \mathrm{SO}_4^{2-}{ }_{(a q)}+\mathrm{H}_2 \mathrm{O}_{(l)}$

Q19. Balance the following equations in basic medium by lon-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) $\mathrm{P}_4(\mathrm{~s})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{PH}_3(\mathrm{~g})+\mathrm{H_2PO}_2^{-}(\mathrm{aq})$
(b) $\mathrm{N}_2 \mathrm{H}_4(\mathrm{l})+\mathrm{ClO}_3^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g})$
(c) $\mathrm{Cl}_2 \mathrm{O}_7(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \rightarrow \mathrm{ClO}_2^{-}(\mathrm{aq})+\mathrm{O}_2(\mathrm{~g})+\mathrm{H}^{+}$

Answer: (a) Oxidation number method :

$P_4$ acts both as an oxidising as well as a reducing agent.
Oxidation number method:
Total decrease in O.N. of $\mathrm{P}_4$ in $\mathrm{PH}_3=3 \times 4=12$
Total increase in O.N. of $\mathrm{P}_4$ in $\mathrm{H}_2 \mathrm{PO}_2^{-}=1 \times 4=4$
Therefore, to balance increases decreases in O.N. multiply $\mathrm{PH}_3$ by 1 and $\mathrm{H}_2 \mathrm{PO}_2^{-}$ by 3 , we have,$\mathrm{P}_4(\mathrm{~s})+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{PH}_3(g)+3 \mathrm{H}_2 \mathrm{PO}_2^{-}(a q)$
To balance $\mathrm{O}$ atoms, multiply $\mathrm{OH}^{-}$by 6 , we have,
$\mathrm{P}_4(\mathrm{~s})+6 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{PH}_3(g)+3 \mathrm{H}_2 \mathrm{PO}_2^{-}(a q)$
To balance $\mathrm{H}$ atoms, add $3 \mathrm{H}_2 \mathrm{O}$ to L.H.S. and $3 \mathrm{OH}^{-}$to the R.H.S., we have,
$\mathrm{P}_4(s)+6 \mathrm{OH}^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{PH}_3(g)+3 \mathrm{H}_2 \mathrm{PO}_2^{-}(a q)+3 \mathrm{OH}^{-}(a q)$
$\mathrm{P}_4(s)+3 \mathrm{OH}^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{PH}_3(g)+3 \mathrm{H}_2 \mathrm{PO}_2^{-}(a q) \dots(i)$
Thus, Eq. (i) represents the correct balanced equation.
Ion electron method. The two half reactions are: Oxidation half reaction:
$\mathrm{P}_4(s) \longrightarrow \mathrm{H}_2 \mathrm{PO}_2^{-}(a q) \dots(ii)$
Balancing $P$ atoms, we have,
$\stackrel{0}{\mathrm{P}_4}(s) \longrightarrow 4 \mathrm{H}_2 \stackrel{+1}{\mathrm{PO}_2^{-}}(a q)$
Balance O.N. by adding electrons,
$\mathrm{P}_4(s) \longrightarrow 4 \mathrm{H}_2 \mathrm{PO}_2^{-}(a q)+4 e^{-}$
Balance charge by adding $8 \mathrm{OH}^{-}$ions,
$\mathrm{P}_4(\mathrm{~s})+8 \mathrm{OH}^{-}(a q) \longrightarrow 4 \mathrm{H}_2 \mathrm{PO}_2^{-}(a q)+4 e^{-} \dots(iii)$
$\mathrm{O}$ and $\mathrm{H}$ get automatically balanced. Thus, Eq. (iii) represents the balanced oxidation half reaction.
Reduction half reaction:
$\stackrel{0}{\mathrm{P}_4}(\mathrm{~s}) \longrightarrow \stackrel{-3}{\mathrm{PH}}_3(\mathrm{~g})$
Balancing $\mathrm{P}$ atoms, we have,
$\mathrm{P}_4(\mathrm{~s}) \longrightarrow 4 \mathrm{PH}_3(\mathrm{~g})$
Balance O.N. by adding electrons,
$\mathrm{P}_4(\mathrm{~s})+12 e^{-} \longrightarrow 4 \mathrm{PH}_3(\mathrm{~g})$
Balance charge by adding $12 \mathrm{OH}^{-}$ions,
$\mathrm{P}_4(\mathrm{~s})+12 e^{-} \longrightarrow 4 \mathrm{PH}_3(g)+12 \mathrm{OH}^{-}(a q)$
Balance $\mathrm{O}$ atoms, by adding $12 \mathrm{H}_2 \mathrm{O}$ to L.H.S. of above equation.
$\mathrm{P}_4(\mathrm{~s})+12 \mathrm{H}_2 \mathrm{O}(l)+12 e^{-} \longrightarrow 4 \mathrm{PH}_3(g)+12 \mathrm{OH}^{-}(a q) \dots(v)$
To cancel out electrons, multiply Eq. (iii) by 3 and add it to Eq. (v), we have,
$4 \mathrm{P}_4(\mathrm{~s})+24 \mathrm{OH}^{-}(a q)+12 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 4 \mathrm{PH}_3(a q)+12 \mathrm{H}_2 \mathrm{PO}_2^{-}(a q)+12 \mathrm{H}_2 \mathrm{O}(l)+12 \mathrm{OH}^{-}(a q)$
$\mathrm{P}_4(g)+3 \mathrm{OH}^{-}(a q)+3 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{PH}_3(a q)+3 \mathrm{H}_2 \mathrm{PO}_2^{-}(a q) \dots(vi)$
Thus, Eq. (vi) represents the correct balanced equation.

(b)

$\mathrm{N}_2 \mathrm{H}_4(\mathrm{l})+\mathrm{ClO}_3^{-}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g})$
Oxidation number method :

Total increase in O.N. of $\mathrm{N}=2 \times 4=8$
Total decreases in O.N. of $\mathrm{Cl}=1 \times 6=6$
Therefore, to balance increase/decrease in O.N. multiply $\mathrm{N}_2 \mathrm{H}_4$ by 3 and $\mathrm{ClO}_3^{-}$ by 4 , we have,
$3 \mathrm{~N}_2 \mathrm{H}_4(l)+4 \mathrm{ClO}_3^{-}(a q) \longrightarrow \mathrm{NO}(g)+\mathrm{Cl}^{-}(a q)$
To balance $N$ and $\mathrm{Cl}$ atoms, multiply NO by 6 and $\mathrm{Cl}^{-}$by 4 , we have,
$3 \mathrm{~N}_2 \mathrm{H}_4(l)+4 \mathrm{ClO}_3^{-}(a q) \longrightarrow 6 \mathrm{NO}(g)+4 \mathrm{Cl}^{-}(a q)$
Balance $\mathrm{O}$ atoms by adding $6 \mathrm{H}_2 \mathrm{O}$,
$3 \mathrm{~N}_2 \mathrm{H}_4(l)+4 \mathrm{ClO}_3^{-}(a q) \longrightarrow 6 \mathrm{NO}(g)+4 \mathrm{Cl}^{-}(a q)+6 \mathrm{H}_2 \mathrm{O}(l) \dots(i)$
$\mathrm{H}$ atoms get automatically balanced and thus Eq. (i) represents the correct balanced equation.
Ion-electron method :
Oxidation half-reaction :
$\left[\mathrm{N}_2 \mathrm{H}_4+8 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{NO}+8 e^{-}+6 \mathrm{H}_2 \mathrm{O}\right] \times 6$
Reduction half-reaction :
$\left[\mathrm{ClO}_3^{-}+6 e^{-}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Cl}^{-}+6 \mathrm{OH}^{-}\right] \times 8$
Net reaction is

$\begin{aligned} & 6 \mathrm{~N}_2 \mathrm{H}_4+8 \mathrm{ClO}_3^{-} \longrightarrow 12 \mathrm{NO}+8 \mathrm{Cl}^{-}+12 \mathrm{H}_2 \mathrm{O} \\ & 3 \mathrm{~N}_2 \mathrm{H}_4+4 \mathrm{ClO}_3^{-} \longrightarrow 6 \mathrm{NO}+4 \mathrm{Cl}^{-}+6 \mathrm{H}_2 \mathrm{O} \end{aligned}$
$\text { Reductant : } \mathrm{N}_2 \mathrm{H}_4 \text {; Oxidant : } \mathrm{ClO}_3^{-}$

(c) $\mathrm{Cl}_2 \mathrm{O}_{7(\mathrm{g})}+\mathrm{H}_2 \mathrm{O}_{2(aq))} \longrightarrow \mathrm{ClO}_{2(aq)}^{-}+\mathrm{O}_{2(\mathrm{g})}+\mathrm{H}_{(a q)}^{+}$

$\mathrm{Cl}_2 \mathrm{O}_7(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \rightarrow \mathrm{ClO}_2^{-}(\mathrm{aq})+\mathrm{O}_2(\mathrm{~g})+\mathrm{H}^{+}$
Oxidation number method :
Net reaction is
$\mathrm{Cl}_2 \mathrm{O}_7+4 \mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{ClO}_2^{-}+5 \mathrm{H}_2 \mathrm{O}+4 \mathrm{O}_2$
Ion-electron method:
Oxidation half-reaction :
$\left[\mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \longrightarrow \mathrm{O}_2+2 e^{-}+2 \mathrm{H}_2 \mathrm{O}\right] \times 4$

Reduction half-reaction :
$\mathrm{Cl}_2 \mathrm{O}_7+8 \mathrm{e}^{-}+3 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{ClO}_2^{-}+6 \mathrm{OH}^{-}$

Net reaction is
$\begin{aligned} & \mathrm{Cl}_2 \mathrm{O}_7+4 \mathrm{H}_2 \mathrm{O}_2+2 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{ClO}_2^{-}+5 \mathrm{H}_2 \mathrm{O}+4 \mathrm{O}_2 \\ & \text { Reductant : } \mathrm{H}_2 \mathrm{O}_2 ; \quad \text { Oxidant : } \mathrm{Cl}_2 \mathrm{O}_7 \end{aligned}$

