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 $0.2905$ then the value of equilibrium constant for the cell reaction is :

e^{\frac{0.32}{0.0295}}

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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}{0.0591}}

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Solution

The correct option is C $10^{\frac{0.32}{0.0295}}$
The expression for the emf of cell is
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0592}{n} l o g \frac{[{Z n}^{2 +}]}{[{F e}^{2 +}]}$
Substitute values in the above expression
$0.2905 V = E_{c e l l}^{0} - \frac{0.0592}{2} l o g \frac{0.01}{0.001} E_{c e l l}^{0} = 0.32 V$
The expression for the equilibrium constant is $E_{c e l l}^{0} = \frac{0.0592}{n} l o g K$
Substitute values in the above expression.
$0.32 V = \frac{0.0592}{2} l o g K l o g K = \frac{0.32}{0.0296} K = 10^{\frac{0.32}{0.0296}} .$

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