Question
Given the standard electrode potentials,
$\begin{array}{l} K ^{+} / K =-2.93 V, Ag ^{+} / Ag =0.80 V \\ Hg ^{2+} / Hg =0.79 V \\ Mg ^{2+} / Mg =-2.37 V, Cr ^{3+} / Cr =-0.74 V\end{array}$
Arrange these metals in their increasing order of reducing power.

Answer

When the oxidation potential of the metal is high i.e. the reduction potential of the metal ion is low then the tendency of that metal to donate electron is high and it is a strong reducing agent. Therefore, on the basis of given standard electrode potential (reduction potential) values, the increaseing order of reducing capacity of these metal will be as follows : Ag < Hg < Cr < Mg < K

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