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

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Solution

Lower reduction potential corresponds to higher reducing power.

The increasing order of the standard reduction potentials is $K^{+} / K < M g^{2 +} / M g < C r^{3 +} / C r < H g^{2 +} / H g < A g^{+} / A g$ .

The increasing order of the reducing power is $A g < H g < C r < M g < K$ .

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Similar questions

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

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

A g, C r

H g

E_{A g^{+} / A g}^{\circ} = + 0.80 V

E_{C r^{3} / C r}^{\circ} = - 0.74 V

E_{H g^{3 +} / H g}^{\circ} = + 0.79 V

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