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Which of the following is not an example of redox reaction?
(d) Hint: Redox reaction = oxidation-reduction reaction Following reactions are the example of redox reaction
Option (d) is not an example of redox reaction because it is an example of double displacement reaction.
\(
\mathrm{BaCl}_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{BaSO}_4+2 \mathrm{HCl}
\)
The more positive the value of \(\mathrm{E}^{\ominus}\), the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.
\(
\mathbf{E}^{\ominus} \text { values: } \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 ; \mathrm{I}_2(\mathrm{~s}) / \mathrm{I}^{-}=+0.54
\)
\(
\mathrm{Cu}^{2+} / \mathrm{Cu}=+0.34 ; \mathrm{Ag}^{+} / \mathrm{Ag}=+0.80 \mathrm{~V}
\)
(d) Hint: High reduction potential value strongest is the oxidisng agent
Since, \(\mathbf{E}^{\ominus}\) of the redox couple \(\mathrm{Ag}^{+} / \mathrm{Ag}\) is the most positive, i.e., \(0.80 \mathrm{~V}\), therefore \(\mathrm{Ag}^{+}\)is the strongest oxidising agent and highest tendency to get reduced.
\(\mathrm{E}^{\ominus}\) values of some redox couples are given below. On the basis of these values choose the correct option.
\(
\begin{array}{r}
\mathbf{E}^{\ominus} \text { values }: \mathrm{Br}_2 / \mathrm{Br}^{-}=+1.90 ; \mathrm{Ag}^{+} / \mathrm{Ag}(\mathrm{s})=+0.80 \\
\mathrm{Cu}^{2+} / \mathrm{Cu}(\mathrm{s})=+0.34 ; \mathrm{I}_2(\mathrm{~s}) / \Gamma=+0.54
\end{array}
\)
(d) Copper will reduce \(\mathrm{Br}_2\), if the \(\mathbf{E}^{\ominus}\) of the redox reaction, \(2 \mathrm{Cu}+\mathrm{Br}_2 \rightarrow \mathrm{CuBr}_2\) is +ve.
Now,
\(
\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 e^{-} ; \quad \mathbf{E}^{\ominus}=-0.34 \mathrm{~V}
\)
\(
\begin{array}{ll}
\mathrm{Br}_2+2 e^{-} \longrightarrow 2 \mathrm{Br}^{-} ; & \mathbf{E}^{\ominus}=+1.90 \mathrm{~V} \\
\hline \mathrm{Cu}+\mathrm{Br}_2 \longrightarrow \mathrm{CuBr}_2 ; & \mathbf{E}^{\ominus}=+1.56 \mathrm{~V}
\end{array}
\)
Since \(\mathbf{E}^{\ominus}\) of this reaction is +ve, therefore, \(\mathrm{Cu}\) can reduce \(\mathrm{Br}_2\) and hence option (d) is correct.
Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
\(
\begin{array}{r}
\mathbf{E}^{\ominus} \text { values : } \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=+0.77 ; \mathrm{I}_2 / \mathrm{I}^{-}=+0.54 ; \\
\mathrm{Cu}^{2+} / \mathrm{Cu}=+0.34 ; \mathrm{Ag}^{+} / \mathrm{Ag}=+0.80 \mathrm{~V}
\end{array}
\)
(a) \(2 \mathrm{Fe}^{3+}+2 e^{-} \longrightarrow 2 \mathrm{Fe}^{2+} ; \mathrm{E}^{\ominus}=+0.77 \mathrm{~V}\)
\(
\frac{2 \mathrm{I}^{-} \longrightarrow \mathrm{I}_2+2 e^{-} ; \mathrm{E}^{\ominus}=-0.54 \mathrm{~V} \text { (sign of } \mathrm{E}^{\ominus} \text { is reversed) }}{2 \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{I}_2 ; \mathrm{E}^{\ominus}{ }_{\text {cell }}=+0.23 \mathrm{~V}}
\)
This reaction is feasible since \(\mathrm{E}^{\ominus}_{\text {cell }}=+\mathrm{ve}\).