Q20. What sorts of informations can you draw from the following reaction?
$(\mathrm{CN})_2(\mathrm{~g})+2 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{CN}^{-}(\mathrm{aq})+\mathrm{CNO}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})$

Answer: Let $x$ be the oxidation number of $\mathrm{C}$.
$\therefore \quad$ O.N. of C in cyanogen, $(\mathrm{CN})_2=2(x-3)=0$ or $x=+3$
or $x=+3$
O.N. of $\mathrm{C}$ in cyanide ion, $\mathrm{CN}^{-}=x-3=-1$
or $x=+2$
O.N. of $\mathrm{C}$ in cyanate ion, $\mathrm{CNO}^{-}=x-3-2=-1$
or $x=+4$
The four informations about the reaction are :
(i) The reaction involves decomposition of cyanogen, $(\mathrm{CN})_2$ in the alkaline medium to cyanide ion, $\mathrm{CN}^{-}$and cyanate ion, $\mathrm{CNO}^{-}$.
(ii) The O.N. of C decreases from +3 in $(\mathrm{CN})_2$ to +2 in $\mathrm{CN}^{-}$ion and increases from +3 in $(\mathrm{CN})_2$ to +4 in $\mathrm{CNO}^{-}$ion. Thus, cyanogen is simultaneously reduced to cyanide ion and oxidised to cyanate ion.
(iii) It is an example of disproportionation redox reaction.
(iv) Cyanogen is a pseudo halogen while cyanide ion is a pseudo halide ion.

Q21. The $\mathrm{Mn}^{3+}$ ion is unstable in solution and undergoes disproportionation to give $\mathrm{Mn}^{2+}, \mathrm{MnO}_2$, and $\mathrm{H}^{+}$ion. Write a balanced ionic equation for the reaction.

Answer: The skeletal equation is:
$\mathrm{Mn}^{3+}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{MnO}_2(s)+\mathrm{H}^{+}(a q) .$
Oxidation half equation:
$\stackrel{+3}{\mathrm{Mn}^{3+}}(a q) \longrightarrow \stackrel{+4}{\mathrm{MnO}_2}(s)$
Balance O.N. by adding electrons,
$\mathrm{Mn}^{3+}(a q) \longrightarrow \mathrm{MnO}_2(\mathrm{~s})+e^{-}$
Balance charge by adding $4 \mathrm{H}^{+}$ions,
$\mathrm{Mn}^{3+}(a q) \longrightarrow \mathrm{MnO}_2(s)+4 \mathrm{H}^{+}(a q)+e^{-}$
Balance $\mathrm{O}$ atoms by adding $2 \mathrm{H}_2 \mathrm{O}$ :
$\stackrel{3+}{\mathrm{Mn}}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{MnO}_2(s)+4 \mathrm{H}^{+}(a q)+e^{-} \dots(i)$
Reduction half equation:
$\stackrel{+3}{\mathrm{Mn}^{3+}} \longrightarrow \stackrel{+2}{\mathrm{Mn}^{2+}}$
Balance O.N. by adding electrons:
$\mathrm{Mn}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q) \dots(ii)$
Adding Eq. (i) and Eq. (ii), the balanced equation for the disproportionation reaction is
$2 \mathrm{Mn}^{3+}(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{MnO}_2(\mathrm{~s})+\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}^{+}(a q)$

Q22. Consider the elements :
Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only postive oxidation state.
(c) Identify the element that exhibits both positive and negattve oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Answer: (a) F: Fluorine is the most electronegative element and shows only -1 oxidation state.
(b) Cs: Because of the presence of single electron in the valence shell, (alkali metals) Cs exhibits an oxidation state of +1 only.
(c) I: Because of the presence of seven electrons in the valence shell, I shows an oxidation state of- 1 in compounds with more electropositive elements (such as $\mathrm{H}, \mathrm{Na}, \mathrm{K}, \mathrm{Ca}$, etc.) and oxidation states of $+3,+5$, +7 in compounds of I with more electronegative elements (such as, O, F, etc.)
(d) Ne: It is an inert gas (with high ionization enthalpy and highly positive electron gain enthalpy) and hence, it exhibits neither negative nor positive oxidation states.

Q23. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

Answer: The skeletal equation is:
$\mathrm{Cl}_2(a q)+\mathrm{SO}_2(a q)+\mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{SO}_4^{2-}(a q)$
Reduction half equation:
$\mathrm{Cl}_2(a q) \longrightarrow \mathrm{Cl}^{-}(a q)$
Balance $\mathrm{Cl}$ atoms,
$\stackrel{0}{\mathrm{Cl}_2}(a q) \longrightarrow \stackrel{-1}{\mathrm{2Cl}}(a q)$
Balance O.N. by adding electrons:
$\mathrm{Cl}_2(a q)+2 e^{-} \longrightarrow 2 \mathrm{Cl}^{-}(a q) \dots(i)$
Oxidation half equation:
$\stackrel{+4}{\mathrm{SO}_2}(a q) \longrightarrow \stackrel{+6}{\mathrm{SO}_4^{2-}}(a q)+2 e^{-}$
Balance O.N. by adding electrons:
$\mathrm{SO}_2(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)+2 e^{-}$
Balance charge by adding $4 \mathrm{H}^{+}$ions:
$\mathrm{SO}_2(a q) \longrightarrow \mathrm{SO}_4^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 e-$
Balance $\mathrm{O}$ atoms by adding $2 \mathrm{H}_2 \mathrm{O}$,
$\mathrm{SO}_2(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow \mathrm{SO}_4^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 e^{-} \dots(ii)$
Adding Eq. (i) and Eq. (ii), we have,
$\mathrm{Cl}_2(a q)+\mathrm{SO}_2(a q)+2 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+\mathrm{SO}_4^{2-}(a q)+4 \mathrm{H}^{+}(a q)$
This represents the balanced redox reaction.

Q24. Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction.
(b) Select three metals that can show disproportionation reaction.

Answer: (a) The non-metals which show disproportionation reaction are $\mathrm{N}, \mathrm{P}, \mathrm{S}, \mathrm{Cl}, \mathrm{Br}$ and I.
$\text { (i) } \quad 2 \mathrm{NO}_{2(a q)}+2 \mathrm{OH}_{(a q)}^{-} \longrightarrow \mathrm{NO}_{2(a q)}^{-}+\mathrm{NO}_{3(a q)}^{-}+\mathrm{H}_2 \mathrm{O}_{(l)}$
$\text { (ii) } \mathrm{P}_{4(s)}+3 \mathrm{OH}_{(a q)}^{-}+3 \mathrm{H}_2 \mathrm{O}_{(l)} \longrightarrow \mathrm{PH}_{3(\mathrm{~g})}+3 \mathrm{H}_2 \mathrm{PO}_{2(a q)}^{-}$
$\text { (iii) } \mathrm{S}_8(s)+12 \mathrm{OH}^{-}{ }_{(\mathrm{aq})} \longrightarrow 4 \mathrm{~S}^{2-}(a q)+2 \mathrm{S}_2 \mathrm{O}_3^{2-}(a q)+6 \mathrm{H}_2 \mathrm{O}(l)+6 \mathrm{H}_2 \mathrm{O}_{(l)}$
$\text { (iv) } \mathrm{Cl}_{2(\mathrm{~g})}+2 \mathrm{OH}_{(a q)}^{-} \longrightarrow \mathrm{Cl}_{(a q)}^{-}+\mathrm{ClO}_{(a q)}^{-}+\mathrm{H}_2 \mathrm{O}_{(l)}$
Bromine and iodine give similar reactions as chlorine.
(b) The metals which undergo disproportionation reaction are:
$\mathrm{Cu}^{+}, \mathrm{Ga}^{+}, \mathrm{In}^{+}$, etc.
(i) $2 \mathrm{Cu}_{(a q)}^{+} \longrightarrow \mathrm{Cu}^{2+}{ }_{(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{s})}$
(ii) $3 \mathrm{Ga}_{\text {(aq })}^{+} \longrightarrow \mathrm{Ga}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{Ga}_{(\mathrm{s})}$
(iii) $3 \operatorname{In}_{(a q)}^{+} \longrightarrow \operatorname{In}_{(a q)}^{3+}+2 \operatorname{In}_{(s)}$

Q25. In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with $10.00 \mathrm{~g}$. of ammonia and $20.00 \mathrm{~g}$ of oxygen?

Answer: The reaction involved in the manufacturing process is:
$\therefore \quad 68 \mathrm{~g}$ of $\mathrm{NH}_3$ will react with $160 \mathrm{~g}$ of $\mathrm{O}_2$ to produce $120 \mathrm{~g}$ of $\mathrm{NO}_{(\mathrm{g})}$
But oxygen which is actually available $(20.0 \mathrm{~g})$ is less than the amount which is required. Therefore, oxygen is the limiting reactant.
Now $160 \mathrm{~g}$ of $\mathrm{O}_2$ will form $\mathrm{NO}=120 \mathrm{~g}$ $20 \mathrm{~g}$ of $\mathrm{O}_2$ will form $\mathrm{NO}=\frac{(120)}{(160)} \times(20)=15 \mathrm{~g}$

Q26. Using the standard electrode potentials, predict if the reaction between the following is feasible:
(a) $\mathrm{Fe}^{3+}(\mathrm{aq})$ and $\mathrm{I}^{-}(\mathrm{aq})$
(b) $\mathrm{Ag}^{+}$(aq) and $\mathrm{Cu}(\mathrm{s})$
(c) $\mathrm{Fe}^{3+}$ (aq) and $\mathrm{Cu}(\mathrm{s})$
(d) $\mathrm{Ag}$ (s) and $\mathrm{Fe}^{3+}$ (aq)
(e) $\mathrm{Br}_2$ (aq) and $\mathrm{Fe}^{2+}$ (aq).

