The **Nernst equation** for a half-cell reaction (M^n+ + n e⁻ → M) at …

Question

The **Nernst equation** for a half-cell reaction (M^n+ + n e⁻ → M) at 298 K can be written as:

Accepted Answer

D. $E = E^0 + \dfrac{0.0591}{n}\log[M^{n+}]$

$E = E^0 - (0.0591/n)\log Q$ where $Q = 1/[M^{n+}]$ for the reduction $M^{n+} + ne^- 	o M$ (since pure solid $M$ has activity 1). So $E = E^0 - (0.0591/n)\log(1/[M^{n+}]) = E^0 + (0.0591/n)\log[M^{n+}]$. Half-cell potential rises with $[M^{n+}]$.

Answer

$E = E^0 / (n\log[M^{n+}])$

Answer

$E = E^0 - \dfrac{0.0591}{n}\log[M^{n+}]$

Answer

$E = E^0 - 0.0591 \, n \log[M^{n+}]$