Qstnid -Ii-2- Given The Standard Electrode Potentials, K<sup>+</sup>/k = -2.93v, Ag<sup>+</sup>/ag = 0.80v, Hg<sup>2+</sup>/hg = 0.79v Mg<sup>2+</sup>/mg = -2.37 V, Cr<sup>3+</sup>/cr = - 0.74v Arrange These Metals In Their Increasing Order Of Reducing Power. - 03: Electrochemistry - Neet-xii-chemistry

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NEET-XII-Chemistry

03: Electrochemistry

page 2

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#2
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.

Ans : The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K⁺/K < Mg²⁺/Mg < Cr³⁺/Cr < Hg²⁺/Hg < Ag⁺/Ag.

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K