Question

The standard reduction potentials of $C{u}^{2+}/Cu$ and $C{u}^{2+}/C{u}^{+}$ are $0.337V$ and $0.153V$ respectively. The standard electrode potential of $C{u}^{+}/Cu$ half-cell is

• A. 0.184 V
• B. 0.887 V
• C. 0.521 V
• D. 0.490 V

Answer 1

Answer: (C)

$E^{\circ }$ is an intensive property :
$E^{\circ }\Delta G^{\circ }=-n{E}^{\circ }F$
(i) $C{u}^{2+}+2e^{-}⟶Cu0.337V-0.674F$
(ii) $C{u}^{2+}+e^{-}⟶C{u}^{+}0.153V-0.153F$
Subtracting (ii) from (i) gives :
$C{u}^{+}+e^{-}⟶Cu\Delta G^{\circ }=-0.521F=-n{E}^{\circ }F$
$\Rightarrow E^{\circ }=0.521V$
$\because n=1$
For the given concentration cell,
the cell reaction are $M⟶M^{2+}$ at left hand electrode.
$M^{2+}⟶M$ at right hand electrode
$\Rightarrow M^{2+}$ (RHS electrode) $⟶M^{2+}$ (LHS electrode)
$E^{\circ }=0$
Applying Nernst equation
$E_{\text{cell }}=0.059=0-\frac{0.059}{2}\log \frac{\left[M^{2+}\right]\text{ at LHS electrode }}{0.001}$
$\Rightarrow \log \frac{\left[M^{2+}\right]\text{ at LHS electrode }}{0.001}=-2$
$\Rightarrow \left[M^{2+}\right]\text{ at LHS electrode }=10^{-2}\times 0.001=10^{-5}M$