Give the Nernst equation for the following reaction: $M_{(aq)}^{n+} + ne^{-} \to M_{(s)}$

For the electrode reaction: $M_{(aq)}^{n+} + ne^{-} \to M_{(s)}$
The Nernst equation for the electrode potential at any concentration is given by:
$E_{(M^{n+} | M)} = E_{(M^{n+} | M)}^{\circ} - \frac{RT}{nF} \ln \frac{[M]}{[M^{n+}]}$
Since the concentration of a pure solid is taken as unity $([M] = 1)$,the equation simplifies to:
$E_{(M^{n+} | M)} = E_{(M^{n+} | M)}^{\circ} - \frac{RT}{nF} \ln \frac{1}{[M^{n+}]}$
Alternatively,using the property of logarithms:
$E_{(M^{n+} | M)} = E_{(M^{n+} | M)}^{\circ} + \frac{RT}{nF} \ln [M^{n+}]$
Where:
$n = \text{number of electrons involved}$
$E_{(M^{n+} | M)} = \text{electrode potential at non-standard concentration}$
$E_{(M^{n+} | M)}^{\circ} = \text{standard reduction potential}$
$R = \text{Gas constant} = 8.314 \ J \ K^{-1} \ mol^{-1}$
$F = \text{Faraday constant} = 96487 \ C \ mol^{-1}$
$T = \text{Temperature in Kelvin}$
$[M^{n+}] = \text{molar concentration of the species } M^{n+}$