Answer: (a) The possible reaction between $\mathrm{Fe}^{3+}{ }_{(a q)}$ and $\mathrm{I}_{(a q)}^{-}$is
$2 \mathrm{Fe}^{3+}{ }_{(a q)}+2 \mathrm{I}_{(a q)}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}{ }_{(a q))}+\mathrm{I}_{2(s)}$
The above redox reaction can be split into the following two half reactions,
Oxidation :
$2 \mathrm{I}_{(\mathrm{a q})}^{-} \longrightarrow \mathrm{I}_{2(s)}+2 e^{-} ; E^{\ominus}_{\mathrm{oxi}}=-0.54 \mathrm{~V} \dots(i)$
Reduction :
$\left[\mathrm{Fe}_{(a q)}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+}{ }_{(a q)}\right] \times 2 ; E^{\ominus}_{\mathrm{red}}=+0.77 \mathrm{~V} \dots(ii)$
Overall reaction :
$2 \mathrm{Fe}_{(a q))}^{3+}+2 \mathrm{I}_{(a q)}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+\mathrm{I}_{2(s)} ; E^{\ominus}=+0.23 \mathrm{~V}$
Since the EMF for the above reaction is positive, therefore, the above reaction is feasible.
(b) The possible reaction between $\mathrm{Ag}_{(a q)}^{+}$ and $\mathrm{Cu}_{(s)}$ is
$\mathrm{Cu}_{(s)}+2 \mathrm{Ag}_{(a q)}^{+} \longrightarrow \mathrm{Cu}^{2+}{ }_{(a q)}+2 \mathrm{Ag}_{(s)}$
The above redox reaction can be split into the following two half reactions, Oxidation :
$\mathrm{Cu}_{(s)} \longrightarrow \mathrm{Cu}_{(a q)}^{2+}+2 e^{-} ; \quad E^{\ominus}_{\mathrm{oxi}}=-0.34 \mathrm{~V}$
Reduction :
$\left[\mathrm{Ag}_{(a q)}^{-}+e^{-} \longrightarrow \mathrm{Ag}_{(s)}\right] \times 2 ; \quad E^{\ominus}_{\mathrm{red}}=+0.80 \mathrm{~V}$
Overall reaction :
$\mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{a q})}^{+} \longrightarrow \mathrm{Cu}_{(\mathrm{a q})}^{2+}+2 \mathrm{Ag}_{(\mathrm{s})} ; E^{\ominus}=+0.46 \mathrm{~V}$
Since the EMF for the above reaction comes out to be positive, therefore, the above reaction is feasible.
(c) There are two probabilities for reaction between $\mathrm{Cu}$ and $\mathrm{Fe}^{3+}$. The reaction between $\mathrm{Fe}^{3+} \text { (aq) }$ and $\mathrm{Cu}(\mathrm{s})$ occurs according to the following equation
(i) $\mathrm{Cu}_{(s)}+2 \mathrm{Fe}^{3+}{ }_{(a q)} \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Fe}^{2+}(a q)$
The above reaction can be split into the following two half reactions, Oxidation:
$\mathrm{Cu}_{(s)} \longrightarrow \mathrm{Cu}^{2+}{ }_{(a q)}+2 e^{-} ; \quad E^{\ominus}_{\mathrm{oxi}}=-0.34 \mathrm{~V}$
Reduction :
$\left[\mathrm{Fe}_{(a q)}^{3-}+e^{-} \longrightarrow \mathrm{Fe}_{(a q)}^{2-}\right] \times 2 ; \quad E^{\ominus}_{\mathrm{red}}=+0.77 \mathrm{~V}$
Overall reaction :
$\mathrm{Cu}_{(s)}+2 \mathrm{Fe}^{3+}{ }_{(a q)} \longrightarrow \mathrm{Cu}^{2+}{ }_{(a q)}+2 \mathrm{Fe}^{2+}{ }_{(a q)};E^{\ominus}=+0.43 \mathrm{~V}$
Since the EMF for the above reaction is positive, therefore, the above reaction is feasible.
(ii) If the reaction between $\mathrm{Fe}^{3+}{ }_{(a q)}$ and $\mathrm{Cu}_{(s)}$ occurs according to the following equation,
$3 \mathrm{Cu}_{(s)}+2 \mathrm{Fe}^{3+}{ }_{(a q)} \longrightarrow 3 \mathrm{Cu}_{(a q)}^{2+}+2 \mathrm{Fe}_{(s)}$
The EMF of the reaction comes out to be negative, i.e., $-0.376 \mathrm{~V}(-0.34 \mathrm{~V}-0.036 \mathrm{~V})$ and hence, this reaction is not feasible.
(d) There are two possibilities for reaction between $\mathrm{Ag}$ and $\mathrm{Fe}^{3+}$ ion.
(i) If the reaction occurs by following equation,
$\mathrm{Ag}_{(s)}+\mathrm{Fe}_{(a q)}^{3+} \longrightarrow \mathrm{Ag}_{(a q)}^{+}+\mathrm{Fe}_{(a q)}^{2+}$
The reaction can be split into the following two half reactions,
Oxidation :
$\mathrm{Ag}_{(s)} \longrightarrow \mathrm{Ag}_{(a q)}^{+}+e^{-} ; \quad E^{\ominus}_{\mathrm{oxi}}=-0.80 \mathrm{~V}$
Reduction:
$\mathrm{Fe}^{3+}{ }_{(a q)}+e^{-} \longrightarrow \mathrm{Fe}_{(a q)}^{2+} ; \quad E^{\ominus}_{\mathrm{red}}=+0.77 \mathrm{~V}$
Overall reaction :
$\mathrm{Ag}_{(s)}+\mathrm{Fe}^{3+}{ }_{(a q)} \longrightarrow \mathrm{Ag}_{(a q)}^{+}+\mathrm{Fe}^{2+}{ }_{(a q) ;} ; E^{\ominus}=-0.03 \mathrm{~V}$
Since the EMF for the above reaction is negative, therefore, the above reaction is not feasible.
(ii) The reaction between $\mathrm{Ag}_{(s)}$ and $\mathrm{Fe}^{3+}{ }_{(a q)}$ may occur according to the following equation,
$3 \mathrm{Ag}_{(s)}+\mathrm{Fe}_{(a q)}^{3+} \longrightarrow 3 \mathrm{Ag}_{(a q)}^{+}+\mathrm{Fe}_{(s)}$
EMF of this reaction comes to be even more negative i.e., $-0.836 \mathrm{~V}$, and hence this redox reaction is also not feasible.
(e) The reaction between $\mathrm{Br}_{2(a q)}$ and $\mathrm{Fe}^{2+}{ }_{(a q)}$ occurs according to the following equation:
$\mathrm{Br}_{2(a q)}+2 \mathrm{Fe}_{(a q)}^{2+} \longrightarrow 2 \mathrm{Br}_{(a q)}^{-}+2 \mathrm{Fe}^{3+}_{(a q)}$
The above reaction can be split into the following two half reactions.
Oxidation :
$\left[\mathrm{Fe}_{(a q)}^{2+} \longrightarrow \mathrm{Fe}_{(a q)}^{3+}+e^{-}\right] \times 2 ; \quad E^{\ominus}_{\mathrm{oxi}}=-0.77 \mathrm{~V}$
Reduction :
$\mathrm{Br}_{2(a q)}+2 e^{-} \longrightarrow 2 \mathrm{Br}_{(a q))^{-}}^{-} ; \quad E^{\ominus}_{\text {red }}=+1.09 \mathrm{~V}$
Overall reaction :
$2 \mathrm{Fe}_{(a q)}^{2+}+\mathrm{Br}_{2(\mathrm{aq})} \longrightarrow 2 \mathrm{Fe}^{3+}{ }_{(\mathrm{aq})}+2 \mathrm{Br}^{-}_{(\mathrm{aq})} ; \quad E^{\ominus}=+0.32 \mathrm{~V}$
Since the EMF for the above reaction is positive, therefore, the above reaction is feasible.

Q27. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of $\mathrm{AgNO}_3$ with silver electrodes
(ii) An aqueous solution $\mathrm{AgNO}_3$ with platinum electrodes
(iii) A dilute solution of $\mathrm{H}_2 \mathrm{SO}_4$ with platinum electrodes
(iv) An aqueous solution of $\mathrm{CuCl}_2$ with platinum electrodes.

