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.

Open in App

Solution

The increasing order of reducing power is as follows:

$A g^{+} | A g < H g^{2 +} | H g < C r^{3 +} | C r < M g^{2 +} | M g < K^{+} | K$
Metal with most positive standard electrode potential has the least reducing power.
Metal with most negative standard electrode potential has maximum reducing power.

Suggest Corrections

Similar questions

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

Given the standard electrode potentials,

$\frac{K^{+}}{K} = - 2.93 V,$

$\frac{A g^{+}}{A g} = 0.80 V$

$\frac{H g^{2 +}}{H g} = 0.79 V,$

$\frac{M g^{2 +}}{M g} = - 2.37 V$

$\frac{C r^{3 +}}{C r} = - 0.74 V$

Arrange these metals in their increasing order of reducing power.

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

Join BYJU'S Learning Program