The emf of the cell, $Z n | Z n^{2 +} (0.01 M) | | F e^{2 +} (0.001 M) | F e$
at $298 K$ is $0.2905 V$ then the value of equilibrium constant for the cell reaction is:

e^{0.32 / 0.0295}

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10^{0.32 / 0.0295}

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10^{0.26.0.0295}

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10^{0.32 / 0.0591}

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Solution

The correct option is A $10^{0.32 / 0.0295}$
we have,
$E = E^{0} - \frac{0.0591}{2} l o g \frac{Z n^{2 +}}{F e^{2 +}}$
$0.2905 = E^{0} - 0.02955 l o g \frac{0.01}{0.001}$
$E_{c e l l}^{0} = 0.32$
At equilibrium condition we have,
$E_{c e l l}^{0} = \frac{0.0591}{2} l o g K$

$\frac{0.32}{0.0295} = l o g K$
$K = 10^{0.32 / 0.0295}$

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