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

0.637 V

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< 0.637 V

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> 0.637 V

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0.889 V

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Solution

The correct option is C $> 0.637 V$
Cathode: $P b^{2 +} + 2 e^{-} \to P b; E_{r e d}^{0} = - 0.126 V$

Anode: $Z n \to Z n^{2 +} + 2 e^{-}; E_{o x d}^{0} = 0.763 V$

All reaction $P b^{2 +} + Z n \to Z n^{2 +} + P b$

$E_{c e l l}^{0} = - 0.126 + 0.763 V E_{c e l l}^{0} = 0.637 V$

Using Nernst Equation

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} l o g \frac{[Z n^{2 +}]}{[P b^{2 +}]}$

$E_{c e l l} = 0.637 - \frac{0.0591}{2} l o g \frac{[0.1]}{[1]}$

$E_{c e l l} = 0.667 V$

Which is greater than $0.637 V$

$E_{c e l l} > 0.637 V$

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The standard reduction potential of Pb and Zn electrodes are −0.126 and −0.763 volts respectively. The e.m.f. of the cell, is:Zn∣∣Zn2+(0.1M)∣∣∣∣Pb2+(1M)∣∣Pb

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