Answer: (i) An aqueous solution of $\mathrm{AgNO}_3$ using platinum electrodes :(P.I.S.A. Based)
Both $\mathrm{AgNO}_3$ and water will ionise in aqueous solution
$\begin{aligned} \mathrm{AgNO}_3(s) & \stackrel{(a q)}{\longrightarrow} \mathrm{Ag}^{+}(a q)+\mathrm{NO}_3^{-}(a q) \\ \mathrm{H}_2 \mathrm{O} & \stackrel{(a q)}{\rightleftharpoons} \mathrm{H}^{+}(a q)+\mathrm{HO}^{-}(a q) \end{aligned}$
At cathode: $\mathrm{Ag}^{+}$ions with less discharge potential are reduced in preference to $\mathrm{H}^{+}$ions which will remain in solution. As a result, silver will be deposited at cathode.
$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \text { (deposited) }$
At anode : An equivalent amount of silver will be oxidised to $\mathrm{Ag}^{+}$ions by releasing electrons.
$\mathrm{Ag}(s) \stackrel{(a q)}{\longrightarrow} \mathrm{Ag}^{+}(a q)+e^{-}$
As result of electrolysis, $\mathrm{Ag}$ from silver anode dissolves as $\mathrm{Ag}^{+}(\mathrm{aq})$ ions while an equivalent amount of $\mathrm{Ag}^{+}$ (aq) ions from the aqueous $\mathrm{AgNO}_3$ solution get deposited on the cathode.
(ii) An aqueous solution of $\mathrm{AgNO}_3$ using platinum electrodes:
In the case, the platinum electrodes are the non-attackable electrodes. On passing current the following changes will occur at the electrodes.
At cathode : $\mathrm{Ag}^{+}$ions will be reduced to $\mathrm{Ag}$ which will get deposited at the cathode.
At anode : Both $\mathrm{NO}_3^{-}$and $\mathrm{OH}^{-}$ions will migrate. But ${\mathrm{OH}^{-}}$ions with less discharge potential will be oxidised in preference to $\mathrm{NO}_3$ ions which will remain in solution.
$\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{OH}+\mathrm{e}^{-} ; 4 \mathrm{OH} \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})$
Thus, as a result of electrolysis, silver is deposited on the cathode while $\mathrm{O}_2$ is evolved at the anode. The solution will be acidic due to the presence of $\mathrm{HNO}_3$.
(iii) A dilute solution of $\mathrm{H}_2 \mathrm{SO}_4$ using platinum electrodes :
On passing current, both acid and water will ionise as follows :
$\mathrm{H}_2 \mathrm{SO}_4(l) \stackrel{(a q)}{\longrightarrow} 2 \mathrm{H}^{+}(a q)+\mathrm{SO}_4^{2-}(a q)$
$\mathrm{H}_2 \mathrm{O} \stackrel{(a q)}{\rightleftharpoons} \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$
At cathode :
$\mathrm{H}^{+}(\mathrm{aq})$ ions will migrate to the cathode and will be reduced to $\mathrm{H}_2$.
$\mathrm{H}^{+}(\mathrm{aq})+\mathrm{e}^{-}\rightarrow \mathrm{H} ; \mathrm{H}+\mathrm{H} \rightarrow \mathrm{H}_2(\mathrm{~g})$
Thus, $\mathrm{H}_2(\mathrm{~g})$ will evolve at cathode.
At anode: $\mathrm{OH}$ ions will be released in preference to $\mathrm{SO}^{2-}_4$ ions because their discharge potential is less. They will be oxidised as follows :
$\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{OH}+\mathrm{e}^{-} ; 4 \mathrm{OH} \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{O}_2(\mathrm{~g})$
Thus, $\mathrm{O}_2(\mathrm{~g})$ will be evolved at anode. The solution will be acidic and will contain $\mathrm{H}_2 \mathrm{SO}_4$.
(iv) An aqueous solution of $\mathrm{CuCl}_2$ using platinum electrodes:
The electrolysis proceeds in the same manner as discussed in the case of $\mathrm{AgNO}_3$ solution. Both $\mathrm{CuCl}_2$ and $\mathrm{H}_2 \mathrm{O}$ will ionise as follows :
$\mathrm{CuCl}_2 \stackrel{(a q)}{\longrightarrow} \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)$
$\mathrm{H}_2 \mathrm{O} \stackrel{(a q)}{\rightleftharpoons} \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$
At cathode: $\mathrm{Cu}^{2+}$ ions will be reduced in preference to $\mathrm{H}^{+}$ions and copper will be deposited at cathode.
$\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}($ deposited)
At anode: $\mathrm{Cl}^{-}$ions will be discharged in preference to $\mathrm{OH}^{-}$ions which will remain in solution.
$\mathrm{Cl}^{-} \rightarrow \mathrm{Cl}+\mathrm{e}^{-} ; \mathrm{Cl}+\mathrm{Cl} \rightarrow \mathrm{Cl}_2$ (g) (evolved)
Thus, $\mathrm{Cl}_2$ will evolve at anode.

Q28. Arrange the following metals in the order in which they displace each other from the solution of their salts.
$\mathrm{Al}, \mathrm{Cu}, \mathrm{Fe}, \mathrm{Mg}$ and $\mathrm{Zn}$.

Answer: We know that,
$E^{\ominus}_{\mathrm{Al}^{3+} / \mathrm{Al}}=-1.66 \mathrm{~V}, E^{\ominus}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}=+0.34 \mathrm{~V},$

$E^{\ominus}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}=-0.44 \mathrm{~V}, E^{\ominus}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}=-2.36 \mathrm{~V} \text { and }E^{\ominus}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}=-0.76 \mathrm{~V}$

Since a metal with lower electrode potential is a stronger reducing agent, therefore, $\mathrm{Mg}$ can displace all the above metals from their aqueous solutions, Al can displace all metals except Mg from the aqueous solutions of their salts. Zn can displace all metals except Mg and Al from the aqueous solutions of their salts while Fe can displace only Cu from the aqueous solution of its salts. Thus, the order in which they can displace each other from the solution of their salts is $\mathrm{Mg}, \mathrm{Al}, \mathrm{Zn}, \mathrm{Fe}, \mathrm{Cu}$.

Q29. Given the standard electrode potentials,
$\begin{aligned} & \mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}, \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V} \\ & \mathrm{Hg}^{2+} / \mathrm{Hg}=0.79 \mathrm{~V} \\ & \mathrm{Mg}^{2+} / \mathrm{Mg}=-2.37 \mathrm{~V} . \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.74 \mathrm{~V} \end{aligned}$
arrange these metals in their increasing order of reducing power.

Answer: Lower the electrode potential, better is the reducing power. Since the electrode potentials increase in the order;
$\mathrm{K}^{+} / \mathrm{K}(-2.93 \mathrm{~V})$,
$\mathrm{Mg}^{2+} / \mathrm{Mg}(-2.37 \mathrm{~V}), \mathrm{Cr}^{3+} / \mathrm{Cr}(-0.74 \mathrm{~V})$,
$\mathrm{Hg}^{2+} / \mathrm{Hg}(0.79 \mathrm{~V}), \mathrm{Ag}^{+} / \mathrm{Ag}(0.80 \mathrm{~V})$,
therefore, reducing power of metals increases in the order, i.e., $\mathrm{Ag}<\mathrm{Hg}<\mathrm{Cr}<\mathrm{Mg}<\mathrm{K}$.

Q30. Depict the galvanic cell in which the reaction $\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$ takes place, Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) Individual reaction at each electrode.

Answer: The given redox reaction is
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})$
Since $\mathrm{Zn}$ gets oxidized to $\mathrm{Zn}^{2+}$ions, and $\mathrm{Ag}^{-}$ gets reduced to $\mathrm{Ag}^{+}$ metal, therefore, oxidation occurs at zinc electrode and reduction occurs at the silver electrode.
Thus, galvanic cell corresponding to the above redox reaction may be depicted as :
$\mathrm{Zn}\left|\mathrm{Zn}^{2+}(a q) \| \mathrm{Ag}^{+}(a q)\right| \mathrm{Ag}$
(i) Since oxidation occurs at the zinc electrode, therefore, electrons accumulate on the zinc electrode and hence, zinc electrode is negatively charged.
(ii) The ions carry current. The electrons flow from $\mathrm{Zn}$ to $\mathrm{Ag}$ electrode while the current flows from $\mathrm{Ag}$ to $\mathrm{Zn}$ electrode.
(iii) The reactions occurring at the two electrodes are:
At Anode: $\mathrm{Zn}(s) \longrightarrow \mathrm{Zn}^{2+}(a q)+2 e^{-}$
At Cathode: $\mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s)$

Exemplar

Q31. The reaction $\mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow \mathrm{ClO}^{-}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})$
represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.

Answer:

$\stackrel{0}{\mathrm{Cl}_2}(\mathrm{g})+\stackrel{-2 +1}{\quad \quad \mathrm{2OH}^{-}(\mathrm{aq})} \rightarrow \stackrel{+1-2}{\quad \quad \mathrm{ClO}^{-}(\mathrm{aq})}+\stackrel{-1}{\quad \mathrm{Cl}^{-}(\mathrm{aq})}+\stackrel{+1 -2}{\quad \mathrm{H}_2{O(l)}}$
In this reaction, $\mathrm{O} . \mathrm{N}$. of $\mathrm{Cl}$ increases from 0 (in $\mathrm{Cl}_2$ ) to +1 (in $\mathrm{ClO}^{-}$) and decreases to – 1 (in $\mathrm{Cl}^{-}$). Therefore, $\mathrm{Cl}_2$ is both oxidized to $\mathrm{ClO}^{-}$and reduced to $\mathrm{Cl}^{-}$. Since $\mathrm{Cl}^{-}$ion cannot act as an oxidizing agent (because it cannot decrease its $\mathrm{O} . \mathrm{N}$. lower than -1), therefore, $\mathrm{Cl}_2$ bleaches substances due to oxidizing action of hypochlorite, $\mathrm{ClO}$ ion.

Q32. $\mathrm{MnO}_4^{2-}$ undergoes disproportionation reaction in acidic medium but $\mathrm{MnO}_4^{-}$ does not. Give reason.

Answer: In $\mathrm{MnO_4}^{-}, \mathrm{Mn}$ is in the highest oxidation state of +7 (i.e.. cannot be oxidized further) and hence it cannot undergo disproportionation.
In contrast, the $\mathrm{O} . \mathrm{N}$. of $\mathrm{Mn}$ in $\mathrm{MnO}_4{ }^{2-}$ is +6.  Therefore, it can increase its O. N. $\mathrm{MnO}_4{ }^{2-}$ from +6 to +7 in $\mathrm{MnO}_4^{-}$ and decreases to +4 in $\mathrm{MnO}_2$. Thus, $\mathrm{MnO}_4{ }^{2-}$ undergoes disproportionation according to the following reaction.

Q33. $\mathrm{PbO}$ and $\mathrm{PbO}_2$ react with $\mathrm{HCl}$ according to following chemical equations :
$\begin{aligned} & 2 \mathrm{PbO}+4 \mathrm{HCl} \longrightarrow 2 \mathrm{PbCl}_2+2 \mathrm{H}_2 \mathrm{O} \\ & \mathrm{PbO}_2+4 \mathrm{HCl} \longrightarrow \mathrm{PbCl}_2+\mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O} \end{aligned}$
Why do these compounds differ in their reactivity?

