In the Nernst Equations,
$E = E^{\circ} - \frac{2.303 R T}{n F} {log}_{10} Q$
$Q$ will be equal to the equilibrium constant $K_{C}$ , then calculate the cell potential $E$ of the cell.

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Solution

Given, $E = E^{0} - \frac{2.303 * R * T}{n F} l o g_{10} Q$
At condition of equilibrium
cell potential of the cell = 0 and Q = $K_{C}$
so, we have,
$0 = E^{0} - \frac{2.303 * R * T}{n F} l o g_{10} K_{C}$

$E^{0} = \frac{2.303 * R * T}{n F} l o g_{10} K_{C}$
putting value of $E^{0}$ in above equation we have
$E = \frac{2.303 * R * T}{n F} l o g_{10} K_{C} - \frac{2.303 * R * T}{n F} l o g_{10} Q$

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