Given E _ Cl _ 2 / Cl ^ - ^ o = 1 .36 V, E _ Cr ^ 3 + / Cr ^ o = -...

Strongest reducing agent has the highest tendency to get oxidised. Higher the oxidation potential, higher is the tendency to get oxidised. Reduction potential of ${Cr}^{3 +}$ is least (${E}_{{Cr}^{3 +} / Cr}^{o} = - 0 .74 V$) which indicates that oxidation potential of $Cr$ is highest (${E}_{Cr / {Cr}^{3 +}}^{o} = 0 .74 V$). Thus, $Cr$ is strongest reducing agent.