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Redox Reactions and Couple

The standard ...

Question

The standard reduction potential of Pb and Zn electrodes are – 0.126 and -0.763 volts respectively. The e.m.f of the cell
$Z n | Z n^{2 +} (0.1 M) | | P b^{2 +} (1 M) | P b$ is

0.637 V

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<0.637 V

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>0.637 V

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0.889

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Solution

The correct option is C
>0.637 V

$E = E^{\circ} - \frac{0.059}{n} l o g \frac{Z n^{+ 2}}{P b^{+ 2}} E^{\circ} = E_{c}^{\circ} - E_{a}^{\circ} = - 0.126 - (- 0.763)$

$\Rightarrow - 0.126 + 0.763 = 0.637$

$\Rightarrow 0.637 - \frac{0.059}{2} l o g \frac{0.1}{1}$

$= 0.637 - \frac{0.059}{2} l o g (10 - 1)$

$= 0.637 - \frac{0.059}{2} (- 1) l o g 10 \Rightarrow 0.637 + \frac{0.059}{2} \times 1 = 0.6374 + 0.0295$

$= 0.6665$

i.e. Answer (c) $[> 0.637]$

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