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.

Question

Philoid · Accepted Answer

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

Given the standard electrode potentials,K+/K = –2.93V, Ag+/Ag = 0.80V,Hg2+/Hg = 0.79VMg2+/Mg = –2.37V. 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 = –2.37V. Cr³⁺/Cr = –0.74V

arrange these metals in their increasing order of reducing power.