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

The Correct Option is C

$E^\circ$ 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 $