Given are the standard electrode potentials of few half-cells. The correct order of these metals in increasing reducing power will be:
$K^{+} / K = - 2.93 V, A g^{+} / A g = 0.80 V$ ,
$M g^{2 +} / M g = - 2.37 V, C r^{3 +} / C r = - 0.74 V$ .

K < M g < C r < A g

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A g < C r < M g < K

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M g < K < C r < A g

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C r < A g < M g < K

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Solution

The correct option is C $A g < C r < M g < K$
Given is the Reduction potentials of the elements. We have to know the relationship between SRP(Standard Reduction Potential) and SOP(Standard Oxidation Potential).
SRP = -(SOP)
More the oxidation potential the element will have more tendency to get oxidized and hence higher reducing power. So, lesser the Reduction potential higher is its reducing power. Hence here, since the decreasing order of Reduction potential is $A g < C r < M g < K$ which is the order of increasing reducing power.
Hence option (B) is correct.

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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

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.

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