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The Nernst Equation & Concentration Cells

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Voltage tracks the reaction quotient

E°cell is a special case — it assumes every solute is 1 M and every gas 1 bar. The Nernst equation corrects that standard voltage for the actual mixture using the reaction quotient Q, built from current concentrations exactly as in equilibrium work (pure solids and liquids omitted). As a cell discharges, reactants fall and products rise, so Q climbs and E sags. When Q finally reaches K the potential hits zero — a dead battery is nothing more than a cell that has reached equilibrium.

The Nernst equation (298 K)
E = E° − (0.0592 / n) log Q
General form E = E° − (RT/nF) ln Q; at 25 °C the constants collapse to 0.0592/n with base-10 log. Q < 1 raises E above E°; Q > 1 lowers it; Q = 1 recovers E = E°.

Reading the shift without a calculator

Often the sign of the correction is all you need. More reactant, or less product, shrinks Q below 1, making log Q negative and −(0.0592/n) log Q positive — so E rises above E°. More product, or less reactant, inflates Q above 1 and E falls below E°. This is Le Châtelier translated into volts: pile up a species the cell consumes and you hand the reaction more driving force, which shows up as a higher voltage.

Worked example

For the cell Fe(s) | Fe²⁺ (0.10 M) || Ag⁺ (0.10 M) | Ag(s), the reaction is Fe(s) + 2 Ag⁺ → Fe²⁺ + 2 Ag(s). Given E°(Ag⁺/Ag) = +0.80 V and E°(Fe²⁺/Fe) = −0.44 V, find E at 298 K.

  1. 1.Silver has the higher reduction potential, so Ag is the cathode and Fe is the anode: E°cell = E°cathode − E°anode = 0.80 − (−0.44) = +1.24 V.
  2. 2.n = 2 (iron releases 2 e⁻; two Ag⁺ each take 1). Build Q from the reaction, leaving out the solids: Q = [Fe²⁺] ÷ [Ag⁺]² = 0.10 ÷ (0.10)² = 0.10 ÷ 0.010 = 10.
  3. 3.Apply Nernst: E = E° − (0.0592/n) log Q = 1.24 − (0.0592/2) log(10).
  4. 4.log(10) = 1 and 0.0592/2 = 0.0296, so E = 1.24 − 0.0296 = 1.21 V.
  5. 5.Q > 1, so E dropped slightly below E° — exactly as the positive correction term predicts.
Answer: E ≈ 1.21 V, just under the standard 1.24 V.

Concentration cells: voltage from nothing but a gradient

Take the same electrode and the same ion on both sides — say two copper strips in two Cu²⁺ solutions of different concentration. The electrodes are identical, so E° = 0: there is no chemical driving force at all. Yet the cell still produces voltage, because nature drives the system toward equal concentrations. The dilute side oxidizes (anode) to make more ions; the concentrated side reduces (cathode) to consume them. With E° = 0 the Nernst equation reduces to E = −(0.0592/n) log Q, and Q = [dilute] ÷ [concentrated]. The bigger the concentration ratio, the bigger the voltage — until the two sides equalize and E dies.

Concentration cell (identical electrodes, 298 K)
E = −(0.0592 / n) log Q, with E° = 0
Q = [ion]dilute ÷ [ion]concentrated < 1, so log Q < 0 and E > 0. The concentrated half-cell is always the cathode; electrons flow to shrink the gradient.
Worked example

A concentration cell has two copper electrodes. One dips in [Cu²⁺] = 1.0 M, the other in [Cu²⁺] = 0.0010 M; n = 2. Identify the cathode and compute E at 298 K.

  1. 1.Identical electrodes ⇒ E° = 0. The system drives toward equal concentration, so the concentrated 1.0 M side is reduced — it is the cathode — and the dilute 0.0010 M side is oxidized (anode).
  2. 2.Build Q as [dilute] ÷ [concentrated] = 0.0010 ÷ 1.0 = 1.0 × 10⁻³.
  3. 3.E = −(0.0592/n) log Q = −(0.0592/2) log(1.0 × 10⁻³) = −(0.0296)(−3).
  4. 4.E = +0.0888 V ≈ +0.089 V.
Answer: The 1.0 M (concentrated) electrode is the cathode; E ≈ +0.089 V, generated purely by the 1000-fold concentration gradient.
Checkpoint

A concentration cell has two silver electrodes, one in 1.0 M Ag⁺ and one in 0.010 M Ag⁺. Which compartment is the cathode?

Checkpoint

A concentration cell has two zinc electrodes: the cathode in [Zn²⁺] = 0.10 M and the anode in [Zn²⁺] = 0.0010 M, with n = 2. What is E at 298 K?

On the exam

On the AP exam a concentration cell is instantly recognizable: same metal on both sides, same ion, E° = 0. Do not panic that the voltage looks like it should be zero — plug into E = −(0.0592/n) log Q and let the concentration ratio do the work. The concentrated half-cell is always the cathode.

Tip

When a question only asks which way E moves, skip the arithmetic. Decide whether the change makes Q larger or smaller: larger Q pulls E down, smaller Q pushes E up. Getting that sign right eliminates most of the multiple-choice options before you compute anything.

Answer the 2 checkpoints as you read.

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