The Nernst Equation

Potentials Depend on Concentration

So far every potential was a standard potential — measured at 1 M concentration. But real cells rarely sit at 1 M. As a cell discharges, reactant ions are consumed and product ions build up, so the cell potential changes. The Nernst equation tells us exactly how.

The key insight: electrode potential depends on the concentration (activity) of the ions involved and on temperature. The Nernst equation quantifies this dependence.

For a single electrode reaction $M^{n+} + ne^- \rightarrow M$:

$$E_{(M^{n+}/M)} = E^\circ_{(M^{n+}/M)} - \frac{RT}{nF}\ln\frac{1}{[M^{n+}]}$$

where $R$ = gas constant, $T$ = temperature (K), $n$ = electrons transferred, $F$ = Faraday constant (96487 C mol$^{-1}$).

The Working Form at 298 K

Substituting $R = 8.314$ J K$^{-1}$ mol$^{-1}$, $T = 298$ K, $F = 96487$ C mol$^{-1}$, and converting $\ln$ to $\log_{10}$ (factor 2.303) gives the exam-ready form:

$$\boxed{E = E^\circ - \frac{0.059}{n}\log Q}$$

For a full cell with reaction $aA + bB \rightarrow cC + dD$ transferring $n$ electrons:

$$E_{cell} = E^\circ_{cell} - \frac{0.059}{n}\log\frac{[\text{products}]}{[\text{reactants}]}$$

where $Q$ is the reaction quotient (product concentrations over reactant concentrations, each raised to its stoichiometric power; pure solids and liquids are taken as 1).

Key Point: The constant $\dfrac{2.303RT}{F}=0.059$ V at 298 K. This number appears in nearly every Nernst numerical - commit it to memory.

[JEE Tip] A common trap: $n$ is the number of electrons in the balanced cell reaction, not in one half-reaction written casually. Balance first, then read off $n$.

Applying the Nernst Equation

For the Daniell cell, Zn + Cu$^{2+}$ → Zn$^{2+}$ + Cu ($n = 2$):

$$E_{cell} = E^\circ_{cell} - \frac{0.059}{2}\log\frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$$

Notice only the ion concentrations appear; the solid metals (Zn, Cu) are taken as 1.

As the cell runs, [Zn$^{2+}$] rises and [Cu$^{2+}$] falls, so $\log Q$ grows and $E_{cell}$ decreases — until it reaches zero, at which point the cell is dead (equilibrium).

Solved Examples

Example 1: Electrode potential at non-standard concentration

Calculate the potential of a Zn$^{2+}$/Zn electrode in 0.1 M Zn$^{2+}$. ($E^\circ=-0.76$ V)

Solution: $E = E^\circ - \dfrac{0.059}{n}\log\dfrac{1}{[\text{Zn}^{2+}]}$, $n=2$. $E = -0.76 - \dfrac{0.059}{2}\log\dfrac{1}{0.1} = -0.76 - 0.0295\times1 = -0.7895$ V.

Answer: −0.79 V.

Example 2: Cell emf at given concentrations

For Zn | Zn$^{2+}$(0.1 M) || Cu$^{2+}$(1.0 M) | Cu, find $E_{cell}$. ($E^\circ_{cell}=1.10$ V)

Solution: $E_{cell}=1.10-\dfrac{0.059}{2}\log\dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}=1.10-0.0295\log\dfrac{0.1}{1.0}$. $=1.10-0.0295(-1)=1.10+0.0295=1.13$ V.

Answer: 1.13 V.

Example 3: Effect of dilution on cathode

How does decreasing [Cu$^{2+}$] affect the Daniell cell emf?

Solution: $E_{cell}=E^\circ-\dfrac{0.059}{2}\log\dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$. Lowering [Cu$^{2+}$] increases $Q$, so $\log Q$ rises and $E_{cell}$ decreases. (A cathode running low on its ion weakens the cell.)

Example 4: Hydrogen electrode at a given pH

Calculate the potential of a hydrogen electrode in contact with a solution of pH = 5 (H$_2$ at 1 bar).

Solution: For 2H$^+$ + 2e$^-$ → H$_2$, $E = E^\circ - \dfrac{0.059}{2}\log\dfrac{p_{H_2}}{[\text{H}^+]^2}$ with $E^\circ=0$, $p_{H_2}=1$. $= -\dfrac{0.059}{2}\log\dfrac{1}{[\text{H}^+]^2}=-0.059\log\dfrac{1}{[\text{H}^+]} = -0.059\times\text{pH}$. $= -0.059\times5 = -0.295$ V.

Answer: −0.295 V.

Example 5: Concentration cell

A concentration cell has the same metal in both half-cells: Cu | Cu$^{2+}$(0.01 M) || Cu$^{2+}$(0.1 M) | Cu. Find $E_{cell}$ (note $E^\circ_{cell}=0$).

Solution: $E_{cell}=0-\dfrac{0.059}{2}\log\dfrac{[\text{Cu}^{2+}]_{anode}}{[\text{Cu}^{2+}]_{cathode}}=-\dfrac{0.059}{2}\log\dfrac{0.01}{0.1}$. $=-0.0295\times(-1)=+0.0295$ V.

Answer: 0.0295 V. (In a concentration cell, the dilute side is the anode and emf comes purely from the concentration difference.)

Example 6: When is the cell dead?

At what value of $E_{cell}$ has a galvanic cell reached equilibrium ("dead battery")?

Solution: At equilibrium no net reaction occurs and no work can be drawn, so $E_{cell}=0$. (Then $Q=K$, the equilibrium constant.)

Example 7: Finding n from the balanced reaction

For 2Al + 3Cu$^{2+}$ → 2Al$^{3+}$ + 3Cu, what value of $n$ goes into the Nernst equation?

Solution: Balance electrons: Al → Al$^{3+}$ + 3e$^-$ (×2 gives 6e$^-$); Cu$^{2+}$ + 2e$^-$ → Cu (×3 gives 6e$^-$). So n = 6 electrons transferred.

Example 8: emf with both concentrations non-standard

For Zn | Zn$^{2+}$(0.5 M) || Cu$^{2+}$(0.05 M) | Cu, $E^\circ_{cell}=1.10$ V. Find $E_{cell}$.

Solution: $Q=\dfrac{0.5}{0.05}=10$. $E_{cell}=1.10-\dfrac{0.059}{2}\log10 = 1.10-0.0295=1.0705$ V.

Answer: ≈ 1.07 V.

Example 9: Increasing emf

For the Daniell cell, how should you change [Cu$^{2+}$] and [Zn$^{2+}$] to increase the emf?

Solution: $E_{cell}=E^\circ-\dfrac{0.059}{2}\log\dfrac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$. To increase $E_{cell}$, make $Q$ smaller: increase [Cu$^{2+}$] and/or decrease [Zn$^{2+}$].

Example 10: emf of a silver concentration cell

Ag | Ag$^+$(0.001 M) || Ag$^+$(0.1 M) | Ag. Find $E_{cell}$ ($n=1$).

Solution: $E_{cell}=-0.059\log\dfrac{0.001}{0.1}=-0.059\log(0.01)=-0.059\times(-2)=+0.118$ V.

Answer: 0.118 V.