#### Question

Given the standard electrode potentials,

K^{+}/K = –2.93V, Ag^{+}/Ag = 0.80V,

Hg^{2+}/Hg = 0.79V

Mg^{2+}/Mg = –2.37V. Cr^{3+}/Cr = –0.74V

Arrange these metals in their increasing order of reducing power.

#### Solution 1

Lower the electrode potential, better is the reducing agent. Since the electrode potentials increase in the oder; K^{+}/K (-2.93 V), Mg^{2+}/Mg (-2.37 V), Cr^{3+}/Cr (-0.74 V), Hg^{2+}/Hg (0.79 V), Ag^{+}/Ag (0.80 V), therefore, reducing power of metals decreases in the same order, i.e., K, Mg, Cr, Hg, Ag

#### Solution 2

The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is Ag < Hg < Cr < Mg < K.

Is there an error in this question or solution?

Solution Given the Standard Electrode Potentials,K+/K = –2.93v, Ag+/Ag = 0.80v,Arrange These Metals in Their Increasing Order of Reducing Power. Concept: Redox Reactions in Terms of Electron Transfer Reactions - Introduction.