Given the standard electrode potentials, K + /K = –2.93V, Ag + /Ag = 0.80V, Hg 2+ /Hg = 0.79V Mg 2+ /Mg = –2.37 V, Cr 3+ /Cr = – 0.74V Arrange these metals in their increasing order of reducing power.

Question

Philoid · Accepted Answer

Reducing power of metals increase with the decrease of reduction potential. Hence, the increasing order of reducing power will be as,

Ag < Hg < Cr < Mg < K

Explanation -

When the reduction potential is lower, the element has more tendency to get oxidized and thus more will be reducing power. The metal that has more negative electrode potential will be the one with more reducing power. Thus, here potassium(K) has the highest reducing power among the given elements.

Given the standard electrode potentials,K+ /K = –2.93V, Ag+ /Ag = 0.80V,Hg2+ /Hg = 0.79VMg2+ /Mg = –2.37 V, Cr3+ /Cr = – 0.74VArrange these metals in their increasing order of reducing power.

Given the standard electrode potentials,
K⁺ /K = –2.93V, Ag⁺ /Ag = 0.80V,

Hg²⁺ /Hg = 0.79V

Mg²⁺ /Mg = –2.37 V, Cr³⁺ /Cr = – 0.74V

Arrange these metals in their increasing order of reducing power.