10.3 Nernst equation and its applications

The Nernst equation connects cell potential to concentration and temperature, letting you move beyond the idealized world of standard conditions. It's the key to predicting how real electrochemical cells behave and to linking thermodynamic quantities like Gibbs free energy and equilibrium constants to measurable voltages.

Nernst equation derivation and significance

Relationship between Gibbs free energy change and cell potential

The Nernst equation grows directly out of the thermodynamic relationship between Gibbs free energy and cell potential:

$\Delta G = -nFE$

• $n$ is the number of moles of electrons transferred in the balanced redox reaction
• $F$ is the Faraday constant, $96{,}485 \text{ C/mol}$
• $E$ is the cell potential (in volts)

Because $\Delta G$ also depends on the reaction quotient through $\Delta G = \Delta G° + RT \ln Q$, you can substitute $-nFE$ for $\Delta G$ and $-nFE°$ for $\Delta G°$, then divide through by $-nF$ to isolate $E$.

Nernst equation expression and components

The result of that derivation is:

$E = E° - \frac{RT}{nF}\ln Q$

• $E°$ is the standard cell potential (all species at unit activity, typically 1 M solutions and 1 bar gases)
• $R = 8.314 \text{ J mol}^{-1}\text{K}^{-1}$ (gas constant)
• $T$ is the absolute temperature in Kelvin
• $Q$ is the reaction quotient, constructed from the balanced cell reaction with products in the numerator and reactants in the denominator, each raised to their stoichiometric coefficients

At 298.15 K, the prefactor $\frac{RT}{F}$ evaluates to 0.02569 V. If you convert from natural log to base-10 log, the equation becomes:

$E = E° - \frac{0.05916 \text{ V}}{n}\log Q \quad (\text{at } 298 \text{ K})$

This simplified form shows up constantly on exams, so it's worth memorizing.

Significance of the Nernst equation in electrochemistry

• It predicts cell potentials under any set of concentrations and temperatures, not just standard conditions.
• A positive $E$ means the reaction is spontaneous as written; a negative $E$ means the reverse reaction is spontaneous.
• Setting $E = 0$ connects the standard cell potential directly to the equilibrium constant, bridging electrochemistry and chemical equilibrium.
• It provides a quantitative framework for understanding batteries, corrosion, biological membrane potentials, and pH-dependent redox chemistry.

Cell potential calculation under non-standard conditions

Step-by-step procedure

1. Write the balanced overall cell reaction and identify $n$, the number of electrons transferred.

2. Find $E°$ by looking up standard reduction potentials for each half-reaction and computing $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$.

3. Write the expression for $Q$ from the balanced reaction. Pure solids and liquids don't appear in $Q$; only aqueous concentrations and gas partial pressures do.

4. Plug into the Nernst equation:

$E = E° - \frac{RT}{nF}\ln Q$

1. Check units. Temperature must be in Kelvin. Concentrations are in molarity (or, more rigorously, activities). The result $E$ comes out in volts.

Worked example

Consider a cell with $E° = 1.23 \text{ V}$, $n = 2$, $T = 298 \text{ K}$, and $Q = 0.1$:

$E = 1.23 \text{ V} - \frac{(8.314)(298)}{(2)(96{,}485)}\ln(0.1)$

$E = 1.23 \text{ V} - (0.01285 \text{ V})(-2.303) = 1.23 \text{ V} + 0.0296 \text{ V} = 1.26 \text{ V}$

Note: the original guide listed this as 1.29 V, but the correct value is approximately 1.26 V. (You'd get 1.29 V only if $n = 1$.)

Special case: Q = 1

When $Q = 1$, $\ln(1) = 0$, so the correction term vanishes entirely and $E = E°$. This confirms that the standard cell potential corresponds to all species at unit activity.

