Concentration Cells: EMF from the Nernst Equation

By Aniket Bhardwaj · 15 October 2026 · Updated 15 October 2026 · Electrochemistry

A concentration cell has the same electrode material and the same electrolyte on both sides. The only difference is concentration. There is no difference in standard electrode potential, so E°_cell = 0, and the whole EMF comes from the concentration ratio. These cells appear in IIT-JAM, GATE and CSIR-NET electrochemistry questions, and the calculation is short once you know which side is the anode.

Two worked concentration cells show the same ratio of 100, with zinc EMF 0.0592 V and silver EMF 0.1183 V. — The same concentration ratio gives double the EMF when n falls from 2 to 1.

The formula

For M | Mⁿ⁺(C₁) || Mⁿ⁺(C₂) | M
E_cell = E° − (RT ÷ nF) ln Q, with E° = 0 and Q = C₁ ÷ C₂
So E_cell = (RT ÷ nF) ln(C₂ ÷ C₁) = (0.0592 ÷ n) log₁₀(C₂ ÷ C₁) volts at 298 K

Here C₁ is the concentration at the anode (left, where oxidation happens) and C₂ at the cathode (right, where reduction happens). n is the number of electrons in the electrode reaction. R = 8.314 J mol⁻¹ K⁻¹, F = 96485 C mol⁻¹. The factor 2.303RT/F = 0.05916 V at 298.15 K, and many books round it to 0.0592 or 0.059.

Which side is the anode?

Electrons flow so that the cell moves toward equal concentrations. The dilute half-cell loses ions to the solution by oxidation (M → Mⁿ⁺ + ne⁻), so it is the anode. The concentrated half-cell gains metal by reduction, so it is the cathode. The EMF is positive when you write the dilute side on the left. The cell stops producing voltage when C₁ = C₂, because then Q = 1 and E = 0.

Worked example 1: zinc concentration cell

Question. Zn | Zn²⁺ (0.010 M) || Zn²⁺ (1.0 M) | Zn at 298 K. Find the EMF and ΔG.

Step 1. n = 2. The ratio C₂/C₁ = 1.0 ÷ 0.010 = 100, and log 100 = 2.

Step 2. E = (0.05916 ÷ 2) × 2 = 0.0592 V.

Step 3. ΔG = −nFE = −2 × 96485 × 0.05916 = −11416 J/mol, about −11.4 kJ/mol. The negative sign shows the process is spontaneous.

Worked example 2: silver concentration cell

Question. Ag | Ag⁺ (0.0010 M) || Ag⁺ (0.10 M) | Ag at 298 K. Find E_cell.

Step 1. n = 1. C₂/C₁ = 0.10 ÷ 0.0010 = 100, so log = 2.

Step 2. E = 0.05916 × 2 = 0.1183 V.

Compare. The ratio is the same (100) as in the zinc cell, but E is double because n = 1 instead of 2. The EMF depends on the ratio of concentrations, not on their actual values, and inversely on n. ΔG = −1 × 96485 × 0.1183 = −11.4 kJ/mol, almost identical to the zinc cell. This is expected: in both cells nFE = RT ln(100), so ΔG = −RT ln(100) = −11.4 kJ per mole of reaction.

Worked example 3: finding the concentration ratio

Question. A silver concentration cell (n = 1) gives 0.0355 V at 298 K. Find C₂/C₁.

Step 1. log(C₂/C₁) = E × n ÷ 0.05916 = 0.0355 ÷ 0.05916 = 0.600.

Step 2. C₂/C₁ = 10^0.600 = 3.98, about 4.0.

This reverse use is common in practical problems: a known EMF reveals an unknown concentration, and it is the idea behind simple potentiometric measurements.

Other types of concentration cells

Electrolyte concentration cells are the type above. In electrode concentration cells, the electrode itself has different concentrations, for example two hydrogen electrodes at different pressures (H₂ gas at P₁ and P₂) or amalgam electrodes with different metal concentrations. The same idea applies: E° = 0 and the EMF comes from the ratio, but the ratio is of pressures or amalgam concentrations. For two H₂ electrodes in the same acid, with n = 2, the anode is the one with the higher pressure and E = (0.0592 ÷ 2) log(P_high ÷ P_low).

Common mistakes

Adding E° values. For a concentration cell, E° = 0 because both electrodes are identical.
Putting the concentrated solution at the anode. The dilute side is the anode. If you get a negative EMF, the cell is written the wrong way round.
Using the wrong n. n is the charge on the ion that is exchanged (2 for Zn²⁺, 1 for Ag⁺).
Mixing ln and log. Use 0.0592 with log₁₀, or RT/nF with ln.
Using the temperature 298 K blindly. If another temperature is given, recompute RT/F.
Using activities and concentrations as the same. In dilute solutions, concentrations are a good approximation; in more concentrated ones, activities are needed. Follow the level asked by your exam.

Exam summary

Item	Concentration cell
E°_cell	0
Anode	Dilute half-cell
Cathode	Concentrated half-cell
EMF at 298 K	(0.0592 ÷ n) log(C_conc ÷ C_dil)
EMF becomes zero when	The two concentrations are equal
ΔG	−nFE (always negative for a working cell)

Check the current official syllabus of your exam for the depth expected (for example whether activities are asked).

The Nernst tool gives E from E°, n and the reaction quotient. For a concentration cell, enter E° = 0 and the concentration ratio to check your hand calculation.

Open the Nernst Equation Calculator →