Chemistry - Electrochemistry Question with Solution | TestHub

$\begin{array}{cccccccc}& {\text{Cd}}^{2+}\left(\text{aq}\right)& +& \text{Fe}\left(\text{s}\right)& \rightleftharpoons & \text{Cd}\left(\text{s}\right)& +& {\text{Fe}}^{2+}\left(\text{aq}\right)\\ \text{At eqm.}& 1-\text{x}& & -& & -& & \text{x}\end{array}$

${\text{At equilibrium, E}}^{\text{o}}=\frac{\text{0.0591}}{2}log\left(\frac{{\text{Fe}}^{2+}}{{\text{Cd}}^{2+}}\right)$

$\text{0.037}=\frac{\text{0.0591}}{2}log\left(\frac{\text{x}}{1-\text{x}}\right)$

$\text{x}=\left[{\text{Fe}}^{2+}\right]\Rightarrow \text{0.947 M}$

$\therefore \left[{\text{Cd}}^{2+}\right]=\text{0.053 M}$