Chemistry - Electrochemistry Question with Solution | TestHub

Here${\mathrm{E}}_{\mathrm{R}\mathrm{P}}^{\mathrm{o}}$of Cr = - 0.41 so its${\mathrm{E}}_{\mathrm{O}\mathrm{P}}^{\mathrm{o}}$is 0.41.
As here${\mathrm{E}}_{\mathrm{O}\mathrm{P}}^{\mathrm{o}}$for$\mathrm{C}{\mathrm{r}}^{2+}/\mathrm{C}{\mathrm{r}}^{3+}$is maximum +0.41 V so$\mathrm{C}{\mathrm{r}}^{2+}$will be easily oxidised to$\mathrm{C}{\mathrm{r}}^{3+}$.
For rest of the metals${\mathrm{E}}_{\mathrm{O}\mathrm{P}}^{\mathrm{o}}$will be negative so they have less tendency to be oxidized then Cr here.