(b) \(
\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 e^{-} ; \mathrm{E}^{\ominus}=-0.34 \mathrm{~V}\left(\text { sign of } \mathrm{E}^{\ominus}\right. \text { is reversed) }
\)
\(
\frac{2 \mathrm{Ag}^{+}+2 e^{-} \longrightarrow 2 \mathrm{Ag} ; \mathrm{E}^{\ominus}=+0.80 \mathrm{~V}}{\mathrm{Cu}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag} ; \mathrm{E}^{\ominus}=+0.46 \mathrm{~V}}
\)
\(
\text { This reaction is feasible since } \mathrm{E}^{\ominus}_{\text {cell }}=+\mathrm{ve} \text {. }
\)
(c) \(
2 \mathrm{Fe}^{3+}+2 e^{-} \longrightarrow 2 \mathrm{Fe}^{2+} ; \mathrm{E}^{\ominus}=+0.77 \mathrm{~V}
\)
\(
\frac{\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+}+2 e^{-} ; \mathrm{E}^{\ominus}=-0.34 \mathrm{~V} \text { (sign of } \mathrm{E}^{\ominus} \text { is reversed) }}{2 \mathrm{Fe}^{3+}+\mathrm{Cu} \longrightarrow 2 \mathrm{Fe}^{2+}+\mathrm{Cu}^{2+} ; \mathrm{E}^{\ominus}=+0.43 \mathrm{~V}}
\)
This reaction is feasible since \(\mathrm{E}^{\ominus}_{\text {cell }}=+\mathrm{ve}\).
(d) \(
\mathrm{Ag} \longrightarrow \mathrm{Ag}^{+}+e^{-} ; \mathrm{E}^{\ominus}=-0.80 \mathrm{~V} \text { (sign of } \mathrm{E}^{\ominus} \text { is reversed) }
\)
\(
\frac{\mathrm{Fe}^{3+}+2 e^{-} \longrightarrow \mathrm{Fe}^{2+} ; \mathrm{E}^{\ominus}=+0.77 \mathrm{~V}}{\mathrm{Ag}+\mathrm{Fe}^{3+} \longrightarrow \mathrm{Ag}^{+}+\mathrm{Fe}^{2+} ; \mathrm{E}^{\ominus}=-0.03 \mathrm{~V}}
\)
This reaction is not feasible since \(\mathrm{E}^{\ominus}_{\text {cell }}=-\) ve. Thus, option (d) is correct.
Thiosulphate reacts differently with iodine and bromine in the reactions given below:
\(
\begin{aligned}
& 2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-} \\
& \mathrm{S}_2 \mathrm{O}_3^{2-}+2 \mathrm{Br}_2+5 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{SO}_4^{2-}+2 \mathrm{Br}^{-}+10 \mathrm{H}^{+}
\end{aligned}
\)
Which of the following statements justifies the above dual behaviour of thiosulphate?
(a) Hint: Gain of electron is reduction and loss of electron is oxidation The two reactions are as follows:
\(+2~ -2 \quad \quad \quad 0 \quad \quad \quad \quad 2.5 \)
\(
2 \mathrm{~S}_2 \mathrm{O}_3^{2-}(\mathrm{aq})+\mathrm{I}_2(\mathrm{~s}) \rightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq})
\)
\(+2-2 \quad \quad \quad \quad 0 \quad \quad \quad \quad \quad \quad \quad +6~ 2-\)
\(
\mathrm{S}_2 \mathrm{O}_3^{2-}(\mathrm{aq})+2 \mathrm{Br}_2(\mathrm{l})+5 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{SO}_4^{2-}(\mathrm{aq})+4 \mathrm{Br}^{-}(\mathrm{aq})+10 \mathrm{H}^{+}(\mathrm{aq})
\)
Bromine being stronger oxidizing agent than \(\mathrm{I}_2\), it oxidises \(\mathrm{S}\) of \(\mathrm{S}_2 \mathrm{O}_3^{2-}\) to \(\mathrm{SO}_4{ }^{2-}\) whereas \(\mathrm{I}_2\) oxidises it only into \(\mathrm{S}_4 \mathrm{O}_6{ }^{2-}\) ion.
