Question

Standard electrode potentials of few half-cell reactions are given below :
$MnO^{-}_{4}+8H^{+}+5e^{-} \to Mn^{2+} +4H_{2}O$ ; $\,E^{\circ}=1.51\,V$
$Cr_{2}O^{2-}_{7}+14H^{+}+6e^{-} \to2Cr^{3+}+7H_{2}O$ ; $\,E^{\circ}=1.33\,V$
$Fe^{3+}+e^{-} \to Fe^{2+}$ ; $\,E^{\circ}=0.77\,V$
$Cl_{2}+2e^{-} \to2Cl$ ; $\,E^{\circ}=1.36\,V $
Based on the above information match the column I with column II and mark the appropriate choice.

	Column I		Column II
(A)	$1 \,mol$ of $MnO^-_4$ to $Mn^{2+}$	(i)	$579000\, C$
(B)	$1 \,mol$ of $Cr_2O_7^{2-}$ to $2Cr^{3+}$	(ii)	$193000 \,C$
(C)	$1 \,mol$ of $Fe^{3+}$ to $Fe^{2+}$	(iii)	$482500\, C$
(D)	$1\,mol$ of $Cl_2$ to $2Cl^-$	(iv)	$96500\ C$

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

Answer 1

Answer: (C)

$MnO^{-}_{4}=96500\times5\,F=482500\,C$
$Cr_{2}O^{2-}_{7}=96500\times6\,F=579000\,C$
$Fe^{3+}=96500\times1\,F=96500\,C$
$Cl_{2}=96500\times2\,F=193000\,C$