Question
$\mathrm{Fe}+\mathrm{Cd}^{2+} \rightarrow \mathrm{Cd}+\mathrm{Fe}^{2+}$ will this reaction take place or not?
$\text{E}^0_\frac{\text{Fe}^{2+}}{\text{Fe}}=-0.44\text{V}$
$\text{E}^0_\frac{\text{Cd}^{2+}}{\text{Cd}}=-0.40\text{V}$
$\text{E}^0_\frac{\text{Fe}^{2+}}{\text{Fe}}=-0.44\text{V}$
$\text{E}^0_\frac{\text{Cd}^{2+}}{\text{Cd}}=-0.40\text{V}$