The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
Hint: Flourine is the most electronegative element thus it never show positive oxidation state Oxidation number of hydrogen is always +1 is a wrong rule since, it is +1 in hydrogen halide, -1 in hydrides and zero in \(\mathrm{H}_2\) molecule.
In which of the following compounds, an element exhibits two different oxidation states.
(b) Hint: Compound in which nitrogen present in cationic part as well as anionic part.
\(\mathrm{NH}_4 \mathrm{NO}_3\) is actually \(\mathrm{NH}_4^{+}\)and \(\mathrm{NO}_3^{-}\). It is an ionic compound.
The oxidation state of nitrogen in the two species is different as shown below.
Let the oxidation state of \(\mathrm{N}\) in \(\mathrm{NH}_4{ }^{+}\)be \(x\).
\(
\begin{aligned}
& x+4 \times (+1)=+1 \\
& x=-3
\end{aligned}
\)
\(\ln \mathrm{NO}_3{ }^{-}\)
Let, oxidation number of \(\mathrm{N~} {\text {in }} \mathrm{NO}_3^{-}\)is \(\mathrm{x}\)
\(
\Rightarrow \mathrm{x}+(3 \times-2)=-1 \text { or } \mathrm{x}-6=-1 \text { or } \mathrm{x}=+5
\)
Which of the following arrangements represent increasing oxidation number of the central atom?
(a) Writing the O.N. of \(\mathrm{Cr}, \mathrm{Cl}\) and \(\mathrm{Mn}\) on each species in the four set of ions, we have,
\(
\stackrel{+3}{\mathrm{Cr} \mathrm{O}_2^{-}}, \stackrel{+5}{\mathrm{Cl}} \mathrm{O}_3^{-}, \stackrel{+6}{\mathrm{Cr}} \mathrm{O}_4^{2-}, \stackrel{+7}{\mathrm{Mn} \mathrm{O}_4^{-}}
\)
\(
\stackrel{+5}{\mathrm{Cl}} \mathrm{O}_3^{-}, \stackrel{+6}{\mathrm{Cr}} \mathrm{O}_4^{2-}, \stackrel{+7}{\mathrm{Mn}} \mathrm{O}_4^{-}, \stackrel{+3}{\mathrm{CrO}_2^{-}}
\)
\(
\stackrel{+3}{\mathrm{Cr}} \mathrm{O}_2^{-}, \stackrel{+5}{\mathrm{Cl}} \mathrm{O}_3^{-}, \stackrel{+7}{\mathrm{Mn}} \mathrm{O}_4^{-}, \stackrel{+6}{\mathrm{Cr} \mathrm{O}_4^{2-}}
\)
\(
\stackrel{+6}{\mathrm{Cr}} \mathrm{O}_4^{2-}, \stackrel{+7}{\mathrm{Mn}} \mathrm{O}_4^{-}, \stackrel{+3}{\mathrm{Cr}} \mathrm{O}_2^{-}, \stackrel{+5}{\mathrm{Cl}} \mathrm{O}_3^{-}
\)
Only in arrangement (a), the O.N. of central atom increases from left to right. Therefore, option (a) is correct.
The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
(d) \(3 d^1 4 s^2\) will exhibit \(+3,3 d^3 4 s^2=+5,3 d^5 4 s^1=+6\) and \(3 d^5 4 s^2=+7\) oxidation numbers.