Answer: (i) $\stackrel{+2}{\mathrm{2PbO}}+4 \mathrm{HCl} \rightarrow \stackrel{+2}{\mathrm{PbCl}_2}+2 \mathrm{H}_2 \mathrm{O}$
(ii) $\stackrel{+4}{\mathrm{PbO}_2}+4 \mathrm{HCl} \rightarrow \stackrel{+2}{\mathrm{PbCl}_2}+\stackrel{0}{\mathrm{Cl}_2}+2 \mathrm{H}_2 \mathrm{O}$
In reaction (i), O.N. of none of the atoms undergo a change. Therefore, it is not a redox reaction. It is an acidbase reaction, because $\mathrm{PbO}$ is a basic oxide which reacts with $\mathrm{HCl}$ acid.
The reaction (ii) is a redox reaction in which $\mathrm{PbO}_2$ gets reduced and acts as an oxidizing agent.

Q34. Nitric acid is an oxidising agent and reacts with $\mathrm{PbO}$ but it does not react with $\mathrm{PbO}_2$. Explain why?

Answer: Nitric acid is an oxidizing agent and reacts with $\mathrm{PbO}$ to give a simple acid- base reaction without any change in oxidation state. In $\mathrm{PbO}_2, \mathrm{~Pb}$ is in +4 oxidation state and cannot be oxidized further, hence no reaction takes place between $\mathrm{PbO}_2$ and $\mathrm{HNO}_3$.
$2 \mathrm{PbO}+4 \mathrm{HNO}_3 \rightarrow \underset{\text { Salt }}{2 \mathrm{~Pb}}\left(\mathrm{NO}_3\right)_2 +\underset{\text { (Acid base reaction) }} {2 \mathrm{H}_2 \mathrm{O}}$

Q35. Write balanced chemical equation for the following reactions:
(i) Permanganate ion $\left(\mathrm{MnO}_4^{-}\right)$reacts with sulphur dioxide gas in acidic medium to produce $\mathrm{Mn}^{2+}$ and hydrogensulphate ion.
(Balance by ion electron method)
(ii) Reaction of liquid hydrazine $\left(\mathrm{N}_2 \mathrm{H}_4\right)$ with chlorate ion $\left(\mathrm{ClO}_3^{-}\right)$in basic medium produces nitric oxide gas and chloride ion in gaseous state.
(Balance by oxidation number method)
(iii) Dichlorine heptaoxide $\left(\mathrm{Cl}_2 \mathrm{O}_7\right)$ in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion $\left(\mathrm{ClO}_2^{-}\right)$and oxygen gas. (Balance by ion electron method)

Answer:

(i) $(\mathrm{2MnO}_4^{-}+5 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \rightarrow \mathrm{5HSO}_4^{-}+\mathrm{2Mn}^{2+}$
Balancing by ion-electron method:
$\stackrel{+7 -2}{\mathrm{MnO}_4^{-}}+\stackrel{+4-2}{\mathrm{SO}_2} \longrightarrow \stackrel{+2}{\mathrm{Mn}^{2+}}+\stackrel{+1+6-2}{\mathrm{HSO}_4^{-}} \text {(Skeletal equation) }$
$\text { Oxidation half: } \quad \mathrm{SO}_2 \longrightarrow \mathrm{HSO}_4^{-}$
$\text { Reduction half: } \quad \mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$
$\text { Oxidation half: } \quad \mathrm{SO}_2 \longrightarrow \mathrm{HSO}_4^{-}+2 e^{-}$
$\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HSO}_4^{-}+3 \mathrm{H}^{+}+2 e^{-} \dots(i)$
$\text { (Add } 2 \mathrm{H}_2 \mathrm{O} \text { molecules to balance } \mathrm{O} \text { atoms) }$
Reduction half:
$\begin{aligned} & \mathrm{MnO}_4^{-}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+} \\ & \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \dots(ii) \end{aligned}$
(Add $4 \mathrm{H}_2 \mathrm{O}$ molecules to balance $\mathrm{O}$ atoms and $\mathrm{H}$ atoms) Add oxidation and reduction half
$\begin{aligned} & {\left[\mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{HSO}_4^{-}+3 \mathrm{H}^{+}+2 e^{-}\right] \times 5} \\ & \frac{\left[\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}\right] \times 2}{2 \mathrm{MnO}_4^{-}+5 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \longrightarrow 5 \mathrm{HSO}_4^{-}+2 \mathrm{Mn}^{2+}} \end{aligned}$
(ii) $3 \mathrm{~N}_2 \mathrm{H}_4+4 \mathrm{ClO}_3^{-} \rightarrow 6 \mathrm{NO}+4 \mathrm{Cl}^{-}+6 \mathrm{H}_2 \mathrm{O}$
Balancing by oxidation number method:

$6 \mathrm{~N}_2 \mathrm{H}_4+8 \mathrm{ClO}_3^{-} \rightarrow 12 \mathrm{NO}+8 \mathrm{Cl}^{-}+12 \mathrm{H}_2 \mathrm{O}$

(iii) $\mathrm{Cl}_2 \mathrm{O}_7(\mathrm{~g})+4 \mathrm{H}_2 \mathrm{O}_2(\mathrm{aq}) \longrightarrow 2 \mathrm{ClO}_2^{-}(\mathrm{aq})+3 \mathrm{H}_2 \mathrm{O}(l)+4 \mathrm{O}_2(\mathrm{~g})+2 \mathrm{H}^{+}(\mathrm{aq})$
Balancing by ion-electron method:
Oxidation half:
$\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{O}_2+2 e^{-}$
Adding $2 \mathrm{H}^{+}$on right side to balance $\mathrm{H}$ atoms and charge.
$\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{O}_2+2 \mathrm{H}^{+}+2 e^{-}$
Reduction half:
$\mathrm{Cl}_2 \mathrm{O}_7+8 e^{-} \longrightarrow 2 \mathrm{ClO}_2^{-}$
Adding $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{H}^{+}$to balance $\mathrm{H}$ and $\mathrm{O}$ atoms
$\mathrm{Cl}_2 \mathrm{O}_7+8 e^{-}+6 \mathrm{H}^{+} \longrightarrow 2 \mathrm{ClO}_2^{-}+3 \mathrm{H}_2 \mathrm{O}$
Adding oxidation and reduction half
$\begin{aligned} & {\left[\mathrm{H}_2 \mathrm{O}_2 \rightarrow \mathrm{O}_2+2 e^{-}+2 \mathrm{H}^{+}\right] \times 4} \\ & \mathrm{Cl}_2 \mathrm{O}_7+8 e^{-}+6 \mathrm{H}^{+} \rightarrow 2 \mathrm{ClO}_2^{-}+3 \mathrm{H}_2 \mathrm{O} \\ & \overline {\mathrm{Cl}_2 \mathrm{O}_7+4 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{ClO}_2^{-}+3 \mathrm{H}_2 \mathrm{O}+4 \mathrm{O}_2+2 \mathrm{H}^{+}} \\ & \end{aligned}$

Q36. Calculate the oxidation number of phosphorus in the following species.
(a) $\mathrm{HPO}_3^{2-}$ and
(b) $\mathrm{PO}_4^{3-}$

Answer: (a) Let the oxidation number of $\mathrm{P}$ in $\mathrm{HPO}_3{ }^{2-}$ be $x$.
So, $+1+x+(-6)=-2; x=+3$
(b) Let the oxidation number of $\mathrm{P}$ in $\mathrm{PO}_4^{3-}$ be $x$.
So, $x+(-8)=-3$; $x=+5$

Q37. Calculate the oxidation number of each sulphur atom in the following compounds:
(a) $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$
(b) $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$
(c) $\mathrm{Na}_2 \mathrm{SO}_3$
(d) $\mathrm{Na}_2 \mathrm{SO}_4$

Answer:(a) Structure of $\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3$ is

Let the oxidation number of sulphur atom is $\mathrm{x}$.
$\begin{aligned} & \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \\ & 2 \times(+1)+2 \mathrm{x}+(-2) \times 3=0 \\ & 2+2 \mathrm{x}-6=0 \\ & 2 \mathrm{x}-4=0 \\ & 2 \mathrm{x}=4 \\ & \mathrm{x}=+2 \end{aligned}$
By conventional method, we can’t find the oxidation number of each sulphur atom.$\text { Thus, the oxidation states of two } \mathrm{S} \text { atoms in } \mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3 \text { are }-2 \text { and }+6 \text {. }$
(b) Structure of $\mathrm{Na}_2 \mathrm{~S}_4 \mathrm{O}_6$ is

$\text { From the left, } S^1=(-2)+(-2)+(-1)=+5$
$\begin{aligned} & \mathrm{S}^2=0 \\ & \mathrm{~S}^3=0 \\ & \mathrm{~S}^4=+5 \end{aligned}$
(iii)

$\begin{aligned} & \mathrm{Na}_2 \mathrm{SO}_3 \\ & \quad 2 \times(+1)+x+3 \times(-2)=0 \\ & \quad x=+4 \end{aligned}$
(iv)

$\begin{aligned} & \mathrm{Na}_2 \mathrm{SO}_4 \\ & \quad+2+x+(-8)=0 \\ & \quad x=+6 \end{aligned}$

Q38. Balance the following equations by the oxidation number method.
(i) $\mathrm{Fe}^{2+}+\mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}$
(ii) $\mathrm{I}_2+\mathrm{NO}_3^{-} \longrightarrow \mathrm{NO}_2+\mathrm{IO}_3^{-}$
(iii) $\mathrm{I}_2+\mathrm{S}_2 \mathrm{O}_3^{2-} \longrightarrow \mathrm{I}^{-}+\mathrm{S}_4 \mathrm{O}_6^{2-}$
(iv) $\mathrm{MnO}_2+\mathrm{C}_2 \mathrm{O}_4^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2$

