Question

The standard reduction potential data at $25^{\circ}C$ is given below.
$E^{\circ}(Fe^{3+}, Fe^{2+}) = + 0.77 V; E^{\circ}(Fe^{2+}, Fe) = - 0.44 V$
$E^{\circ}(Cu^{2+}, Cu ) = + 0.34 V; E^{\circ}(Cu^{+}, Cu) = + 0.52 V$
$E^{\circ}[O_{2(g)} + 4H^{+} + 4e^{-} \to 2H_{2}O] = + 1.23 V;$
$E^{\circ}[O_{2(g)} + 2H_{2}O + 4e^{-} \to 4OH^{-}] = + 0.40 V$
$E^{\circ} (Cr^{3+}, Cr) = - 0.74 V; E^{\circ}(Cr^{2+}, Cr) = - 0.91 V$
Match $E^{\circ}$ of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists:

List I	List II
P. $E^{\circ}(Fe^{3+}, Fe)$	1. $-0.18 V$
Q. $E^{\circ}(4H_{2}O \to 4H^{+} + 4OH^{-})$	2. $-0.4 V$
R. $E^{\circ}(Cu^{2+} + Cu \to 2Cu^{+})$	3. $-0.04 V$
S. $E^{\circ}(Cr^{3+}, Cr^{2+})$	4. $-0.83 V$

• A. $(P)-(4), (Q)- (1), (R)-(2), (S)-(3)$
• B. $(P)-(2), (Q)- (3), (R)-(4), (S)-(1)$
• C. $(P)-(1), (Q)- (2), (R)-(3), (S)-(4)$
• D. $(P)-(3), (Q)- (4), (R)-(1), (S)-(2)$

Answer 1

Answer: (D)

$Fe^{3+} +e^{-} \to Fe^{2+} ;\Delta G^{\circ}_{1} = -1F \times 0.77 $
$Fe^{2+} +2e^{-} \to Fe; \Delta^{\circ}G_{2} = +2F \times 0.44$
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$Fe^{3+} +3e^{-} \to Fe; \Delta G^{\circ}_{3} = - 3F \times E^{\circ}_{Fe^{3+}/Fe}$
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$\Delta G^{\circ}_{3} =\Delta G^{\circ}_{1} + \Delta G^{\circ}_{2}$
$-3F \times E^{\circ}_{Fe^{3+}/Fe} = - 0.77F +0.88F$
$- 3E^{\circ}_{Fe^{3+}/Fe} = 0.11V$
$Q: 2H_{2}O \to O_{2} +4H^{+} +4e^{-} ; E^{\circ} = - 1.23V$
$O_{2} +2H_{2}O +4e^{-} \to 4OH^{-} ;E^{\circ} = + 0.40V$
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$4H_{2}O \to 4H^{+} +4OH^{-} ; E^{\circ} = - 0.83V$
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$R : Cu^{2+} +2e^{-} \to Cu ;E^{\circ} = +0.34V$
$2Cu^{2+} \to 2Cu^{+} +2e^{-} ;E^{\circ} = - 0.52V$
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$Cu^{2+} +Cu \to 2Cu^{+} ;E^{\circ} = - 0.18V$
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$S : Cr^{3+} +3e^{-} \to Cr ;\Delta G^{\circ}_{1} = +3F \times 0.74$
$Cr \to Cr^{2+} +2e^{-} ; \Delta G^{\circ}_{2} = - 2F \times 0.91$
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$Cr^{3+} +e^{-} \to Cr^{2+} ; \Delta G^{\circ}_{3} = - 1F \times E^{\circ}_{Cr^{3+}/Cr^{2+}}$
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$\Delta G^{\circ}_{3} =\Delta G^{\circ}_{1} +\Delta G^{\circ}_{2}$
$-1F \times E^{\circ}_{Cr^{3+}/Cr^{2+}} =2.22F -1.82F =0.4F$
$E^{\circ}_{Cr^{3+}/Cr^{2+}} = - 0.4V$