Identify disproportionation reaction
(d) Reactions in which the same substance is oxidized as well as reduced are called disproportionation reactions. Writing the O.N. of each element above its symbol in the given reactions,
(a) \(
\stackrel{-4+1}{\mathrm{CH}_4}+2 \stackrel{0}{\mathrm{O}}_2 \longrightarrow \stackrel{+4-2}{\mathrm{CO}_2}+2 \stackrel{+1}{\mathrm{H}_2} \stackrel{-2}{\mathrm{O}}
\)
(b) \(
\stackrel{-4+1}{\mathrm{C} \mathrm{H}_4}+\stackrel{0}{4 \mathrm{Cl}_2} \longrightarrow \stackrel{+4-1}{\mathrm{CCl}_4}+4 \stackrel{+1-1}{\mathrm{HCl}}
\)
(c) \(2 \stackrel{0}{\mathrm{F}_2}+2 \stackrel{-2+1}{\mathrm{OH}^{-}} \longrightarrow \stackrel{-1}{2 \mathrm{F}^{-}}+\stackrel{+2-1}{\mathrm{OF}_2}+\stackrel{+1}{\mathrm{H}_2} \stackrel{-2}{\mathrm{O}}\)
(d) \(\stackrel{+4-2}{2 \mathrm{NO}_2}+\stackrel{-2+1}{2 \mathrm{OH}^{-}}\longrightarrow \stackrel{+3-2}{\mathrm{NO}_2^{-}}+\stackrel{+5-2}{\mathrm{NO}_3^{-}}+\stackrel{+1}{\mathrm{H}_2} \stackrel{-2}{\mathrm{O}}\)
Thus, in reaction (d), \(\mathrm{N}\) is both oxidized as well as reduced since the \(0 . \mathrm{N}\). of \(\mathrm{N}\) increases from +4 in \(\mathrm{NO}_2\) to +5 in \(\mathrm{NO}_3^{-}\)and decreases from +4 in \(\mathrm{NO}_2\) to +3 in \(\mathrm{NO}_2^{-}\).
Which of the following elements does not show disproportionation tendency?
(c) Being the most electronegative element, F can only be reduced and hence it always shows an oxidation number of-1. Further, due to the absence of d-orbitals, it cannot be oxidized and hence it does not show +ve oxidation numbers. In other words, F cannot be simultaneously oxidized as well as reduced and hence does not show disproportionation reactions. Thus, option (c) is correct.
Which of the following statement(s) is/are not true about the following decomposition reaction.
\(
2 \mathrm{KClO}_3 \rightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2
\)
(a,d) The answer is the option (a) and (d)
Explanation: The statements above are not true because it is evident in the reaction that, potassium remains in the same oxidation state, and oxygen is being oxidised.
Write the oxidation number of each element above its symbol, then
\(
\stackrel{+1+5 -2}{\mathrm{2K} \mathrm{ClO}_3} \longrightarrow 2 \stackrel{+1-1}{\mathrm{KCl}}+3 \stackrel{0}{\mathrm{O}}_2
\)
(a) The O.N. of K does not change, K. undergoes neither reduction nor oxidation. Thus, option (a) is not correct.
(b) The \(\mathrm{O} . \mathrm{N}\). of chlorine decreases from +5 in \(\mathrm{KClO}_3\) to -1 in \(\mathrm{KCl}\), hence, \(\mathrm{Cl}\) undergoes reduction.
(c) Since, O.N. of oxygen increases from – 2 in \(\mathrm{KClO}_3\) to 0 in \(\mathrm{O}_2\), oxygen is oxidized.
(d) This statement is not correct because \(\mathrm{Cl}\) is undergoing reduction and \(\mathrm{O}\) is undergoing oxidation.
Identify the correct statement (s) in relation to the following reaction:
\(
\mathrm{Zn}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_2+\mathrm{H}_2
\)
(c,d)
\(
\stackrel{0}{\mathrm{Zn}}+\stackrel{+1}{\mathrm{H}} \stackrel{-1}{\mathrm{Cl}} \longrightarrow \stackrel{+2}{\mathrm{Z}} \stackrel{-1}{\mathrm{Cl}}_2+\stackrel{0}{\mathrm{H}_2}
\)
(a) The O.N. of \(\mathrm{Zn}\) increases from 0 to +2 (in \(\mathrm{ZnCl}_2\) ) and therefore, \(\mathrm{Zn}\) acts as a reductant and not as an oxidant. Hence, option (a) is not correct.
