Question

The equilibrium constant for the disproportionation reaction
$2 Cu ^{+}( aq ) \longrightarrow Cu ( s )+ Cu ^{2+}( ag )$ at $25{ }^{\circ} C$
$\left( E ^{\circ} Cu ^{+} / Cu =0 \cdot 52 \,V , E ^{\circ} Cu ^{2+} / Cu =0 \cdot 16\, V \right)$ is

• A. $6 \times 10^{4}$
• B. $6 \times 10^{6}$
• C. $1.2 \times 10^{6}$
• D. $ 1.2 \times 10^{-6}$

Answer 1

Answer: (C)

The reaction
$2 Cu ^{+}( aq ) \longrightarrow Cu ( s )+ Cu ^{2+}( aq ) $
$ E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.0592}{1} \log \frac{\left[ Cu ^{2+}\right]}{\left[ Cu ^{+}\right]^{2}}$
At equilibrium $ E _{\text {cell }}=0 $
$\therefore E_{\text {cell }}^{\circ}=0.0592 \log K_{c}$
or, $ \log K_{c}=\frac{0.52-0.16}{0.0592}$
$\therefore K_{c}=1.2 \times 10^{6}$