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The Nernst Equation and Cell Potential

How does a cell's voltage change with concentration, and how is it linked to K and ΔG?

IntermediateRedox & ElectrochemistryLast reviewed 6 October 2026
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What is it?

Standard cell potentials, E°E°, apply only when every dissolved species is at 1 mol/L (and gases at 1 bar). Real cells are rarely standard, and as a battery runs, the concentrations change. The Nernst equation gives the actual cell potential EE:

E=E°−RTnFln⁡QE = E° - \frac{RT}{nF}\ln Q

where nn is the number of electrons transferred, F=96 485F = 96\,485 C/mol is the Faraday constant and QQ is the reaction quotient. At 25 °C this becomes:

E=E°−0.0592 Vnlog⁡QE = E° - \frac{0.0592\ \text{V}}{n}\log Q

Key idea

As a cell runs, products build up and reactants are used, so Q increases and E falls. When QQ reaches KK, the reaction is at equilibrium and E=0E = 0: the battery is flat.

Why does it matter?

  • Batteries. The Nernst equation explains why a battery’s voltage drops as it is used.
  • Measuring concentrations. pH meters and ion-selective electrodes measure a voltage that depends on concentration through the Nernst equation.
  • Biology. Nerve cells hold different ion concentrations inside and outside; the resulting voltage (membrane potential) is a concentration cell.

How does it work?

1. Using the Nernst equation

  1. Write the balanced overall equation and find E°cellE°_\text{cell} and nn.
  2. Write QQ (products over reactants; leave out solids).
  3. Substitute into E=E°−0.0592 Vnlog⁡QE = E° - \frac{0.0592\ \text{V}}{n}\log Q.

2. Concentration cells

Two half-cells of the same metal in solutions of different concentration still give a voltage, even though E°=0E° = 0. Electrons flow so as to make the concentrations equal: the dilute side is oxidized (anode) and the concentrated side is reduced (cathode).

3. E°, ΔG° and K

Three quantities describe how far a reaction “wants” to go, and each can be found from the others:

ΔG°=−nFE°ΔG°=−RTln⁡Klog⁡K=nE°0.0592 V  (at 25 °C)\small\begin{aligned} &\Delta G° = -nFE° \\[4pt] &\Delta G° = -RT\ln K \\[4pt] &\log K \\[4pt] &= \frac{nE°}{0.0592\ \text{V}}\ \ \text{(at 25 °C)} \end{aligned}
E°cellE°_\text{cell}ΔG°\Delta G°KKReaction at standard conditions
positivenegativegreater than 1spontaneous, products favoured
zerozero1at equilibrium
negativepositiveless than 1non-spontaneous, reactants favoured

Think of it like this

A cell is like water flowing between two tanks. E°E° is the height difference when both tanks start at a standard level. As water flows, the upper tank empties and the lower one fills, so the height difference (E) shrinks until the levels are equal and the flow stops (equilibrium, E = 0).

More precisely

The value 0.0592 V comes from RTFln⁡10\frac{RT}{F}\ln 10 at 298.15 K; at other temperatures use the full form. Because log⁡K=nE°/0.0592\log K = nE°/0.0592, a standard potential of only about 0.2 V for a 2-electron reaction already gives K≈107K \approx 10^{7}: small voltages correspond to very large equilibrium constants. Strictly, QQ uses activities, so the Nernst equation is most accurate for dilute solutions.

Visualise it

Graph of cell potential E against log Q for the zinc–copper cell. E equals 1.10 volts when log Q is zero (standard conditions) and falls in a straight line with slope minus 0.0296 volts per unit of log Q, reaching zero at log Q of about 37, where Q equals K and the cell is flat.
For the Zn–Cu cell, E falls by 0.0296 V for every tenfold increase in Q, reaching 0 when Q = K.

Worked example

Worked example: A non-standard Zn–Cu cell

Question: Calculate EE at 25 °C for Zn(s)+CuX2+(aq)→ZnX2+(aq)+Cu(s)\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)} when [ZnX2+]=1.0[\ce{Zn^2+}] = 1.0 mol/L and [CuX2+]=0.010[\ce{Cu^2+}] = 0.010 mol/L.

