The equilibrium constant for the reaction given below at 298 K is if $E_{c e l l} = 0.2905 V$ at 298 K is:

$Z n (s) + F e^{2 +} (a q) 0.01 M \to Z n^{2 +} 0.1 M (a q) + F e (s)$

e^{0.32 / 0.0295}

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10^{0.595 / 0.76}

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10^{0.0250 / 0.32}

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10^{0.32 / 0.295}

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Solution

The correct option is A $e^{0.32 / 0.0295}$
According to Nernst equation,
$E = E^{o} - \frac{0.059}{n} l o g Q$

$E_{c e l l} = E_{c e l l}^{o} + \frac{0.059}{2} l o g \frac{[F e^{2 +}]}{[Z n^{2 +}]}$

$0.2905 = E_{c e l l}^{o} + \frac{0.059}{2} l o g \frac{0.01}{0.10}$

$∴ E_{c e l l}^{o} = 0.32$

Now, $E_{c e l l}^{o} = \frac{0.059}{2} l o g_{10} K$

$∴ 0.32 = \frac{0.059}{2} l o g_{10} K$

$K = 10^{0.32 / 0.0295}$

Hence, option A is correct.

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