Question

Voltage of the cell $Pt,$ $H_2$($1$ atm) $HOCN$ $\left({10}^{-3}M\right)||A{g}^{+}\left(0.8M\right)|Ag\left(s\right)$ is $0.98$V. Calculate the $K_a$ for $HOCN$. Neglect $\left[H^{+}\right]$ because of oxidation of $H_2\left(g\right)$.
Given: $\frac{2.303RT}{F}=0.06$.

Answer 1

$A{g}^{+}+e^{-}\to Ag$

$E{}_{A{g}^{+}/Ag}=E^o{}_{A{g}^{+}/Ag}-\frac{0.0591}{1}\log \frac{1}{\left[A{g}^{+}\right]}$

$=0.8-0.0591\log \frac{1}{0.8}$

$E{}_{A{g}^{+}/Ag}=0.794$

$E_{H_2/H^{+}}+E_{A{g}^{+}/Ag}=0.982$

$E_{H_2/H^{+}}=0.188$

$\frac{1}{2}{H}_2\to H^{+}+e$

$0.188=0-0.059\log \left[H^{+}\right]$

$\left[H^{+}\right]=6.6\times {10}^{-4}$

$C\alpha =6.6\times {10}^{-4}$

$\alpha =\frac{6.6\times {10}^{-4}}{1.3\times {10}^{-3}}=0.5$

$K_a=\frac{C{\alpha }^2}{1-\alpha }$

$K_a=6.74\times {10}^{-4}$

We get$,$ $K_a=6.74\times {10}^{-4}$.