The reduction potentials of some of the reduction reactions are:
$\begin{array}{ll} {Mg}^{2 +} + 2 e^{-} \to Mg; E^{\circ} = - 2.37 V \\ {Au}^{3 +} + 3 e^{-} \to Au; E^{\circ} = 1.46 V \\ {Pb}^{2 +} + 2 e^{-} \to Pb; E^{\circ} = - 0.13 V \end{array}$
The order of ions reduced at the cathode will be

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${Mg}^{2 +} > {Au}^{3 +} > {Pb}^{2 +}$

${Au}^{3 +} > {Mg}^{2 +} > {Pb}^{2 +}$

${Mg}^{2 +} > {Pb}^{2 +} > {Au}^{3 +}$

${Au}^{3 +} > {Pb}^{2 +} > {Mg}^{2 +}$

answer is A.

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Detailed Solution

The cathodic processes (i.e. reduction) take place in the order of decreasing reduction potentials.

With reference to hydrogen, which has a global reduction potential of zero in electrochemistry, reduction potential (Eo) is defined as the propensity of a chemical species to be reduced by receiving an electron.

The relationship between SRP(Standard Reduction Potential) and SOP(Standard Oxidation Potential).

SRP = -(SOP)

More the oxidation potential the element will have more tendency to get oxidized, hence higher reducing power. So, lesser the Reduction potential, higher is its reducing power. Hence here, since the decreasing order of Reduction potential is Ag<Cr<Mg<K which is the order of increasing reducing power.

Hence option (B) is correct.

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