Question

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

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

Answer 1

Answer: (C)

0.521 V

Answer 2

$E^\circ$ is an intensive property :
\hspace25mm E^\circ \, \, \, \, \, \, \, \Delta G^\circ = - n E^\circ F
(i) $Cu^{2+} +2e^- \longrightarrow Cu \, \, \, \, \, \, \, \, \, 0.337\, V \, \, \, \, \, \, \, -0.674\, F$
(ii) $Cu^{2+} +e^- \longrightarrow Cu^+ \, \, \, \, \, \, \, \, \, 0.153\, V \, \, \, \, \, \, \, -0.153\, F$
Subtracting (ii) from (i) gives :
$\, \, \, \, \, Cu^+ +e^- \longrightarrow Cu \, \, \, \, \, \, \, \, \, \Delta G^\circ = -0.521\, F = - n E^\circ F$
\Rightarrow \hspace10mm E^\circ = 0.521\, V
\because \hspace15mm n = 1
Solutions ( Nos. 13 to 14) For the given concentration cell,
the cell reaction are $M \longrightarrow M^{2+}$ at left hand electrode.
\hspace30mm M^{2+} \longrightarrow M at right hand electrode
$\Rightarrow M^{2+} (RHS\, electrode) \longrightarrow M^{2+} (LHS\, electrode)$
\hspace80mm E^\circ = 0
Applying Nemst equation
$\, \, \, \, \, E_{cell}=0.059=0-\frac{0.059}{2}log\frac{[M^{2+}]at\, LHS\, electrode}{0.001}$
$\Rightarrow log\frac{[M^{2+}]at\, LHS\, electrode}{0.001}=-2$
$\Rightarrow \, \, \, \, [M^{2+}]at\, LHS\, electrode = 10^{-2} \times 0.001 = 10^{-5}\, M$