JEE Main · 2004 · 24 Apr 2004 · Offline · Redox Reactions and Electrochemistry

The ease of moving from $\mathrm{M}{}^{2+}$ to $\mathrm{M}{}^{3+}$ is read from $E_{\mathrm{M}{}^{3+}/\mathrm{M}{}^{2+}}^{\circ }$: the more negative this reduction potential, the more readily the $+2$ ion parts with an electron to become $+3$. Chromium's value of $-0.41\text{V}$ is the lowest of the four, so $\mathrm{Cr}{}^{2+}$ is oxidised to $\mathrm{Cr}{}^{3+}$ most easily and indeed behaves as a strong reducing agent.