Question

In which direction can the reaction, $2Hg\left(l\right)+2A{g}^{+}(aq.)\rightleftharpoons 2Ag\left(s\right)+H{g}_2^{2+}(aq.)$
proceed spontaneously at the following concentrations of the ions participating in the reactions (i) and (ii)?
(i) $\left[A{g}^{+}\right]={10}^{-4}mol{L}^{-1}$ and $\left[H{g}_2^{2+}\right]={10}^{-1}mol{L}^{-1}$
(ii) $\left[A{g}^{+}\right]={10}^{-1}mol{L}^{-1}$ and $\left[H{g}_2^{2+}\right]={10}^{-4}mol{L}^{-1}$
Given: $E_{H{g}_2^{2+}/Hg}^{\circ }=0.79V;E_{A{g}^{+}/Ag}^{\circ }=0.80V$.

Answer 1

(i) $Q=\frac{\left[H{g}_2^{2+}\right]}{[A{g}^{+}{]}^2}=\frac{{10}^{-1}}{[{10}^{-4}{]}^2}={10}^7$
$E^{\circ }=E_{A{g}^{+}/Ag}^{\circ }-E_{H{g}_2^{2+}/Hg}^{\circ }$
$=0.80-0.79=0.01V$
$E=E^{\circ }-\frac{0.059}{n}\log Q$
$=0.01-\frac{0.059}{2}\log {10}^7$
$=-0.1965V$
Negative value shows that the reaction will proceed from right to left, i.e., in backward direction.
(ii) $Q=\frac{\left[H{g}_2^{2+}\right]}{[A{g}^{+}{]}^2}=\frac{{10}^{-4}}{[{10}^{-1}{]}^2}={10}^{-2}$
$n=2$
$E^{\circ }=0.01volt$
$E=E^{\circ }-\frac{0.059}{n}{\log }_{10}Q$
$0.01-\frac{0.059}{2}{\log }_{10}{10}^{-2}$
$=0.01+0.059V$
$=0.069V$
Since, the value of cell potential is positive, the reaction was proceed spontaneously in forward direction.