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

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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Similar questions

Given the standard electrode potentials,
$K^{+} / K = - 2.93 V, A g^{+} / A g = 0.80 V, H g^{2 +} / H g$
$= 0.79 V$
$M g^{2 +} / M g = - 2.37 V . C r^{3 +} / C r = - 0.74 V$
Arrange these metals in their increasing order of reducing power.

Q. Given are the standard electrode potentials of few half-cells. The correct order of these metals in increasing reducing power will be:

K^{+} / K = - 2.93 V, A g^{+} / A g = 0.80 V

M g^{2 +} / M g = - 2.37 V, C r^{3 +} / C r = - 0.74 V

Q. Given the standard reduction potentials,

E_{K^{+} / K}^{\circ} = - 2.93 V, E_{A g^{+} / A g}^{\circ} = + 0.80 V, E_{H g^{2 +} / H g}^{\circ} = 0.79 V

E_{M g^{2 +} / M g}^{\circ} = - 2.37 V, E_{C r^{3 +} / C r}^{\circ} = - 0.74 V

.
The correct increasing order of reducing power is:

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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.

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.

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.