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Most electrochemistry questions start from a table of standard potentials. Concentration cells do not. Both electrodes are made of the same metal dipped in solutions of the same ion, so the table gives you nothing. The voltage comes only from the difference in concentration. This makes them a clean test of whether you understand the Nernst equation or have just memorised it. Here is how to set them up, with worked numbers and the slips to avoid.
Where do concentration cells appear in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG)-2026 syllabus both list the Nernst equation and its applications. The CBSE Chemistry curriculum for 2026-27 includes a Class 12 practical on the "variation of cell potential in Zn/Zn2+|| Cu2+/Cu with change in concentration of electrolytes". The equation itself is in the NCERT Class 12 chapter on Electrochemistry. Check the documents for your own exam year.
Why does a cell with identical electrodes give any voltage?
Students often expect zero, and research shows why. Sanger and Greenbowe (1997)1098-2736(199704)34:4%3C377::AID-TEA7%3E3.0.CO;2-O) interviewed college chemistry students after they had been taught electrochemistry, including concentration cells. One new misconception they found was that cell potentials do not depend on ion concentrations. They also noted that most students with misconceptions could still calculate cell potentials correctly.
The physical picture is simple. Ions tend to move from high to low concentration. In a concentration cell the only way to even out the two solutions is through the electrodes. On the dilute side, the metal dissolves and adds ions (oxidation, so this is the anode). On the concentrated side, ions plate out as metal and are removed (reduction, so this is the cathode). The cell runs until the two concentrations are equal, and then E = 0.
How do you calculate the EMF?
Start from the general Nernst equation, E = E° − (0.059/n) log Q at 298 K. For a concentration cell E° = 0, because the same couple sits on both sides. Q is the ion concentration at the anode divided by that at the cathode, so
E = (0.059/n) log (c at cathode / c at anode).
The 0.059 is 2.303RT/F at 298 K. Using the NIST value of R (8.314 J mol⁻¹ K⁻¹) and the NIST value of F (96 485 C mol⁻¹), it works out to 0.0592 V at 298.15 K.
Worked example 1: a copper cell. Cu | Cu²⁺(0.010 M) || Cu²⁺(0.10 M) | Cu
n = 2, since Cu²⁺ + 2e⁻ → Cu.
E = (0.059/2) log (0.10/0.010) = 0.0295 × 1 = 0.0295 V.
ΔG = −nFE = −2 × 96 487 × 0.0295 ≈ −5.7 kJ per mole of copper transferred. The negative sign confirms that the cell as written runs on its own.
Worked example 2: a hydrogen cell. Pt | H2(1 bar) | H⁺(pH 3) || H⁺(pH 1) | H2(1 bar) | Pt
Here n = 1 and the concentrations are 10⁻³ M and 10⁻¹ M. E = 0.059 log (10⁻¹/10⁻³) = 0.059 × 2 = 0.118 V. In pH form, E = 0.059 × (pH at anode − pH at cathode). Each unit of pH difference is worth 0.059 V at 298 K.
How does this connect to the CBSE practical?
In the practical you change the concentrations in a Daniell cell and record the voltmeter reading. NCERT writes the Nernst equation for this cell as
E = E° − (0.059/2) log ([Zn²⁺]/[Cu²⁺]), with E° = 1.10 V.
Here are some sample calculations, not measured readings:
[Zn²⁺] = [Cu²⁺] = 1 M: log 1 = 0, so E = 1.10 V.
[Zn²⁺] = 0.10 M, [Cu²⁺] = 0.010 M: log 10 = 1, so E = 1.10 − 0.0295 = 1.07 V.
[Zn²⁺] = 1 M, [Cu²⁺] = 0.001 M: log 1000 = 3, so E = 1.10 − 0.0885 = 1.01 V.
A graph of E against log([Zn²⁺]/[Cu²⁺]) should be a straight line with slope about −0.0295 V. Your measured values will differ a little, because real solutions are not ideal and the room is rarely at 298 K. Say so in your record and in the viva.
Which mistakes should you avoid?
Giving a concentration cell a nonzero E°. Same couple on both sides means E° = 0.
Using the wrong n. For Ag⁺ it is 1, so a tenfold difference gives 0.059 V. For Cu²⁺ it is 2, so the same difference gives 0.0295 V.
Putting the concentrated side at the anode. Electrons flow from the dilute side.
Thinking the cell runs forever. It stops when the concentrations become equal.
Using 0.059 at any temperature. It applies only at 298 K. At other temperatures use 2.303RT/F.
For cell notation and the link between EMF and Gibbs energy, see our post on the Nernst equation step by step.
A practice routine
Write five concentration cells with Ag⁺, Cu²⁺, Zn²⁺, H⁺ and Fe²⁺. For each, mark the anode before you calculate anything.
Calculate E for a tenfold and a hundredfold difference. Check that the answer changes by 0.059/n for every factor of ten.
For the Daniell cell, predict whether E rises or falls when you dilute each solution, then confirm with the equation.
Explain in one sentence why the cell stops working.
When a problem combines Nernst, ΔG and K, set it up one step at a time. The Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see which concentration goes where in Q.
Try it now: find E for Ag | Ag⁺(0.001 M) || Ag⁺(0.1 M) | Ag, then check that you get 0.118 V.
