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Electrochemical Series

Given the sta...

Question

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.

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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 = –2.93V, 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.

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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Electrochemical Series

Standard XII Chemistry

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