Electrochemistry - Result Question 41
####43. Given:
[2009]
(i) $Cu^{2+}+2 e^{-} \to Cu, E^{0}=0.337 V$
(ii) $Cu^{2+}+e^{-} \to Cu^{+}, E^{\circ}=0.153 V$
Electrode potential, $E^{\circ}$ for the reaction, $Cu^{+}+e^{-} \to Cu$, will be :
(a) $0.90 V$
(b) $0.30 V$
(c) $0.38 V$
(d) $0.52 V$
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Solution:
- (d) $Cu^{2+}+2 e^{-} \to Cu ; \Delta G^{0}=-n E^{o} F$
$=-2 \times F \times 0.337=-0.674 F$
$Cu^{+} \to Cu^{2+}+e^{-} ; \Delta G^{0}=-n E^{0} F$
$=-1 \times F \times-0.153=0.153 F$
On adding eqn (i) & (ii)
$Cu^{+}+e^{-} \to Cu$;
$\Delta G^{0}=-0.521 F=-n E^{\circ} F ;$
Here $n=1 \quad \therefore E^{\circ}=+0.52 V$
