Question

The equilibrium constant for the disproportionation reaction
$2C{u}^{+}(aq)⟶Cu(s)+C{u}^{2+}(ag)$ at $25{}^{\circ }C$
$\left(E^{\circ }C{u}^{+}/Cu=0\cdot 52V,E^{\circ }C{u}^{2+}/Cu=0\cdot 16V\right)$ 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
$2C{u}^{+}(aq)⟶Cu(s)+C{u}^{2+}(aq)$
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0592}{1}\log \frac{\left[C{u}^{2+}\right]}{{\left[C{u}^{+}\right]}^2}$
At equilibrium $E_{\text{cell }}=0$
$\therefore E_{\text{cell }}^{\circ }=0.0592\log K_c$
or, $\log K_c=\frac{0.52-0.16}{0.0592}$
$\therefore K_c=1.2\times 10^6$