Given E Cl 2 / Cl - o = 1 .36 V ,   E Cr 3 + / Cr o =   - 0 .74 V E Cr 2 O 7 2 - / Cr 3 + o = 1 …

Strongest reducing agent has the highest tendency to get oxidised. Higher the oxidation potential, higher is the tendency to get oxidised. Reduction potential of ${\mathrm{Cr}}^{3+}$ is least (${\mathrm{E}}_{{\mathrm{Cr}}^{3+}/\mathrm{Cr}}^{\mathrm{o}}=\mathrm{ }-0.74\mathrm{V}$) which indicates that oxidation potential of $\mathrm{Cr}$ is highest (${\mathrm{E}}_{\mathrm{Cr}/{\mathrm{Cr}}^{3+}}^{\mathrm{o}}=\mathrm{ }0.74\mathrm{V}$). Thus, $\mathrm{Cr}$ is strongest reducing agent.