Zn $| Z n^{2 +} (a = 0.1 M) ∥ F e^{2 +} (a = 0.01 M) |$ $F e$ . The emf of the above cell is $0.2905 V$ .

Equilibrium constant for the cell reaction is:

10^{0.32 / 0.0591}

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10^{0.32 / 0.0295}

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10^{0.26 / 0.0295}

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e^{0.32 / 0.295}

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Solution

The correct option is A $10^{0.32 / 0.0295}$
For calculation of standard emf:

According to Nernst equation $E = E_{o} - \frac{0.0591}{n} l o g K_{c}$

There is a transformation of two moles of electrons.

$0.2905 = E_{o} - \frac{0.0591}{2} l o g \frac{0.1}{0.01}$

$E_{o} = 0.2905 + 0.0295 = 0.32$

For calculation of equlibrium constant: $E_{o} = \frac{0.0591}{n} l o g K$

$0.32 = \frac{0.0591}{2} l o g K$

$K = 10^{0.32 / 0.0295}$

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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 :

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