Answer: (i) $\mathrm{Fe}^{2+}+\mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}$
$\stackrel{+2}{\mathrm{Fe}^{2+}}+\mathrm{H}^{+}+\stackrel{+6 -2}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}} \longrightarrow \stackrel{+3}{\mathrm{Cr}^{3+}}+\stackrel{+3}{\mathrm{Fe}^{3+}}+\mathrm{H}_2 \mathrm{O}$
(a) Balance the increase and decrease in O.N.
$\stackrel{+2}{\mathrm{6Fe}^{2+}}+\mathrm{H}^{+}+\stackrel{+6 -2}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}} \longrightarrow \stackrel{+3}{\mathrm{2Cr}^{3+}}+\stackrel{+3}{\mathrm{6Fe}^{3+}}+\mathrm{H}_2 \mathrm{O}$
(b) Balancing $\mathrm{H}$ and $\mathrm{O}$ atoms by adding $\mathrm{H}^{+}$and $\mathrm{H}_2 \mathrm{O}$ molecules
$6 \mathrm{Fe}^{2+}+14 \mathrm{H}^{+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H}_2 \mathrm{O}$

(ii) $\mathrm{I}_2+\mathrm{NO}_3^{-} \longrightarrow \mathrm{NO}_2+\mathrm{IO}_3^{-}$

Total increase in O.N. $=5 \times 2=10$
Total decrease in O.N. $=1$
To equalize O.N. multiply $\mathrm{NO}_3^{-}$, by 10
$\mathrm{I}_2+10 \mathrm{NO}_3^{-} \longrightarrow 10 \mathrm{NO}_2+\mathrm{IO}_3^{-}$
Balancing atoms other than $\mathrm{O}$ and $\mathrm{H}$
$\mathrm{I}_2+10 \mathrm{NO}_3^{-} \longrightarrow 10 \mathrm{NO}_2+2 \mathrm{IO}_3^{-}$
Balancing $\mathrm{O}$ and $\mathrm{H}$
$\mathrm{I}_2+10 \mathrm{NO}_3^{-}+8 \mathrm{H}^{+} \rightarrow 10 \mathrm{NO}_2+2 \mathrm{IO}_3^{-}+4 \mathrm{H}_2 \mathrm{O}$
(iii) $\mathrm{I}_2+\mathrm{S}_2 \mathrm{O}_3^{2-} \longrightarrow \mathrm{I}^{-}+\mathrm{S}_4 \mathrm{O}_6^{2-}$
Balance all atoms other than $\mathrm{H}$ and $\mathrm{O}$.
$\mathrm{I}_2+2 \mathrm{~S}_2 \mathrm{O}_3^{2-} \rightarrow 2 \mathrm{I}^{-}+\mathrm{S}_4 \mathrm{O}_6^{2-}$
The oxidation number of I changes from 0 to -1 . The change in the oxidation number of one I atom is 1.
Total change in the oxidation number for $2 \mathrm{I}$ atoms is 2.
The oxidation number of $S$ changes from 2 to 2.5. The change in the oxidation number of one $\mathrm{S}$ atom is 0.5.
Total change in the oxidation number for $4 \mathrm{~S}$ atoms is 2.
The increase in the oxidation number is balanced with decrease in the oxidation number.
$\mathrm{O}$ atoms are balanced. This is the balanced chemical equation.
(iv) $\mathrm{MnO}_2+\mathrm{C}_2 \mathrm{O}_4^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_2$
Total increase in O.N. $=1 \times 2=2$
Total decrease in O.N. $=2$
Now, in order to equalize O.N., we will multiply $\mathrm{CO}_2$ with 2 to give us,
$\mathrm{MnO}_2+\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow \mathrm{Mn}^{2+}+2 \mathrm{CO}_2$
Now, we will balance $\mathrm{H}$ and $\mathrm{O}$ by adding $2 \mathrm{H}_2 \mathrm{O}$ on the right side, and $4 \mathrm{H}^{+}$on the left side of equation.
$\mathrm{MnO}_2+\mathrm{C}_2 \mathrm{O}_4^{2-}+4 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+2 \mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$

Q39. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.
$\text { (i) } 3 \mathrm{HCl}(\mathrm{aq})+\mathrm{HNO}_3(\mathrm{aq}) \longrightarrow \mathrm{Cl}_2(\mathrm{~g})+\mathrm{NOCl}(\mathrm{g})+2 \mathrm{H}_2 \mathrm{O}(l)$
$\text { (ii) } \mathrm{HgCl}_2 \text { (aq) }+2 \mathrm{KI} \text { (aq) } \longrightarrow \mathrm{HgI}_2 \text { (s) }+2 \mathrm{KCl} \text { (aq) }$
$\text { (iii) } \mathrm{Fe}_2 \mathrm{O}_3 \text { (s) }+3 \mathrm{CO} \text { (g) } \xrightarrow{\Delta} 2 \mathrm{Fe} \text { (s) }+3 \mathrm{CO}_2 \text { (g) }$
$\text { (iv) } \mathrm{PCl}_3 \text { (l) }+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow 3 \mathrm{HCl} \text { (aq) }+\mathrm{H}_3 \mathrm{PO}_3 \text { (aq) }$
$\text { (v) } 4 \mathrm{NH}_3+3 \mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{~N}_2(\mathrm{~g})+6 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$

Answer:(i) Writing the O.N. of on each atom,

Here, O.N. of $\mathrm{Cl}$ increases from -1 (in $\mathrm{HCl}$ ) to 0 (in $\mathrm{Cl}_2$ ). Therefore, $\mathrm{Cl}^{-}$is oxidized and hence $\mathrm{HCl}$ acts as a reducing agent.
The O.N. of $\mathrm{N}$ decreases from +5 (in $\mathrm{HNO}_3$ ) to +3 (in NOCl) and therefore, $\mathrm{HNO}_3$ acts as an oxidizing agent. Thus, reaction (i) is a redox reaction.

(ii) Here O.N. of none of the atoms undergo a change and therefore, this is not a redox reaction.

(iii) Here, O.N. of Fe decreases from +3 (in $\mathrm{Fe}_2 \mathrm{O}_3$ ) to 0 (in $\mathrm{Fe}$ ) and therefore, $\mathrm{Fe}_2 \mathrm{O}_3$ acts as an oxidizing agent. $\mathrm{O} . \mathrm{N}$. of $\mathrm{C}$ increases from +2 (in $\mathrm{CO}$ ) to +4 (in $\mathrm{CO}_2$ ) and therefore, $\mathrm{CO}$ acts as a reducing agent. Thus, this is a redox reaction.

(iv) Here O.N. of none of the atoms undergo a change and therefore, it is not a redox reaction.

(v) In the below reaction, oxidising agent is $\mathrm{O}_2$ and the reducing agent is $\mathrm{NH}_3$.
Here. O.N. of $\mathrm{N}$ increases from -3 (in $\mathrm{NH}_3$ ) to 0 in ( $\mathrm{N}_2$ ) and therefore, $\mathrm{NH}_3$ acts as a reducing agent. O.N. of $\mathrm{O}$ decreases from 0 (in $\mathrm{O}_2$ ) to -2 (in $\mathrm{H_2}, \mathrm{O}$ ) and therefore, $\mathrm{O}_2$ acts as an oxidizing agent.
Thus, reaction (v) is a redox reaction.

Q40. Balance the following ionic equations
(i) $\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{H}^{+}+\mathrm{I}^{-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{I}_2+\mathrm{H}_2 \mathrm{O}$
(ii) $\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{Fe}^{2+}+\mathrm{H}^{+} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}$
(iii) $\mathrm{MnO}_4^{-}+\mathrm{SO}_3^{2-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{SO}_4^{2-}+\mathrm{H}_2 \mathrm{O}$
(iv) $\mathrm{MnO}_4^{-}+\mathrm{H}^{+}+\mathrm{Br}^{-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{Br}_2+\mathrm{H}_2 \mathrm{O}$

Answer:

Dividing the equation into two half reactions:
Oxidation half reaction: $\mathrm{I}^{-} \rightarrow \mathrm{I}_2$
Reduction half reaction: $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow \mathrm{Cr}^{3+}$
Balancing oxidation and reduction half reactions separately as:
Oxidation half reaction
$\begin{array}{lll} \mathrm{I}^{-} & \longrightarrow & \mathrm{I}_2 \\ 2 \mathrm{I}^{-} & \longrightarrow & \mathrm{I}_2 \\ 2 \mathrm{I}^{-} & \longrightarrow & \mathrm{I}_2+2 e^{-} \dots(i) \end{array}$
Reduction half reaction
$\begin{aligned} & \mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow \mathrm{Cr}^{3+} \\ & \mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+} \\ & \mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+} \\ \end{aligned}$
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \dots(ii)$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text { (acidic medium) }$
To balance the electrons, multiply eq. (i) by 3 and add to eq. (ii)
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}$

(ii) Step-1: Separate the equation into two half reactions.
The oxidation number of various atoms are shown below:

