Question

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

Answer 1

Hint: Electrode potential in electrochemistry, according to IUPAC, is the electromotive force of a cell built up between the two electrodes. The reduction potential is related to the reducing power of the elements.

Complete step by step answer:
The electrode potential is the measure of the energy per unit charge which is used to drive the oxidation-reduction reactions. It is calculated in terms of two half-reactions, an oxidation half-reaction, and a reduction half-reaction.
$Reduced\quad species\quad \longrightarrow \quad Oxidized\quad species\quad +\quad n{ e }^{ - }$
$Oxidized\quad species\quad +\quad n{ e }^{ - }\quad \longrightarrow \quad Reduced\quad species$
The electrode potential is measure with a reference electrode which is known as the Standard Hydrogen Electrode (SHE) which has the electrode potential as 0 V. A negative value of the potential means that the compound has a greater tendency to get oxidize and reduce the other compound, and a positive value means that the compound has a greater tendency to get reduced and oxidize the other compound.
This means that lower the reduction potential, the higher will be the reducing power. Thus, the increasing order of the standard reduction potentials is ${K}^{+}/K < {Mg}^{2+}/Mg < {Cr}^{3+}/Cr < {Hg}^{2+}/Hg < {Ag}^{+}/Ag$.
Therefore, the increasing order of reducing power is $Ag < Hg < Cr < Mg < K$.

Note: The electrode potential cannot be determined individually but in a reaction with some other electrode. And the electrode potential depends upon the concentration of the substances, the temperature, and the pressure in the case of a gas electrode.