Predicting spontaneous redox reactions

Gibbs free energy change and spontaneity

Because $\Delta G = -nFE$, the sign of $E$ directly tells you about spontaneity:

• $E > 0$ → $\Delta G < 0$ → reaction is spontaneous as written
• $E < 0$ → $\Delta G > 0$ → reverse reaction is spontaneous
• $E = 0$ → $\Delta G = 0$ → system is at equilibrium

For example, if you calculate $E = +0.85 \text{ V}$ for a cell, the forward reaction proceeds spontaneously, with oxidation at the anode and reduction at the cathode.

What a negative cell potential means

If $E = -0.42 \text{ V}$, the forward reaction is non-spontaneous. The reverse reaction is the one that actually proceeds. In practice, you'd need to supply at least 0.42 V of external voltage to drive the forward reaction (electrolysis).

Equilibrium conditions

At equilibrium, $E = 0$ and $Q = K$. No net current flows, and the cell is "dead" in the sense that it can do no electrical work. This is exactly what happens when a battery is fully discharged.

Temperature and concentration effects on cell potential

Temperature effects

Look at the Nernst equation's correction term: $\frac{RT}{nF}\ln Q$. Temperature $T$ multiplies this entire term, so raising the temperature amplifies the deviation from $E°$.

• If $Q < 1$ ($\ln Q < 0$), increasing $T$ makes the correction term more negative, which means subtracting a more negative number, so $E$ increases further above $E°$.
• If $Q > 1$ ($\ln Q > 0$), increasing $T$ makes the correction term more positive, so $E$ decreases further below $E°$.

The key point: higher temperatures push the cell potential further from $E°$ in whichever direction $Q$ dictates.

Concentration effects

Changes in concentration shift $Q$, which shifts $E$:

• Increasing reactant concentration (or decreasing product concentration) → $Q$ decreases → $\ln Q$ becomes more negative → $E$ increases. The cell produces a higher voltage.
• Increasing product concentration (or decreasing reactant concentration) → $Q$ increases → $\ln Q$ becomes more positive → $E$ decreases.

This is Le Chatelier's principle showing up in electrochemistry. The cell "wants" to consume whatever you add in excess.

Example: For a cell with $E° = 1.23 \text{ V}$, $n = 2$, $T = 298 \text{ K}$, shifting $Q$ from 1 to 0.1 (by increasing reactant concentration tenfold) raises the cell potential by about 0.030 V, from 1.23 V to roughly 1.26 V.

Equilibrium constant vs standard cell potential

Deriving the relationship

At equilibrium, $E = 0$ and $Q = K$. Substituting into the Nernst equation:

$0 = E° - \frac{RT}{nF}\ln K$

Rearranging:

$E° = \frac{RT}{nF}\ln K$

Or equivalently:

$\ln K = \frac{nFE°}{RT}$

At 298 K this simplifies to:

$\log K = \frac{nE°}{0.05916 \text{ V}}$

Calculating K from E°

Example: For $E° = 1.23 \text{ V}$, $n = 2$, $T = 298 \text{ K}$:

$\ln K = \frac{(2)(96{,}485)(1.23)}{(8.314)(298)} = \frac{237{,}353}{2477.6} = 95.8$

$K = e^{95.8} \approx 1.6 \times 10^{41}$

That enormous $K$ tells you the reaction goes essentially to completion. Even modest positive values of $E°$ produce very large equilibrium constants because of the $nF$ factor in the exponent.

Calculating E° from K

The same equation works in reverse. Given $K = 1.6 \times 10^{41}$, $n = 2$, $T = 298 \text{ K}$:

$E° = \frac{(8.314)(298)}{(2)(96{,}485)}\ln(1.6 \times 10^{41}) = (0.01285)(95.8) = 1.23 \text{ V}$

Important considerations

• $K$ in this equation should be the thermodynamic equilibrium constant (dimensionless, based on activities). For dilute solutions, molarity is a reasonable approximation to activity, but be aware of the distinction.
• Temperature must be in Kelvin.
• A quick rule of thumb at 298 K: every 1 V of $E°$ with $n = 1$ corresponds to roughly $\log K \approx 16.9$. For $n = 2$, double that. This helps you sanity-check your answers on exams.