In this case, chromium undergoes reduction, oxidation number decreases from +6 (in $\mathrm{Cr}_2 \mathrm{O}_7^{2-}$ ) to +3 (in $\mathrm{Cr}^{3+}$ )
$\mathrm{Fe}^{2+}(\mathrm{O} . \mathrm{N} .=+2)$ changes to $\mathrm{Fe}^{3+}(\mathrm{O} \cdot \mathrm{N} .=+3)$. The species undergoing oxidation and reduction are:
Oxidation: $\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}$
Reduction: $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \longrightarrow \mathrm{Cr}^{3+}$
Step-2: Balance each half reaction separately as:
$\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}$
(a) Balance all atoms other than $\mathrm{H}$ and $\mathrm{O}$. This step is not needed, because, it is already balanced.
(b) The oxidation number on left is +2 and on right is +3 . To account for the difference, the electron is added to the right as:
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}$
(c) Charge is already balanced.
(d) No need to add $\mathrm{H}$ or $\mathrm{O}$.
The balanced half equation is:
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} \dots(i)$
Consider the second half equation $\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow \mathrm{Cr}^{3+}$
(a) Balance the atoms other than $\mathrm{H}$ and $\mathrm{O}$.
$\mathrm{Cr}_2 \mathrm{O}_7^{2-} \rightarrow 2 \mathrm{Cr}^{3+}$
(b) The oxidation number of chromium on the left is +6 and on the right is +3 . Each chromium atom must gain three electrons. Since there are two $\mathrm{Cr}$ atoms, add $6 \mathrm{e}^{-}$on the left.
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}$
(c) Since the reaction takes place in acidic medium add $14 \mathrm{H}^{+}$on the left to equate the net charge on both sides.
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{e}^{-}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}$
(d) To balance $\mathrm{FI}$ atoms, add $7 \mathrm{H}_2 \mathrm{O}$ molecules on the right.
$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{e}^{-}+14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \dots(ii)$
This is the balanced half equation.
Step-3: Now add up the two half equations. Multiply eq. (i) by 6 so that electrons are balanced.
$\begin{aligned} & {\left[\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+e^{-}\right] \times 6} \\ & \mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 e^{-}+14 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\ & \overline {6 \mathrm{Fe}^{2+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+} \longrightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}} \\ & \end{aligned}$
The balanced equation is:
$6 \mathrm{Fe}^{2+}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+} \longrightarrow 6 \mathrm{Fe}^{3+}+2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}$

(iii) Dividing the equation into two half reactions: Oxidation half reaction: $\mathrm{SO}_3^{2-} \longrightarrow \mathrm{SO}_4^{2-}$
Reduction half reaction: $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$

As step (i), we will be generating the unbalanced skeleton i.e.,
$\mathrm{MnO}_4^{-}+\mathrm{SO}_3^{2-}+\mathrm{H}+\rightarrow \mathrm{Mn}^{2+}+\mathrm{SO}_4^{2-}+\mathrm{H}_2 \mathrm{O}$
Here in Mn undergoes reduction and S undergoes oxidation.
As step (ii) we will have to Identify the reactants which will get oxidized or reduced, and then write the half-cell reaction for them. This will be done along with the electron transfers as it will carefully be making the equal numbers of atoms of oxidized and reduced redox couples.
Thus, oxidation is $\mathrm{SO}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}+2 e^{-}$and reduction is $\mathrm{MnO}_4^{-}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}$
As step (iii) we will balance the number of atoms in the oxidized and reduced redox couples. But as we can see that the atoms are balanced, so the reaction will be the same as the previous step.
Thus, oxidation is $\mathrm{SO}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}+2 e^{-}$and reduction is $\mathrm{MnO}_4^{-}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}$
Now, as step (iv) for acidic solutions, we will balance the charge by adding $\mathrm{H}+$ on the required side of the given equation.
Thus, oxidation is $\mathrm{SO}_3^{2-} \rightarrow \mathrm{SO}_4^{2-}+2 e^{-}+2 \mathrm{H}^{+}$and reduction is $\mathrm{MnO}_4^{-}+5 e^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}$
As part of the step (v) we will balance the oxygen atoms. As there are seven oxygen atoms in reduction reaction, so we will be adding water molecule to RHS of the reaction for balancing the oxygen.
Thus, oxidation is $\mathrm{SO}_3^{2-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{SO}_4^{2-}+2 e^{-}+2 \mathrm{H}^{+}$ and
reduction is $\mathrm{MnO}_4^{-}+5 e^{-}+8 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O}$
Now as part of the step (vi) we will be making the electron lost and gain, equal in both the reduction and oxidation reaction.
Thus, oxidation is $5 \mathrm{SO}_3^{2-}+5 \mathrm{H}_2 \mathrm{O} \rightarrow 5 \mathrm{SO}_4^{2-}+10 e^{-}+10 \mathrm{H}^{+}$and
reduction is $2 \mathrm{MnO}_4^{-}+10 e^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}$
Now, as part of the step (vii) we will combine the two given equations by adding them. This will ensure that all the products and reactants remain on the same side.
$\begin{aligned} & 2 \mathrm{MnO}_4^{-}+10 e^{-}+16 \mathrm{H}^{+}+5 \mathrm{SO}_3^{2-}+5 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{SO}_4^{2-}+ 10 e^{-}+10 \mathrm{H}^{+} \end{aligned}$
Now, as step (viii) we will simplify the equation. This will be done by eliminating the similar terms present on both of the sides of the equation.
$2 \mathrm{MnO}_4^{-}+6 \mathrm{H}^{+}+5 \mathrm{SO}_3^{2-} \rightarrow 2 \mathrm{Mn}^{2+}+3 \mathrm{H}_2 \mathrm{O}+5 \mathrm{SO}_4^{2-}$
We will now verify whether all the charges are balanced.
So, LHS $=2 \times-1+6+5 \times-2=-6$ and the RHS $=2 \times+2+3 \times 0+5 \times-2=-6$
Equal charges on both sides imply towards a balanced equation.

(iv) 

Dividing the equation into two half reactions:
Oxidation half reaction: $\mathrm{Br} \longrightarrow \mathrm{Br}_2$
Reduction half reaction: $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$
Balancing oxidation and reduction half reactions separately as:
Oxidation half reaction
$\begin{aligned} & \mathrm{Br}^{-} \longrightarrow \mathrm{Br}_2 \\ & 2 \mathrm{Br}^{-} \longrightarrow \mathrm{Br}_2 \\ & 2 \mathrm{Br}^{-} \longrightarrow \mathrm{Br}_2+2 e^{-} \dots(i) \end{aligned}$
Reduction half reaction
$\begin{aligned} & \mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+} \\ & \mathrm{MnO}_4^{-}+5 e \longrightarrow \mathrm{Mn}^{2+} \end{aligned}$
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}$
$\mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \dots(ii)$
To balance the electrons, multiply eq. (i) by 5 and eq. (ii) by 2 and add
$2 \mathrm{MnO}_4^{-}+10 \mathrm{Br}^{-}+16 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{Br}_2+8 \mathrm{H}_2 \mathrm{O}$

Q41. Explain redox reactions on the basis of electron transfer. Give suitable examples.

Answer:As we know that, the reactions
$\begin{aligned} & 2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NaCl}(\mathrm{s}) \\ & 4 \mathrm{Na}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Na}_2 \mathrm{O}(\mathrm{s}) \end{aligned}$
are redox reactions because in each of these reactions sodium is oxidised due to the addition of either oxygen or more electronegative element to sodium. From our knowledge of chemical bonding we also know that, sodium chloride and sodium oxide are ionic compounds and perhaps better written as
$\mathrm{Na}^{+} \mathrm{Cl}^{-}(\mathrm{s}) \text { and }\left(\mathrm{Na}^{+}\right)_2 \mathrm{O}^{2-}(\mathrm{s}) \text {. }$
Develpment of charges on the species produced suggests us to rewrite the above reaction in the following manner

For convenience, each of the above processes can be considered as two separate steps, one involving the loss of electrons and other the gain of electrons. As an illustration, we may further elaborate one of these, say, the formation of sodium chloride.
$\begin{aligned} & 2 \mathrm{Na}(\mathrm{s}) \rightarrow 2 \mathrm{Na}^{+}(\mathrm{g})+2 \mathrm{e}^{-} \\ & \mathrm{Cl}_2(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}(\mathrm{g}) \end{aligned}$
Each of the above steps is called a half reaction, which explicitly shows involvement of electrons. Sum of the half reactions gives the overall reaction:
$2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_2(\mathrm{~g}) \rightarrow 2 \mathrm{Na}^{+} \mathrm{Cl}^{-}(\mathrm{s}) \text { or } 2 \mathrm{NaCl}(\mathrm{s})$
The given reactions suggest that half reactions that involved loss of electrons are oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions.
It may be out of context to mention here that the new way of defining oxidation and reduction has been achieved only by establishing a correlation between the behaviour of species as per the classical idea and their interplay in electron-transfer change.
In the given reaction, sodium, which is oxidised, acts a reducing agent because it donates electron to each of the elements interacting with it and thus helps in reducing them. Chlorine and oxygen are reduced and act as oxidising agents because these accept electrons from sodium.
To summarise, we may mention that
Oxidation: Loss of electron(s) by any species.
Reduction: Gain of electron(s) by any species.
Oxidising agent: Acceptor of electron(s).
Reducing agent: Donor of electron(s).

Q42. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for $E^{\ominus}$ value).
(i) $\mathrm{Cu}+\mathrm{Zn}^{2+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$
(ii) $\mathrm{Mg}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
(iii) $\mathrm{Br}_2+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-}$
(iv) $\mathrm{Fe}+\mathrm{Cd}^{2+} \longrightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$

Answer: As we know that,
$\begin{aligned} & E^{\ominus}{ }_{\mathrm{Cu}^{2+} / \mathrm{Cu}}=0.34 \mathrm{~V}, E^{\ominus}{ }_{\mathrm{Zn}^{2+} / \mathrm{Zn}}=-0.76 \mathrm{~V}, \\ & E^{\ominus}{ }_{\mathrm{Mg}^{2+} / \mathrm{Mg}}=-2.37 \mathrm{~V}, E^{\ominus}{ }_{\mathrm{Fe}^{2+} / \mathrm{Fe}}=-0.74 \mathrm{~V}, \\ & E^{\ominus}{ }_{\mathrm{Br}_2 / \mathrm{Br}^{-}}=+1.08 \mathrm{~V}, E^{\ominus}{ }_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}=+1.36 \mathrm{~V} \\ & E^{\ominus}{ }_{\mathrm{Fe}^{2+} / \mathrm{Fe}}=-0.74 \mathrm{~V}, E^{\ominus}{ }_{\mathrm{Cd}^{2+} / \mathrm{Cd}}=-0.44 \mathrm{~V} \end{aligned}$

(a)
$\begin{aligned} & E^{\ominus}{ }_{\mathrm{Cu}^{2+} / \mathrm{Cu}}=0.34 \mathrm{~V} \text { and } E^{\ominus}{ }_{\mathrm{Zn}^{2+} / \mathrm{Zn}}=-0.76 \mathrm{~V} \\ & \mathrm{Cu}+\mathrm{Zn}^{2+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Zn} \end{aligned}$
In the given cell reaction, $\mathrm{Cu}$ is oxidised to $\mathrm{Cu}^{2+}$, therefore, $\mathrm{Cu}^{2+} / \mathrm{Cu}$ couple acts as anode and $\mathrm{Zn}^{2+}$ is reduced to $\mathrm{Zn}$, therefore, $\mathrm{Zn}{ }^{2+} / \mathrm{Zn}$ couple acts as cathode.
$\begin{aligned} E^{\ominus} \text { cell } & =\mathrm{E}^{\circ} \text { cathode }-\mathrm{E}^{\circ} \text { anode } \\ E^{\ominus} \text { cell } & =-0.74-(+0.34)=-1.10 \mathrm{~V} \end{aligned}$
Positive value of $E^{\ominus}$ cell indicates that the reaction will not occur.

