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Given the Standard Electrode Potentials,K+/K = –2.93v, Ag+/Ag = 0.80v,Arrange These Metals in Their Increasing Order of Reducing Power. - Chemistry

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.

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Solution 1

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.

Solution 2

Lower the electrode potential, better is the reducing agent. Since the electrode potentials increase in the oder; K+/K (-2.93 V), Mg2+/Mg (-2.37 V), Cr3+/Cr (-0.74 V), Hg2+/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

Concept: Redox Reactions in Terms of Electron Transfer Reactions - Introduction
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 8 Redox Reactions
Q 29 | Page 275
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