Given: $E^{\circ} ({Sn}^{2 +}, {Sn}^{4 +} ∣ Pt) = 0.15 V, E^{\circ} ({Hg}_{2}^{2 +}, {Hg}^{2 +} ∣ Pt) = 0.92 V$ and $E^{\circ} ({Pb}^{2 +}, H^{+} ∣ {PbO}_{2}) = 1.45 V$ Based on this data, which of the following is correct?

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${Sn}^{4 +}$ is a stronger oxidizing agent than ${Pb}^{4 +}$

${Sn}^{2 +}$ is a stronger reducing agent than ${Hg}^{2 +}$

${Hg}^{2 +}$ is a stronger oxidizing agent than ${Pb}^{4 +}$

${Pb}^{2 +}$ is a stronger reducing agent than ${Sn}^{2 +}$

answer is B.

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Detailed Solution

Larger the reduction potential, larger the oxidation tendency of the ion being reduced. Hence, the oxidizing tendency follows the order ${Pb}^{4 +} > {Hg}^{2 +} > {Sn}^{4 +}$

Lesser the reduction potential, larger the reduction tendency of the ion formed in the reduction reaction. Hence, the reducing tendency follows the order ${Sn}^{2 +} > {Hg}^{2 +} > {Pb}^{2 +}$

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