(b) $\mathrm{Mg}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
$E^{\ominus}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}=-2.37 \mathrm{~V} \text { and } E^{\ominus}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}=-0.74 \mathrm{~V}$
In the given cell reaction, $\mathrm{Mg}$ is oxidised to $\mathrm{Mg}^{2+}$ hence, $\mathrm{Mg}^{2+} / \mathrm{Mg}$ couple acts as anode and $\mathrm{Fe}^{2+}$ is reduced to $\mathrm{Fe}$ hence, $\mathrm{Fe}^{2+} / \mathrm{Fe}$ couple acts as cathode.
$\begin{aligned} & E^{\ominus}_{\text {cell }}=E^{\ominus}_{\text {cathode }}-E^{\ominus}_{\text {anode }} \\ & E^{\ominus}_{\text {cell }}=-0.74-(-2.37)=+1.63 \mathrm{~V} \end{aligned}$
Positive value of $E^{\ominus}$ cell 5 indicates that the reaction will occur.
(c) $\mathrm{Br}_2+2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-}$
$E^{\ominus}_{\mathrm{Br}_2 / \mathrm{Br}^{-}}=+1.08 \mathrm{~V}$ and $E^{\ominus}{ }_{\mathrm{Cl}_2 / \mathrm{Cl}^{-}}=+1.36 \mathrm{~V}$
In the given cell reaction, $\mathrm{Cl}^{-}$is oxidised to $\mathrm{Cl}_2$ hence, $\mathrm{Cl}^{-} / \mathrm{Cl}_2$ couple acts as anode and $\mathrm{Br}_2$ is reduced to $\mathrm{Br}^{-}$hence; $\mathrm{Br}^{-} / \mathrm{Br}_2$ couple acts as cathode.
$\begin{aligned} & E^{\ominus} \text { cell }=E^{\ominus} \text { cathode }-E^{\ominus} \text { anode } \\ & E^{\ominus} \text {cell }=+1.08-(+1.36)=-0.28 \mathrm{~V} \end{aligned}$
Negative value of $E^{\ominus}$ cell indicates that the reaction will occur.
(d)
$\begin{aligned} & \mathrm{Fe}+\mathrm{Cd}^{2+} \rightarrow \mathrm{Cd}+\mathrm{Fe}^{2+} \\ & E^{\ominus}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}=-0.74 \mathrm{~V} \text { and } E^{\ominus}_{\mathrm{Cd}^{2+} / \mathrm{Cd}}=-0.44 \mathrm{~V} \end{aligned}$
In the given cell reaction, $\mathrm{Fe}$ is oxidised to $\mathrm{Fe}^{2+}$ hence, $\mathrm{Fe}^{2+} / \mathrm{Fe}$ couple acts as anode and $\mathrm{Cd}^{2+}$ is reduced to $\mathrm{Cd}$ hence, $\mathrm{Cd}^{2+} / \mathrm{Cd}^{-}$couple acts as cathode.
$\begin{aligned} & E^{\ominus} \text { cell }=E^{\ominus}_{\text {cathode }}-E^{\ominus}{ }_{\text {anode }} \\ & E^{\ominus} \text { cell }=-0.44-(-0.74)=+0.30 \mathrm{~V} \end{aligned}$
Positive value $E^{\ominus}$ cell indicates that the reaction will occur.

Q43. Why does fluorine not show disproportionation reaction?

Answer:
In a disproportionation reaction, the same species is simultaneously oxidised as well as reduced. Therefore, for such a redox reaction to occur, the reacting species must contain an element which has atleast three oxidation states.
The element, in reacting species, is present in an intermediate state while lower and higher oxidation states are available for reduction and oxidation to occur (respectively).
Fluorine is the strongest oxidising agent. It does not show positive oxidation state. That is why fluorine does not show disproportionation reaction.

Q44. Write redox couples involved in the reactions (i) to (iv) given below.
(i) $\mathrm{Cu}+\mathrm{Zn}^{2+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Zn}$
(ii) $\mathrm{Mg}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Fe}$
(iii) $\mathrm{Br}_2+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 \mathrm{Br}^{-}$
(iv) $\mathrm{Fe}+\mathrm{Cd}^{2+} \longrightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$

Answer: (i) $\mathrm{Cu}^{2+} / \mathrm{Cu}$ and $\mathrm{Zn}^{2+} / \mathrm{Zn}$
(ii) $\mathrm{Mg}^{2+} / \mathrm{Mg}$ and $\mathrm{Fe}^{2+} / \mathrm{Fe}$
(iii) $\mathrm{Br}_2 / \mathrm{Br}^{-}$and $\mathrm{Cl}_2 / \mathrm{Cl}^{-}$
(iv) $\mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Cd}^{2+} / \mathrm{Cd}$

Q45. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.
$\mathrm{NaClO}_4, \mathrm{NaClO}_3, \mathrm{NaClO}, \mathrm{KClO}_2, \mathrm{Cl}_2 \mathrm{O}_7, \mathrm{ClO}_3, \mathrm{Cl}_2 \mathrm{O}, \mathrm{NaCl}, \mathrm{Cl}_2, \mathrm{ClO}_2 \text {. }$
Which oxidation state is not present in any of the above compounds?

Answer: Suppose that the oxidation number of chlorine in these compounds be $\mathrm{x}$.
O.N. of $\mathrm{Cl}$ in $\mathrm{NaClO}_4 \therefore+1+\mathrm{x}+4(-2)=0$ or, $\mathrm{x}=+7$
O.N. of $\mathrm{Cl}$ in $\mathrm{NaClO}_3 \therefore+1+\mathrm{x}+3(-2)=0$ or, $\mathrm{x}=+5$
O.N. of $\mathrm{Cl}$ in $\mathrm{NaClO} \therefore+1+\mathrm{x}+1(-2)=0$ or, $\mathrm{x}=+1$
O.N. of $\mathrm{Cl}$ in $\mathrm{KClO}_2 \therefore+1+\mathrm{x}+2(-2)=0$ or, $\mathrm{x}=+3$
O.N. of $\mathrm{Cl}$ in $\mathrm{Cl}_2 \mathrm{O}_7 \therefore+2 \mathrm{x}+7(-2)=0$ or, $\mathrm{x}=+7$
O.N. of $\mathrm{Cl}$ in $\mathrm{ClO}_3 \therefore \mathrm{x}+3(-2)=0$ or, $\mathrm{x}=+6$
O.N. of $\mathrm{Cl}$ in $\mathrm{Cl}_2 \mathrm{O} \therefore 2 \mathrm{x}+1(-2)=0$ or, $\mathrm{x}=+1$
O.N. of $\mathrm{Cl}$ in $\mathrm{NaCl} \therefore+1+\mathrm{x}=0$ or, $\mathrm{x}=-1$
O.N. of $\mathrm{Cl}$ in $\mathrm{Cl}_2 \therefore 2 \mathrm{x}=0$ or, $\mathrm{x}=0$
O.N. of $\mathrm{Cl}$ in $\mathrm{ClO}_2 \therefore \mathrm{x}+2(-2)=0$ or, $\mathrm{x}=+4$
None of these compounds have an oxidation number of +2 .
Increasing order of oxidation number of chlorine is : $-1,0,+1,+3,+4,+5,+6,+7$
Therefore, the increasing order of oxidation number of $\mathrm{Cl}$ in compounds is
$\mathrm{NaCl}<\mathrm{Cl}_2<\mathrm{NaClO}<\mathrm{KClO}_2<\mathrm{ClO}_2<\mathrm{NaClO}_3<\mathrm{ClO}_3<\mathrm{Cl}_2 \mathrm{O}_7$

Q46. Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.

Answer:Measure the electrode potential of the given species by connecting the redox couple of the given species with standard hydrogen electrode. If it is positive, the electrode of the given species acts as reductant and if it negative, it acts as an oxidant.
Find the electrode potentials of the other given species in the same way, compare the values and determine their comparative strength as a reductant or oxidant.
e.g., measurement of standard electrode potential of $\mathrm{Zn}^{2+} / \mathrm{Zn}$ electrode using SHE as a reference electrode.

The EMF of the cell comes out to be $0.76 \mathrm{~V}$. (reading of voltmeter is $0.76 \mathrm{~V}$ ). $\mathrm{Zn}^{2+} / \mathrm{Zn}$ couple acts as anode and SHE acts as cathode.
$\begin{aligned} & \therefore E^{\ominus}_{\text {cell }}=0.76=E^{\ominus}_{\text {cathode }}-E^{\ominus}_{\text {anode }} \\ & 0.76=0-E^{\ominus} \text { anode } \\ & E^{\ominus}_{\text {anode }}=-0.76 \mathrm{~V} \\ & E^{\ominus}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}=-0.76 \mathrm{~V} . \end{aligned}$