Arrange $A g, C r$ and $H g$ metals in the increasing order of reducing power. Given:
$E_{A g^{+} / A g}^{\circ} = + 0.80 V$
$E_{C r^{3} / C r}^{\circ} = - 0.74 V$
$E_{H g^{3 +} / H g}^{\circ} = + 0.79 V$

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Solution

As the reduction potential increases, tendency to get reduced increases and hence oxidising power increase. Similarity as reduction potential decreases tendency to get oxidised increases and hence reducing power increases. Therefore, Increasing order of reducing power can be written as

$A g < H g < C r$

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Similar questions

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

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

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

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