The standard reduction potentials (in V) of a few metal ion/metal electrodes are given below.

The reducing strength of the metals follows the order:

A g > C u > P b > C r

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C r > P b > C u > A g

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P b > C r > A g > C u

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C r > A g > C u > P b

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Solution

The correct option is B $C r > P b > C u > A g$
The lower the value of reduction potential, the higher is the standard reduction potential.

$C r^{3 +} / C r = - 0.74;$ $C u^{2 +} / C u = + 0.34;$
$P b^{2 +} / P b = - 0.13;$ $A g^{+} / A g = + 0.8.$

The correct order of reducing strength is: $C r > P b > C u > A g$ .

Hence, option $B$ is correct.

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Similar questions

P b + C u (N O_{3})_{2} \to P b (N O_{3})_{2} + C u

C u + A g (N O_{3})_{2} \to C u (N O_{3})_{2} + A g

Z n + P b (N O_{3})_{2} \to Z n (N O_{3})_{2} + P b

E_{C u / C u^{+ +}}^{\circ} = - 0.34

E_{C r / C r^{+ + +}}^{\circ} = 0.74

E_{P b / P b^{+ +}}^{\circ} = 0.126

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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