Question

# 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

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.