The emf of the cell $Z n | Z n^{2 +} (0.1 M) | | F e^{2 +} | F e (0.01 M)$ is $0.2905 V$ . Equilibrium constant for the cell reaction is

10^{\frac{32}{0.591}}

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10^{\frac{0.32}{0.0295}}

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10^{\frac{0.26}{0.0295}}

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10^{\frac{0.32}{0.295}}

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Solution

The correct option is B $10^{\frac{0.32}{0.0295}}$
$E = E^{\circ} - \frac{0.0591}{n} l o g Q$
$0.2905 = E^{\circ} - \frac{0.0591}{n} l o g [\frac{0.1}{0.01}]$
$E^{\circ} = 0.2905 + 0.0295 = + 0.32 V$
Therefore $0.32 = \frac{0.0591}{2} l o g K$
$K = 10^{\frac{0.32}{0.0295}}$

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