Given are the standard electrode potentials of a few half-cells. The correct order of these metals in increasing reducing power will be :
${{K}^{+}}/K$ = -2.93V,$A{{g}^{+}}/Ag$ = 0.80 V
$M{{g}^{2+}}/Mg$ = -2.37V, $C{{r}^{3+}}/Cr$ = -0.74V.
(A) K < Mg < Cr < Ag
(B) Ag < Cr < Mg < K
(C) Mg < K < Cr < Ag
(D) Cr < Ag < Mg < K

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Hint: The metal which can be easily oxidized will have the highest reducing power as it can reduce the complementary ion/atom to its reduced state by providing electrons. Higher the value of oxidation potential higher the reducing power of the metal.

Complete step by step answer:
Standard electrode potential is defined as the potential of a cell measured under standard conditions i.e. the pressure is 1atm, temperature is 298 K and solids/liquids are in their pure state.
The reduction potential is considered as the standard electrode potential. The metal which can easily release an electron will have high oxidation potential and the respective metal ion will have low reduction potential as accepting an electron is unfavorable.
Hence, we can conclude that the metal with the lowest standard electrode potential will have the highest oxidation potential as well as the reducing power.
When we compare the standard electrode potentials of the 4 metals, the increasing order of reducing power is Ag < Cr < Mg < K.
This is because $A{{g}^{+}}/Ag$ has an electrode potential of 0.80V, $C{{r}^{3+}}/Cr$ has an electrode potential of -0.74 V, $M{{g}^{2+}}/Mg$ has an electrode potential of -2.73V and ${{K}^{+}}/K$has an electrode potential of -2.93V.
Therefore, the correct answer is option (B).

Note: The oxidation potential of an electrode is the negation of the reduction potential. The electrode potential of standard hydrogen electrode is considered as the basis for calculation of electrode potential and hence its potential is 0V.