Given the information:

${\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Fe}\left(\mathrm{s}\right)\to \mathrm{Cd}\left(\mathrm{s}\right)+{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right);E^{\circ }=0.04\mathrm{V}$

If excess of iron powder is added to $0.1 \mathrm{M} {\mathrm{CdCl}}_2$ solution at 298 K, the concentration of ${\mathrm{Cd}}^{2+}$ ions at equilibrium will be

Let x be the concentration of ${\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)$ at equilibrium, we will have

$\begin{array}{l}{\mathrm{Cd}}^{2+}\left(\mathrm{aq}\right)+\mathrm{Fe}\left(\mathrm{s}\right)\to \mathrm{Cd}\left(\mathrm{s}\right)+{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)\\ 0.1\mathrm{M}-x\end{array}$

At equilibrium, $\mathrm{E}=0$, and hence

$E^{\circ }=\frac{RT}{2F}\ln \frac{\left[{\mathrm{Fe}}^{2+}\right]}{\left[{\mathrm{Cd}}^{2+}\right]}$.

Thus $0.04\mathrm{V}=\frac{0.059\mathrm{V}}{2}\log \left(\frac{x}{0.1\mathrm{M}-x}\right)$

$\log \left(\frac{x}{0.1\mathrm{M}-x}\right)=\frac{2\times 0.04}{0.059}=1.356$

or $\frac{x}{0.1\mathrm{M}-x}={10}^{1.356}=22.7$

This gives $x=\frac{22.7\times 0.1\mathrm{M}}{23.7}=0.096 \mathrm{M}$

Hence  ${\left[{\mathrm{Cd}}^{2+}\right]}_{\mathrm{eq}}=0.1\mathrm{M}-0.096\mathrm{M}=0.004 \mathrm{M}$