ElectrochemistryCourse level: College Extension

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

Matching half-reactions with different activities

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.

The standard cell potential is zero, but the unequal activities make the actual cell potential nonzero.

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.

For matching Cu²⁺/Cu half-cells, oxidation on the dilute side produces Cu²⁺ and electrons; reduction on the concentrated side consumes Cu²⁺. Write the net transfer as Cu²⁺(high activity) → Cu²⁺(low activity). Its reaction quotient is alow/ahigh, and the two identical standard potentials cancel. The driving force is the ion chemical-potential difference; no difference in the identity of the metals is required.

copper concentration cell. Read this visual alongside the section; the text alternative describes its scientific content and assumptions.
For matching metal/metal-ion cells: the dilute-ion side is the anode, and the concentrated-ion side the 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.

At 25 °C, E = (0.0592/2)log₁₀(ahigh/alow) for Cu²⁺/Cu. Approximating activities by concentrations, 1.00 M versus 0.00100 M gives log ratio = 3 and E = 0.0888 V. For Ag⁺/Ag the electron number is one, so the same ratio gives twice that voltage. This is a ratio comparison, not a difference-of-concentrations formula. Reversing the cell's written orientation changes the sign of the measured voltage.

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

Sensors and differential-concentration effects

Electrodes can respond approximately logarithmically to ion activity and therefore serve as concentration sensors. A glass pH electrode develops an ion-selective membrane potential; it should not be described as a hydrogen-ion reduction involving one electron. Its ideal potential response has magnitude about 59.2 mV per pH unit at 25 °C because the sensed ion has unit charge, with sign depending on measurement convention.

Differential oxygen availability can also create electrochemical differences on a metal surface, contributing to localized corrosion. Such gas- and oxygen-reduction systems need their own half-reactions; the simple “dilute metal-ion side is the anode” rule is specific to the matching metal/metal-ion cells derived here, not every device called a concentration cell.

Concentration ratio sets a small voltage

A concentration cell uses identical electrode materials and half-reactions, so E°cell=0. Its voltage comes from different ion activities; for the same metal ion on both sides, E=(0.0592/n)log(Chigh/Clow) at 25 °C under ideal conditions. The dilute side is the anode, supplying ions until the difference shrinks.

The concentration substitution assumes sufficiently ideal behavior. At high ionic strength, unequal activity coefficients and liquid-junction potentials can affect a real reading. The simplest textbook formula neglects the junction contribution. With closed reservoirs and adequate electrode material, operation reduces the activity difference until E tends to zero. Finite electrode exhaustion, a broken ionic path or other practical limits can stop current first; equal concentrations are the ideal matching-cell endpoint, not the only possible cause of a stopped real circuit.

A tenfold ratio gives 0.0592/n V at 25 °C.

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

In the ideal matching-metal-ion model with sufficient electrode material, equal ion activities remove the concentration driving force; practical depletion can stop a real cell earlier.

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 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.

    0 of 4 steps revealed.

    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.

      0 of 3 steps revealed.

      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.

        0 of 3 steps revealed.

        Worked example 4

        Two Cu/Cu²⁺ half-cells contain 0.10 M and 1.0 M Cu²⁺ at 25 °C. Estimate E and identify the anode.

        Try it first: Cu²⁺ reduction uses n=2; the concentration ratio is ten.

          0 of 3 steps revealed.

          Common mistakes

          Common mistakes and better checks
          MistakeWhy it's wrongCheck instead
          Concluding that no voltage is possible because E° = 0.The Nernst term supplies the entire potential when concentrations differ.Compute E, not E°.
          Making the concentrated half-cell the anode.Oxidation increases ion concentration, so it must happen on the dilute side.Ask which side needs to gain ions to equalize.
          Writing Q as concentrated over dilute.That gives a negative E for a spontaneous cell.Put the anode compartment ion on top, as the product.
          Expecting a large voltage from a large ratio.The response is logarithmic, so a thousandfold ratio still yields under a tenth of a volt.Estimate 0.0592/n volts per factor of ten.
          Thinking the cell stops because an electrode is consumed.The ideal thermodynamic endpoint is equal ion activity, but a finite electrode or another circuit limitation can stop a real cell earlier.Check Q, not electrode mass.

          Scroll horizontally if needed to read all columns.

          After the lesson · Question bank

          Independent practice

          Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.

          28 topic questions across 3 difficulty levels.

          Loading practice questions…

          What you should now know

          Matching metal/metal-ion half-cells have E°cell = 0 but can generate voltage from unequal ion activities. In the ideal solution model at 25 °C, the dilute metal-ion side oxidizes and the concentrated side reduces, with E = (0.0592/n)log(chigh/clow) in the spontaneous orientation. Operation reduces that difference. Real readings may also depend on activity coefficients and junction potentials, and practical depletion can stop current before the ideal equal-activity endpoint.

          • 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
          • For the spontaneous matching-metal-ion orientation, E = (0.0592/n)log(chigh/clow) > 0 at 25 °C under the ideal model
          • Voltages are small because the dependence is logarithmic
          • The cell dies when the two concentrations become equal

          Your next step

          Lesson: Not marked completed

          Marking a lesson completed records that you finished reading it. Practice establishes mastery separately. Saved on this device.

          Try a short set of different questions to check what you can do independently.

          Watch and Learn

          An optional video to support this lesson.

          Loading video resources…

          Supplementary learning resources

          More resources for this topic →

          External resources belong to their respective creators. Chem Help is not affiliated with or endorsed by these providers. Links open the original provider; their terms apply.

          Sources and further reading

          • Chemistry 2e, Section 17.6: Corrosion
            OpenStax · Rice University · Chapter 17.6
            View source

            Lesson-specific rights and provenance need review.

            The current online Chemistry 2e edition lists CC BY-NC-SA 4.0. The source version used for this lesson and its applicable rights remain unverified.

            Recorded use: further reading. The source record also indicates adaptation or summary; exact passages and applicable terms still need verification.

            Recorded attribution / license claim (applicability unverified): Access for free at openstax.org

            Source rights and review status

            Citation coverage needs review: This link is to the Corrosion section. The exact source passage for concentration cells has not been established.