  1. E°=0.34 V−(−0.76 V)=1.10 VE° = 0.34\ \text{V} - (-0.76\ \text{V}) = 1.10\ \text{V}; n=2n = 2.

  2. Q=[ZnX2+][CuX2+]=1.00.010=100Q = \dfrac{[\ce{Zn^2+}]}{[\ce{Cu^2+}]} = \dfrac{1.0}{0.010} = 100

  3. Nernst:

    E=1.10 V−0.0592 V2log⁡100=1.10 V−0.0592 V=1.04 V\small\begin{aligned} &E = 1.10\ \text{V} \\[4pt] &\quad - \frac{0.0592\ \text{V}}{2}\log 100 \\[4pt] &= 1.10\ \text{V} - 0.0592\ \text{V} \\[4pt] &= 1.04\ \text{V} \end{aligned}

  4. Less CuX2+\ce{Cu^2+} (a reactant) means a slightly lower voltage.

Worked example: A concentration cell

Question: Two copper electrodes dip into 0.0010 mol/L and 1.0 mol/L CuX2+\ce{Cu^2+}. Calculate the cell potential at 25 °C.

  1. E°=0E° = 0 V (same half-reaction); n=2n = 2. The dilute side is the anode: Q=[CuX2+]dilute[CuX2+]concentrated=0.00101.0Q = \dfrac{[\ce{Cu^2+}]_\text{dilute}}{[\ce{Cu^2+}]_\text{concentrated}} = \dfrac{0.0010}{1.0}.

  2. Nernst:

    E=0−0.0592 V2log⁡(1.0×10−3)=0.089 V\small\begin{aligned} &E = 0 - \frac{0.0592\ \text{V}}{2}\log(1.0 \\[4pt] &\quad \times 10^{-3}) \\[4pt] &= 0.089\ \text{V} \end{aligned}

Worked example: From E° to K

Question: Calculate KK at 25 °C for the Zn–Cu reaction (E°=1.10E° = 1.10 V, n=2n = 2).

log⁡K=2×1.10 V0.0592 V=37.16K=1037.16=1.5×1037\small\begin{aligned} &\log K = \frac{2 \times 1.10\ \text{V}}{0.0592\ \text{V}} \\[4pt] &= 37.16 \\[4pt] &K = 10^{37.16} = 1.5 \times 10^{37} \end{aligned}

The reaction goes essentially to completion.

Worked example: ΔG from E

Question: Calculate ΔG\Delta G for the non-standard Zn–Cu cell above (E=1.04E = 1.04 V).

ΔG=−nFE=−2×96 485 Cmol×1.04 V=−2.01×105 J/mol=−201 kJ/mol\small\begin{aligned} &\Delta G = -nFE \\[4pt] &= -2 \times 96\,485\ \tfrac{\text{C}}{\text{mol}} \times 1.04\ \text{V} \\[4pt] &= -2.01 \times 10^{5}\ \text{J/mol} \\[4pt] &= -201\ \text{kJ/mol} \end{aligned}

(1 C × 1 V = 1 J.)

Common mistake

Common mistake: Using the wrong n

nn is the number of electrons transferred in the balanced overall equation: 2 for Zn + Cu²⁺, not 1.

Common mistake: Getting Q upside down

QQ is products over reactants, like KK. For Zn+CuX2+→ZnX2++Cu\ce{Zn + Cu^2+ -> Zn^2+ + Cu}, Q=[ZnX2+]/[CuX2+]Q = [\ce{Zn^2+}]/[\ce{Cu^2+}]. Solids (Zn, Cu) are left out.

Common mistake: Mixing ln and log

RTnFln⁡Q\frac{RT}{nF}\ln Q and 0.0592 Vnlog⁡Q\frac{0.0592\ \text{V}}{n}\log Q are the same at 25 °C. Don’t use 0.0592 with ln.

Notation note

  • E°E°: standard potential; EE: actual potential. FF = 96 485 C/mol.
  • 1 V = 1 J/C, so nFEnFE comes out in J/mol.

