Nernst Equation

Subtract the term $(\frac{RT}{zF})\cdot{}\ln{} Q$ from the standard potential E°: $E = E$° $- (\frac{RT}{zF})\cdot{}\ln{} Q$. Here $R =\frac{ 8.314 J}{(mol\cdot{}K)}$, T is the absolute temperature, z the number of transferred electrons, $F =\frac{ 96 485 C}{mol}$ the Faraday constant and Q the reaction quotient. At 25 °C the prefactor simplifies to about 0.0592 V when switching to the base-10 logarithm: $E = E$° $- (\frac{0.0592}{z})\cdot{}\log{} Q$. Example copper electrode with E° = +0.34 V, $[Cu^{2+}] =\frac{ 0.01 mol}{L}$, $z = 2$: $E = 0.34 - (0.0296)\cdot{}\log{}(\frac{1}{0.01}) = 0.34 - 0.0592 = 0.28 V$.

z is the number of electrons transferred per formula turnover in the redox reaction considered. In the reduction of $Cu^{2+}$ to Cu two electrons are taken up, so $z = 2$; for Ag⁺ to Ag only one, $z = 1$. A common mistake is to confuse z with a stoichiometric coefficient from the reaction equation. z always follows from the change in oxidation number, that is from the electron balance of the half-equation. Because z sits in the denominator of the Nernst term, a z twice as large halves the influence of concentration on the potential. Therefore determine z carefully before you substitute.

Because in the Nernst equation the term $(\frac{RT}{zF})\cdot{}\ln{} Q$ is subtracted from the standard potential. If Q increases, ln Q grows, and a larger amount is subtracted, so E falls. A large reaction quotient means that the products dominate over the reactants; the reaction is already well advanced and has less driving force, which the smaller potential reflects. Conversely a small Q, with reactants dominating, gives a higher potential. At equilibrium Q equals the equilibrium constant K and the potential becomes zero; then the cell delivers no voltage any more. In this way the Nernst equation links concentration and cell voltage.

The standard potential E° is a tabulated reference value that holds under standard conditions: all concentrations $\frac{1 mol}{L}$, gas pressures 1 bar and usually 25 °C. The actual potential E deviates from it as soon as the real concentrations differ from the standard values. The Nernst equation calculates exactly this deviation through the term $(\frac{RT}{zF})\cdot{}\ln{} Q$. If all concentrations are $\frac{1 mol}{L}$, then Q equals one, ln Q equals zero, and E corresponds exactly to E°. In practice concentrations are rarely at standard values, so you need the Nernst equation to determine the real potential of a concentration cell or battery.

A concentration cell consists of two identical electrodes in the same electrolyte but with different concentrations. Because the electrodes are chemically identical, the standard potential E° is zero; the voltage arises solely from the Nernst term. According to $E = -(\frac{RT}{zF})\cdot{}\ln{} Q$ the voltage depends only on the concentration ratio of the two half-cells. The more dilute side becomes the negative electrode, the more concentrated one the positive. The current flow equalizes the concentrations until both are equal and the voltage becomes zero. Such cells strikingly show that a concentration difference alone, without a chemical reaction, can create a measurable voltage.