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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. - Chemistry

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प्रश्न

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.

टीपा लिहा
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उत्तर

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.

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Redox Reactions in Terms of Electron Transfer Reactions - Introduction
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