Concentration Cells

Core Concept

A concentration cell is a galvanic cell whose two half-cells use the same electrode and ion but at different concentrations. There is no difference in standard potentials, so E° = 0; the driving force is purely the concentration difference, which makes electrons flow from the more dilute side to the more concentrated side until the concentrations equalize. Its potential is given by the Nernst equation with E° = 0: E = −(RT/nF) ln Q. Concentration cells are the basis of pH meters, ion-selective electrodes, and biological membrane potentials.

Key Ideas

• Same chemistry, different concentration. Both electrodes are identical (e.g., Cu | Cu²⁺), so E° = 0.
• Dilution drives flow. The dilute half-cell is the anode (oxidation, producing more ions); the concentrated half-cell is the cathode (reduction, consuming ions).
• Spontaneous equalization. The cell runs until both concentrations are equal — at which point Q = 1, ln Q = 0, and E = 0.
• Nernst with E° = 0. E = −(RT/nF) ln Q = (RT/nF) ln(c_high/c_low).
• Concentration → voltage. Measuring E lets you read an unknown concentration (the principle of a pH meter).

Equations and Variables

E = −(RT/nF) ln Q (with E° = 0); at 25 °C, E = −(0.0592 V/n) log Q

How It Works

1. Set up two half-cells of the same metal/metal-ion system with different ion concentrations.
2. The dilute side tends to generate ions (oxidation at the anode); the concentrated side consumes them (reduction at the cathode).
3. Electrons flow through the wire from dilute to concentrated; ions migrate through the salt bridge.
4. As ions move, the concentrations converge, Q → 1, and E decays to zero.
5. The measured E before equalization encodes the concentration ratio (Nernst equation).

Worked Example

A concentration cell has Cu electrodes in 1.0 M and 0.0010 M Cu²⁺ solutions at 25 °C. Find E.

• n = 2 (Cu²⁺ + 2 e⁻ ⇌ Cu).
• Q = c_dilute / c_concentrated = 0.0010 / 1.0 = 1.0 × 10⁻³.
• E = −(0.0592 V/2) log Q = −0.0296 × log(1.0 × 10⁻³)
• = −0.0296 × (−3.0) = +0.089 V

The cell produces about 89 mV purely from the 1000-fold concentration difference. As ions migrate, the two concentrations converge and the voltage drops to zero.

How it works

1. Set up two half-cells of the same metal/metal-ion system with different ion concentrations.
2. The dilute side tends to generate ions (oxidation at the anode); the concentrated side consumes them (reduction at the cathode).
3. Electrons flow through the wire from dilute to concentrated; ions migrate through the salt bridge.
4. As ions move, the concentrations converge, Q → 1, and E decays to zero.
5. The measured E before equalization encodes the concentration ratio (Nernst equation).

Common confusions

• "A concentration cell has no voltage because E° = 0." — Wrong. E° = 0, but E ≠ 0 as long as the concentrations differ; the Nernst term supplies the voltage.
• "Electrons flow from concentrated to dilute." — Wrong. The dilute side oxidizes (anode), sending electrons to the concentrated side.
• "The cell runs until one side is empty." — Wrong. It runs until the concentrations equalize; then E = 0.
• "The electrodes must be different metals." — Wrong. Concentration cells use the same electrode on both sides.
• "Q = c_high/c_low." — Wrong. For the cell as written (dilute → concentrated), Q = c_low/c_high, which is < 1 so E comes out positive.

Quick review

• Concentration cell: same electrode, different ion concentrations.
• E° = 0; E = −(0.0592/n) log(c_low/c_high) at 25 °C.
• Dilute = anode, concentrated = cathode.
• Voltage fades to zero as concentrations equalize.
• Underpins pH meters and ion-selective electrodes.