Question

Standard electrode potentials of few half cell reaction are given below:

$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2 +} + 4H_{2}O;Eº = 1.51VCr_{2}O_{7}^{2 -} + 14H^{+} + 6e^{-} \rightarrow 2Cr^{3 +} + 7H_{2}O;Eº = 1.33VFe^{3 +} + e^{-} \rightarrow Fe^{2 +};Eº = 0.77V$

$Cl_{2} + 2e^{-} \rightarrow 2Cl^{-};Eº = 1.36V$

Based on the above information match the column I with column II an mark the appropriate choice.

	Column I		Column II
(A)	1 mol of MnO4− to Mn2+	(i)	579000 C
(B)	1 mol of CrO72− to 2Cr3+	(ii)	193000 C
(C)	1 mol of Fe3+ to Fe2+	(iii)	482500 C
(D)	1 mol of Cl2 to 2Cl−	(iv)	96500 C

• A. (A) $\rightarrow$ (i), (B) $\rightarrow$ (ii), (C) $\rightarrow$ (iii), (D) $\rightarrow$ (iv)
• B. (A) $\rightarrow$ (ii), (B) $\rightarrow$ (iii), (C) $\rightarrow$ (i), (D) $\rightarrow$ (iv)
• C. (A) $\rightarrow$ (iii), (B) $\rightarrow$ (i), (C) $\rightarrow$ (iv), (D) $\rightarrow$ (ii)
• D. (A) $\rightarrow$ (iv), (B) $\rightarrow$ (ii), (C) $\rightarrow$ (iii), (D) $\rightarrow$ (i)

Answer 1

Answer: (C)

(A) $\rightarrow$ (iii), (B) $\rightarrow$ (i), (C) $\rightarrow$ (iv), (D) $\rightarrow$ (ii)

Answer 2

$MnO_{4}^{-} = 96500 \times 5F = 482500C$

$Cr_{2}O_{7}^{2 -} = 96500 \times 6F = 579000C$

$Fe^{3 +} = 96500 \times 1F = 96500C$

$Cl_{2} = 96500 \times 2F = 193000C$