Given the standard electrode potentials, ...

NCERT Class XI Chemistry Redox Reactions Solutions

Question : 29 of 30

Marks: +1, -0

Given the standard electrode potentials,

\mathrm{K}^{+}/\mathrm{K}

= –2.93 V,

\mathrm{Ag}^{+}/\mathrm{Ag}

= 0.80 V,

\mathrm{Hg}^{2+}/\mathrm{Hg}

= 0.79 V,

\mathrm{Mg}^{2+}/\mathrm{Mg}

= –2.37 V,

\mathrm{Cr}^{3+}/\mathrm{Cr}

= –0.74 V. Arrange these metals in increasing order of their reducing power.

Solution:

Lower the electrode potential, better is the reducing power. Since the electrode potentials increase in the order;

\mathrm{K}^{+}/\mathrm{K}

(– 2.93 V),

\mathrm{Mg}^{2+}/\mathrm{Mg}

(– 2.37 V),

\mathrm{Cr}^{3+}/\mathrm{Cr}

(– 0.74 V),

\mathrm{Hg}^{2+}/\mathrm{Hg}

(0.79 V),

\mathrm{Ag}^{+}/\mathrm{Ag}

(0.80 V), therefore, reducing power of metals increases in the order, i.e., Ag < Hg < Cr < Mg < K.

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