Given the standard electrode potentials,
$K^{+} / K = - 2.93 V A g^{+} / A g = 0.80 V H g^{2 +} / H g = 0.79 V M g^{2 +} / M g = - 2.73 V C r^{3 +} / C r = - 0.74 V$
The increasing order of reducing power is:

A g < C r < H g < K < M g

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M g < K < H g < C r < A g

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A g < H g < C r < M g < K

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K < M g < C r < H g < A g

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Solution

The correct option is C $A g < H g < C r < M g < K$
The given data is standard reduction potential of species. The oxidation potential would have reverse sign of the given values. Higher the oxidation potential more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be $A g < H g < C r < M g < K$ .

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Similar questions

K^{+} + e^{-} \to K; E^{0} = - 2.93 V A g^{+} + e^{-} \to A g; E^{0} = 0.80 V H g_{2}^{2 +} + e^{-} \to 2 H g; E^{0} = 0.79 V M g^{2 +} + e^{-} \to M g; E^{0} = - 2.37 V C r^{3 +} + e^{-} \to C r; E^{0} = - 0.74 V

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

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.

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

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