Concentration Cells and Voltage Decay

A concentration cell is a galvanic cell in which both half-cells are made of the same materials but have different concentrations, so that E° = 0 and the voltage arises only from the concentration difference; the cell stops working when the two concentrations become equal. In brief: a copper concentration cell with 1.0 M and 0.010 M Cu²⁺ gives E = (0.0592/2) log(1.0/0.010) = 0.059 V, and the voltage falls to zero as the concentrations equalize.

A concentration cell shows that a potential can come from a difference in concentration alone. This is the same principle that gives rise to the potentials across the membranes of living cells and to some types of corrosion. This article explains how a concentration cell works, shows how to calculate its voltage and the final state, describes how the voltage of an ordinary cell decays as it discharges and gives practical examples.

How does a concentration cell work?

Consider two copper electrodes, one in a dilute solution of copper(II) sulfate and the other in a concentrated one, connected by a wire and a salt bridge:

Cu(s) | Cu2+(0.010 M) || Cu2+(1.0 M) | Cu(s)

The system tends to equalize the concentrations. In the dilute half-cell, copper is oxidized (Cu → Cu²⁺ + 2e⁻), which increases the concentration of Cu²⁺: this is the anode. In the concentrated half-cell, Cu²⁺ is reduced to Cu (Cu²⁺ + 2e⁻ → Cu), which decreases its concentration: this is the cathode. The two half-reactions are the reverse of each other, so the standard cell potential is E° = 0, and the overall reaction is Cu²⁺(conc) → Cu²⁺(dilute). The driving force is the increase of entropy as the concentrations equalize.

How do you calculate the voltage?

With E° = 0, the Nernst equation reduces to:

E = −(0.0592 / n) log Q = (0.0592 / n) log([ion]concentrated / [ion]dilute)

Here Q = [dilute] / [concentrated] is the reaction quotient for the transfer of the ion from the concentrated to the dilute side. The anode is always in the more dilute solution, and the cathode in the more concentrated one.

Voltage for different concentration ratios

The voltages of concentration cells are small, tens of millivolts for modest ratios, since the potential depends on the logarithm of the ratio.

Worked examples

Example 1: a copper concentration cell

Find E for Cu | Cu²⁺(0.010 M) || Cu²⁺(1.0 M) | Cu at 25 °C. n = 2. E = (0.0592/2) log(1.0/0.010) = 0.0296 × log 100 = 0.0296 × 2 = 0.0592 V.

Example 2: a silver concentration cell

A cell has Ag⁺ at 0.10 M and at 0.0010 M. n = 1. E = 0.0592 × log(0.10/0.0010) = 0.0592 × log 100 = 0.118 V. The silver electrode in the 0.0010 M solution is the anode.

Example 3: the final state

In the copper cell of Example 1, each half-cell contains 1.00 L of solution. As the cell discharges, x mol of Cu²⁺ is formed in the dilute side and removed in the concentrated side. The cell stops when 0.010 + x = 1.0 − x, so x = 0.495 mol. Both compartments finish with [Cu²⁺] = 0.505 M, and E = 0. A total of 2 × 0.495 = 0.99 mol of electrons has passed, which is 0.99 × 96,485 = 95,500 C. This is a considerable charge, although the voltage is very small, so the energy delivered is limited.

Example 4: pH concentration cell

Two hydrogen electrodes (1 bar H₂) are immersed in solutions of pH 2.00 and pH 5.00. The potential difference is E = 0.0592 × (5.00 − 2.00) = 0.178 V. The electrode in the less acidic solution (pH 5.00) is the anode, as it has the lower H⁺ concentration.

Why does the voltage of a cell decay?

The same logic applies to any galvanic cell, not only to concentration cells. When a cell discharges, reactants are used up and products accumulate, so Q increases, and the Nernst equation shows that E falls. The Zn–Cu cell, initially with 1.0 M of each ion in 1.0 L, serves as an example:

• After 90 % of the Cu²⁺ is used, [Cu²⁺] = 0.10 M and [Zn²⁺] = 1.9 M. Q = 19, so E = 1.10 − 0.0296 × log 19 = 1.10 − 0.038 = 1.06 V.
• After 99.9 % is used, [Cu²⁺] = 0.0010 M and Q ≈ 2000. E = 1.10 − 0.0296 × 3.30 = 1.00 V.
• At equilibrium, Q = K ≈ 10³⁷ and E = 0.

In theory the voltage remains near E° until nearly all the reactants are consumed, and then falls sharply. This explains why many batteries give a nearly constant voltage during most of their life, and then die quickly. In real batteries the internal resistance, which increases as the cell discharges, and the changes in the electrode surface cause an additional drop. Voltage decay may also come from polarization at the electrodes, the growth of products on the surface or a drying out of the electrolyte.

Where do concentration effects matter?

• Biology: the membrane potential of nerve and muscle cells is determined by the concentration differences of K⁺, Na⁺ and Cl⁻ across the membrane.
• Corrosion: differential aeration cells form when part of a metal is in contact with a high oxygen concentration (the cathode), and another part with a low concentration (the anode). The low-oxygen region, such as the bottom of a crevice or under a water drop, corrodes.
• Analytical chemistry: ion-selective electrodes measure the concentration of an ion by the potential it generates against a reference of known concentration.
• Energy: salinity gradient power, in which the mixing of fresh water and seawater is used to generate electricity, is under development.
• Desalination and dialysis: processes that involve the movement of ions across membranes between solutions of different concentrations.

Who studied concentration cells?

Walther Nernst’s work in 1889 on the potential of an electrode in a solution of different concentrations was the theoretical foundation, and he used concentration cells to test his equation. The discovery of the resting potential of nerve fibers and its explanation in terms of ion concentration differences by Alan Hodgkin and Andrew Huxley in the 1940s and 1950s (Nobel Prize in Physiology or Medicine, 1963) is a biological application of the same principle.

Which mistakes are common?

• Giving the concentration cell a non-zero E°: both half-reactions are the same, so E° = 0.
• Placing the anode in the concentrated solution: it is always in the dilute one.
• Forgetting n: the coefficient 0.0592/n depends on the charge of the ion.
• Thinking that a large charge means a large voltage: the voltage depends on the ratio, not on the amounts.
• Assuming that a cell stops because the electrodes are used up: in a concentration cell the voltage drops to zero by equalization.