(b) The O.N. of \(\mathrm{Cl}\) does not change and therefore, it neither acts as a reductant nor an oxidant. Hence, option (b) is not correct.
(c) The O.N. of \(\mathrm{H}\) decreases from +1 in \(\mathrm{H}^{+}\)to 0 in \(\mathrm{H}_2\). Therefore, \(\mathrm{H}^{+}\)acts an oxidant. This option is correct.
(d) Zinc acts as reductant because its \(O . N\). changes from 0 to +2 . This option is correct.
The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds.
(c,d) The answer is the option (c) and (d)
Explanation: Atoms having electronic configurations such as in Option (c) and (d) will exhibit more than one oxidation state in the compound because in option (c) electron can be removed from \(4 \mathrm{~s}\) as well as \(3 \mathrm{~d}\). Similarly, in option (d) electron can be removed from \(3 p\) and \(3 s\) both.
Identify the correct statements with reference to the given reaction
\(
\mathrm{P}_4+3 \mathrm{OH}^{-}+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3+3 \mathrm{H}_2 \mathrm{PO}_2^{-}
\)
(c,d)
\(
\stackrel{0}{\mathrm{P}_4}+\stackrel{-2+1}{\mathrm{3OH}}+\stackrel{+1-2}{\mathrm{3{H_2}O}} \rightarrow \stackrel{-3+1}{\mathrm{PH_3}}+ \stackrel{-2+1-2}{\mathrm{3H}_2 \mathrm{PO}_2^{-}}
\)
Because \(\mathrm{O} . \mathrm{N}\). of \(\mathrm{P}\) increases from \(0\left(\mathrm{P}_4\right)\) to \(+1\left(\mathrm{H}_2 \mathrm{PO}_2^{-}\right)\)and decreases from \(0\left(\mathrm{P}_4\right)\) to \(-3\left(\mathrm{PH}_3\right)\), therefore, \(\mathrm{P}\) has undergone both oxidation as well as reduction. So, option (c) is correct. Option (d) is also correct because \(\mathrm{O} . \mathrm{N}\). of \(\mathrm{H}\) remains +1 in all the compounds and hence hydrogen is undergoing neither oxidation nor reduction.
Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
(a, b) All electrodes which have negative electrode potential are stronger reducing agents than \(\mathrm{H}_2\) gas and hence acts as anodes when connected to standard hydrogen electrodes. Thus, \(\mathrm{Al}^{3+} / \mathrm{Al}\left(\mathrm{E}^{\ominus}=1.66 \mathrm{~V}\right)\) and \(\mathrm{Fe}^{2+} / \mathrm{Fe}\left(\mathrm{E}^{\ominus}=0.44 \mathrm{~V}\right)\) act as the anode.
Match Column I with Column II for the oxidation states of the central atoms.
\(
\begin{array}{|l|l|}
\hline \text { Column I } & \text { Column II } \\
\hline \text { (i) } \mathrm{Cr}_2 \mathrm{O}_7^{2-} & (a) +3 \\
\hline \text { (ii) } \mathrm{MnO}_4^{-} & (b) +4 \\
\hline \text { (iii) } \mathrm{VO}_3^{-} & (c) +5 \\
\hline \text { (iv) } \mathrm{FeF}_6^{3-} & (d) +6 \\
\hline & (e)+7 \\
\hline
\end{array}
\)
(i) Hint: The oxidation state of \(\mathrm{Mn}\) in \(\mathrm{MnO}_4^{-}\)is +7 .
