Question

In the given reaction,
$2 Cu ^{+}(a q) \rightleftharpoons Cu ^{2+}(a q)+ Cu (s)$
$E_{ Cu ^{+} / Cu }^{\circ}=0.6\, V \text { and } E_{ Cu ^{2+} / Cu }^{\circ}=0.41\, V$
Find out the equilibrium constant.

• A. $2.76 \times 10^{2}$
• B. $2.76 \times 10^{4}$
• C. $2.76 \times 10^{6}$
• D. $2.76 \times 10^{8}$

Answer 1

Answer: (C)

Right hand cell reaction,
$Cu ^{+}+e^{-} \longrightarrow Cu$
Left hand cell reaction,
$Cu ^{+}\longrightarrow Cu ^{2+}+e^{-}$
Cell reactions, $2 Cu ^{+} \longrightarrow Cu + Cu ^{2+}$
Cell potential $E_{\text {cell }}^{\circ} =E_{ Cu ^{+} / Cu }^{\circ}-E_{ Cu ^{2+} / Cu ^{+}}^{\circ} $
$=0.60-0.41=0.19\, V$
$-n F E^{\circ}=-R T \ln K_{ eq }$
$\log K_{ eq } =\frac{n E^{\circ}}{(2.303 R T / F)}=\frac{2 \times 0.19\, V }{0.059\, V }=6.44$
$K_{ eq } =10^{6.44}=2.76 \times 10^{6}$