Question

Given:
(i) $Cu ^{2+}+2 e ^{-} \rightarrow Cu ,\, E ^{ o }=0.337\, V$
(i) $Cu ^{2+}+ e ^{-} \rightarrow Cu ^{+},\, E ^{ o }-0.153\, V$
Electrode potential $E ^{\circ}$ for the reaction,
$Cu ^{+}+ e ^{-} \rightarrow Cu$, will be :

• A. 0.52 V
• B. 0.90 V
• C. 0.30 V
• D. 0.38 V

Answer 1

Answer: (A)

The option marked is incorrect.
The correct option is $B $ i.e $0.52 V$
Solution:
$\Delta G ^{\circ}=- nFE { }^{\circ}$
So for the reaction, $C u^{2+}+2 e^{-} \rightarrow C u$
$\Delta G ^{\circ}=2 \times F \times 0.337 \ldots \ldots \ldots .(1)$
for the reaction, $Cu ^{+} \rightarrow C u^{2+}+e^{-}$
$\Delta G ^{\circ}=1 \times F \times 0.153 \ldots \ldots \ldots .(2)$
So adding both the equations we get,
$C u^{+}+e^{-} \rightarrow C u ; \Delta G ^{\circ}=-0.521 F$
$\Delta G ^{\circ}=- nFE$
$\Delta E ^{\circ}=0.52 \,V$