(i) Oxidation number of \(\mathrm{Cr}\) in \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\)
\(
\begin{aligned}
& 2 \mathrm{x}+7(-2)=-2 \\
& 2 \mathrm{x}-14=-2 \\
& 2 \mathrm{x}=+12 \\
& \mathrm{x}=+6
\end{aligned}
\)
(ii) Oxidation number of \(\mathrm{Mn}\) in \(\mathrm{MnO}_4^{-}\)
\(
\begin{aligned}
& x+4(-2)=-1 \\
& x-8=-1 \\
& x=+7
\end{aligned}
\)
(iii) Oxidation number of \(\mathrm{V}\) in \(\mathrm{VO}_3^{-}\)
\(
\begin{aligned}
& x+3(-2)=-1 \\
& x-6=-1 \\
& x=+5
\end{aligned}
\)
(iv) Oxidation number of \(\mathrm{Fe}\) in \(\mathrm{FeF}_6^{3-}\)
\(
\begin{aligned}
& x+6(-1)=-3 \\
& x-6=-3 \\
& x=+3
\end{aligned}
\)
Match the items in Column I with relevant items in Column II.
\(
\begin{array}{|l|l|}
\hline \text { Column I } & \text { Column II } \\
\hline \text { (i) Ions having positive charge } & (a) +7 \\
\hline \text { (ii) The sum of oxidation number of all atoms in a neutral molecule } & (b) -1 \\
\hline \text { (iii) Oxidation number of hydrogen ion. }\left(\mathrm{H}^{+}\right) & (c) +1 \\
\hline \text { (iv) Oxidation number of fluorine in } \mathrm{NaF} & (d) 0 \\
\hline \text { (v) lons having negative charge } & \text { (e) Cation } \\
\hline & \text { (f) Anion } \\
\hline
\end{array}
\)
(a) Hint: lons having a positive charge is known as cation
(i) lons having positive charge \(\rightarrow\) Cation
(ii) The sum of oxidation number of all atoms in a neutral molecule \(\rightarrow\) Zero
(iii) Oxidation number of hydrogen ion \(\left(\mathrm{H}^{+}\right) \rightarrow+1\)
(iv) Oxidation number of fluorine in \(\mathrm{NaF} \rightarrow-1\)
(v) lons having negative charge \(\rightarrow\) Anion
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Among halogens fluorine is the best oxidant.
Reason (R) : Fluorine is the most electronegative atom.
(b) Hint: Higher the reduction potential value, the stronger the oxidising agent. Among halogens, \(F_2\) is the best oxidant because it has the highest reduction potential value \(\mathrm{E}^{\ominus}\)
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason (R) : The oxygen of peroxide is in -1 oxidation state and it is converted to zero oxidation state in \(\mathrm{O}_2\) and -2 oxidation state in \(\mathrm{H}_2 \mathrm{O}\).
(a) Hint: Oxidation-reduction reaction, also called redox reaction, A redox couple is defined as pair of compounds or elements having together the oxidised and reduced forms of it and taking part in an oxidation or reduction.
\(
\text { half-reaction. } \mathrm{So} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+} \text { and } \mathrm{Cu}^{2+} / \mathrm{Cu} \text { are redox couples. }
\)
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent.
Reason (R) : Oxidation state of manganese changes from +2 to +7 during the reaction.
(c) Hint: \(\mathrm{Mn}\) oxidation state in \(\mathrm{KMnO}_4\) is +7 .
The reaction is as follows:
\(
10 \mathrm{KI}+2 \stackrel{+7}{\mathrm{KM}} \mathrm{nO}_4+8 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \stackrel{+2}{\mathrm{M}} \mathrm{nSO}_4+6 \mathrm{~K}_2 \mathrm{SO}_4+8 \mathrm{H}_2 \mathrm{O}+5 \mathrm{I}_2
\)
The oxidation state of \(\mathrm{Mn}\) decreases from +7 to +2 . Hence, the Assertion is true but the reason is false.
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Assertion (A) : Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell.
Reason (R) : In the representation \(\mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^{\ominus}\) and \(\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\ominus}, \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) and \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) are redox couples.
(b) Redox couple is the combination of oxidized and reduced form of a substance. In the representation \(\mathrm{E}^{\ominus}_{ \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}\) and \(\mathrm{E}^{\ominus}_{ \mathrm{Cu}^{2+} / \mathrm{Cu}, \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}\)
and \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) are redox couples
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