Worked example 1
A cell is Cu(s) | Cu²⁺(0.0010 M) || Cu²⁺(1.00 M) | Cu(s). Find E at 25 °C and identify the anode.
Try it first: Ask which side must dissolve metal to move toward equal concentrations.
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Electrochemistry & Applications of ThermodynamicsElectrochemistryContent level: Core 24 min
Concentration Cells
What you'll be able to do: Explain how a cell built from two identical half-cells can produce voltage, and calculate its potential from the concentration ratio.
Best after: The Nernst Equation and Nonstandard Conditions
Introduction
Two copper electrodes, two copper solutions, one cell. Standard potentials say the voltage should be zero, yet a meter reads a real number. Concentration cells are the cleanest possible test of whether you actually understand the Nernst equation.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Explain why a concentration cell has E° = 0 yet produces voltage
- Identify which half-cell is the anode in a concentration cell
- Calculate the potential from the concentration ratio
- Predict how the potential changes as the cell operates
Lesson
A cell with no chemistry difference
Build a cell with a copper electrode in 0.0010 M Cu²⁺ and another copper electrode in 1.0 M Cu²⁺, joined by a salt bridge. Because the same half-reaction appears on both sides, E°cathode - E°anode = 0.34 - 0.34 = 0. Standard conditions predict nothing, yet the cell drives current.
E° = 0 does not mean E = 0. Standard values assume equal, 1 M concentrations.
The driving force is dilution
Nature moves toward uniform concentration because that state has higher entropy. The cell provides an electrochemical route to that mixing: the dilute side dissolves metal to raise its concentration, and the concentrated side deposits metal to lower its own. The reaction is literally Cu²⁺(concentrated) → Cu²⁺(dilute).
Cu²⁺(1.0 M) → Cu²⁺(0.0010 M)
Which side is the anode
Oxidation raises the ion concentration where it occurs, so it must occur in the dilute half-cell. The dilute side is therefore the anode and loses electrode mass, while the concentrated side is the cathode and gains mass. This is worth memorizing because it feels backwards to many students.
Dilute = anode. Concentrated = cathode.
The working equation
Since E° = 0, the Nernst equation reduces to E = -(0.0592/n) log Q, with Q equal to the dilute concentration over the concentrated one. That ratio is less than 1, its log is negative, and the leading minus sign makes E positive, as it must be for a spontaneous cell.
E = -(0.0592 / n) log([dilute] / [concentrated])
Magnitudes are small
Because the dependence is logarithmic, even a thousandfold concentration difference with n = 2 produces only about 0.089 V. Concentration cells make excellent sensors but poor power sources.
factor of 10 in ratio → 0.0592/n volts
Where they show up
A pH electrode is a concentration cell responding to hydrogen ion activity, delivering 0.0592 V per pH unit at 25 °C for its one-electron process. Differential aeration cells, where a metal surface sits under drops of water with different dissolved oxygen levels, are concentration cells that cause pitting corrosion under gaskets and paint chips.
Key ideas
Definition
Concentration cell
A galvanic cell with identical electrodes and electrolytes differing only in concentration.
Rule
Standard potential
E° = 0 because the two half-cells are the same couple.
Rule
Anode
The dilute half-cell, where oxidation raises the concentration.
Rule
Potential
E = -(0.0592/n) log([dilute]/[concentrated]).
Rule
Endpoint
The cell dies when the concentrations become equal, not when an electrode is used up.
Equation
Concentration cell potential
E = -(0.0592 / n) log([dilute] / [concentrated])
- n = electrons in the shared half-reaction
Equation
Per decade response
ΔE = 0.0592 / n volts per tenfold ratio
- ΔE = voltage change per factor of ten
Worked examples
Worked example 2
A silver concentration cell reads 0.118 V, with the concentrated side at 1.00 M. Find the dilute concentration. For silver, n = 1.
Try it first: Note that n = 1 doubles the sensitivity compared with a copper cell.
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Worked example 3
The copper cell in the first example is allowed to run for a long time. What happens to the voltage and why?
Try it first: Track both concentrations as electrons flow.
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Common mistakes
Concluding that no voltage is possible because E° = 0.
Why it's wrong: The Nernst term supplies the entire potential when concentrations differ.
Check instead: Compute E, not E°.
Making the concentrated half-cell the anode.
Why it's wrong: Oxidation increases ion concentration, so it must happen on the dilute side.
Check instead: Ask which side needs to gain ions to equalize.
Writing Q as concentrated over dilute.
Why it's wrong: That gives a negative E for a spontaneous cell.
Check instead: Put the anode compartment ion on top, as the product.
Expecting a large voltage from a large ratio.
Why it's wrong: The response is logarithmic, so a thousandfold ratio still yields under a tenth of a volt.
Check instead: Estimate 0.0592/n volts per factor of ten.
Thinking the cell stops because an electrode is consumed.
Why it's wrong: It stops when the concentrations match; both electrodes are the same element.
Check instead: Check Q, not electrode mass.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
A concentration cell uses the same redox couple on both sides, so the two standard potentials cancel and E° = 0. The entire voltage comes from the Nernst correction term, driven by the difference in concentration between the two half-cells. The system behaves like any spontaneous dilution: the more dilute half-cell is the anode, where metal dissolves and raises the concentration, and the more concentrated half-cell is the cathode, where ions plate out and lower it. The potential is E = -(0.0592/n) log([dilute]/[concentrated]), which is always positive, and it depends only on the ratio, so a tenfold difference gives 0.0592/n volts. As the cell runs, the two concentrations converge; when they are equal, Q = 1, E = 0, and the cell is dead even though neither electrode has been consumed. Concentration cells are the working principle behind ion-selective electrodes, including the pH meter, and they explain certain forms of localized corrosion.
- Concentration cells have E° = 0 and get all their voltage from the Nernst term
- The dilute half-cell is the anode; the concentrated half-cell is the cathode
- E = -(0.0592/n) log([dilute]/[concentrated]), always positive
- Voltages are small because the dependence is logarithmic
- The cell dies when the two concentrations become equal
Sources and further reading
- Chemistry 2e, Section 17.6: CorrosionOpenStax · Rice University · Chapter 17.6View source
Access for free at openstax.org License