Remember this

Remember this

  • Nernst: E=E°−RTnFln⁡QE = E° - \frac{RT}{nF}\ln Q; at 25 °C, E=E°−0.0592 Vnlog⁡QE = E° - \frac{0.0592\ \text{V}}{n}\log Q.
  • As the cell runs, Q rises and E falls; at equilibrium Q = K and E = 0.
  • Concentration cells: same electrodes, different concentrations, E° = 0 but E ≠ 0.
  • ΔG°=−nFE°=−RTln⁡K\Delta G° = -nFE° = -RT\ln K; log⁡K=nE°/0.0592\log K = nE°/0.0592 V at 25 °C.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    State the Nernst equation.
    Answer

    E=E°−RTnFln⁡QE = E° - \dfrac{RT}{nF}\ln Q

  2. Question
    Give the Nernst equation at 25 °C with log.
    Answer

    E=E°−0.0592 Vnlog⁡QE = E° - \dfrac{0.0592\ \text{V}}{n}\log Q

  3. Question
    Why does a battery's voltage fall as it is used?
    Answer

    Products build up and reactants are used, so Q increases and E decreases.

  4. Question
    What is E when the cell reaction reaches equilibrium?
    Answer

    Zero: Q = K and the battery is flat.

  5. Question
    What is a concentration cell?
    Answer

    A cell with the same electrode reaction on both sides but different concentrations; E° = 0 but E is not zero.

  6. Question
    In a concentration cell, which side is the anode?
    Answer

    The more dilute side (it is oxidized, raising its concentration).

  7. Question
    Give the relationship between ΔG° and E°.
    Answer

    ΔG°=−nFE°\Delta G° = -nFE°, with F = 96 485 C/mol.

  8. Question
    Give the relationship between E° and K at 25 °C.
    Answer

    log⁡K=nE°0.0592 V\log K = \dfrac{nE°}{0.0592\ \text{V}}

  9. Question
    A cell has E° = 1.10 V and n = 2. What is K at 25 °C?
    Answer

    log K = 2 × 1.10 ÷ 0.0592 = 37.16, so K = 1.5 × 10³⁷.

  10. Question
    If E° is positive, what are the signs of ΔG° and the size of K?
    Answer

    ΔG° negative and K greater than 1: spontaneous, products favoured.

Quiz

7 questions

  1. Question 1EasyAs a galvanic cell discharges, what happens to Q and E?
    Show answer

    Answer: Q increases, E decreases

    Products accumulate, so Q rises; by the Nernst equation E = E° − (0.0592/n) log Q, E falls until it reaches zero at equilibrium.

  2. Question 2EasyWhat is the cell potential when the cell reaction is at equilibrium?
    Show answer

    Answer: 0 V

    At equilibrium Q = K, there is no driving force and E = 0: the battery is flat.

  3. Question 3MediumFor Zn + Cu²⁺ → Zn²⁺ + Cu (E° = 1.10 V), what is E when [Zn²⁺] = 1.0 M and [Cu²⁺] = 0.010 M?
    Show answer

    Answer: 1.04 V

    Q = 1.0 ÷ 0.010 = 100; E = 1.10 V − (0.0592 V ÷ 2) × log 100 = 1.10 V − 0.0592 V = 1.04 V.

  4. Question 4MediumA silver concentration cell has [Ag⁺] = 0.0010 M on one side and 0.10 M on the other. What is E at 25 °C? (n = 1)
    Show answer

    Answer: 0.118 V

    E° = 0; Q = 0.0010 ÷ 0.10 = 0.010; E = 0 − 0.0592 V × log(0.010) = +0.118 V.

  5. Question 5MediumA reaction has E° = 0.46 V and n = 2. What is ΔG°?
    Show answer

    Answer: −88.8 kJ/mol

    ΔG° = −nFE° = −2 × 96 485 C/mol × 0.46 V = −88 770 J/mol = −88.8 kJ/mol.

  6. Question 6HardFor E° = 0.46 V and n = 2 at 25 °C, what is K?
    Show answer

    Answer: 3.5 × 10¹⁵

    log K = 2 × 0.46 ÷ 0.0592 = 15.54, so K = 10¹⁵·⁵⁴ = 3.5 × 10¹⁵. 3.5 × 10⁷ forgets n = 2.

  7. Question 7HardA reaction has K = 1.0 × 10⁵ and n = 1 at 25 °C. What is E°?
    Show answer

    Answer: 0.30 V

    E° = (0.0592 V ÷ n) × log K = 0.0592 V × 5.0 = 0.296 V ≈ 0.30 V. K greater than 1 means E° is positive.

Notes and downloads

  • Worksheet

    The Nernst Equation Worksheet

    8 questions on the Nernst equation, concentration cells and converting between E°, ΔG° and K. Answer key included.

    IntermediateFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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