Concentration Cells

Introduction

A cell can generate voltage even when its two electrodes are the same metal. If the solutions touching them have different ion activities, the chemical-potential difference drives a net transfer that tends to reduce the imbalance. Concentration cells expose the thermodynamic origin of voltage particularly clearly.

Core explanation

Consider M(s) M²⁺(low activity) M²⁺(high activity) M(s). Both electrodes involve M²⁺+2e⁻⇌M(s) and therefore have the same standard reduction potential. Its standard cell potential is zero. Their actual reduction potentials differ because the Nernst relation contains ion activity. For the reduction, E=E°+(RT/2F)ln a(M²⁺). The higher-activity side has the more positive reduction potential and acts as cathode in the spontaneous galvanic direction.

At the low-activity side, metal oxidizes: M(s)→M²⁺(low)+2e⁻, increasing ion activity there. At the high-activity side, M²⁺(high)+2e⁻→M(s), decreasing ion activity there. Adding the reactions gives net transfer of M²⁺ from high to low activity in the thermodynamic accounting, with electrons passing through the external wire and ions moving through the connecting solution. As the activities become equal, ideal concentration-cell voltage tends toward zero.

Subtracting electrode potentials gives Ecell=(RT/nF)ln(a high/a low) for the simple same-ion cell when n is the electron count of that electrode reaction. At 298 K, Ecell≈(0.05916 V/n)log₁₀(a high/a low). For a tenfold ratio with a divalent metal, E≈0.0296 V. The ratio must be of activities, not automatically raw molarities, and the two sides must be compared on compatible standards.

The salt bridge or liquid connection introduces a possible liquid-junction potential because different ions move at different rates. A measured voltage can therefore differ from the ideal electrode-concentration result. Using suitable bridge electrolyte and experimental design can reduce this contribution but not always eliminate it. A solution concentration cell can also be constructed using different gas pressures or other chemical potentials, not only two metal-ion molarities.

The cell's reversible electrical work comes from mixing or reducing the chemical-potential difference, not from consuming a different anode metal. This is why E°cell=0 yet Ecell can be positive away from standard equal-activity conditions. The finite total energy depends on how much material can redistribute before activities equalize.

If one uses concentration in place of activity for moderately concentrated electrolytes, nonideal activity coefficients can affect the result. A tenfold analytical concentration ratio may not equal a tenfold activity ratio. Temperature also appears in the Nernst slope, so a numerical 0.05916 V coefficient is specific to about 298 K.

Step-by-step reasoning

1. Write identical reduction half-reactions and note E°cell=0. 2. Use Nernst to compare their actual reduction potentials. 3. Label high-activity side cathode and low-activity side anode for the simple metal-ion case. 4. Compute E from the activity ratio. 5. Check junction potential and nonideality before comparing with measured EMF.

Visual explanation

Draw identical metal electrodes in two beakers. Shade the high-ion-activity beaker densely with ion dots and the low one sparsely. Mark metal dissolution on the low side and deposition on the high side. Draw an arrow labeled “difference shrinks” as the cell operates.

Real-world analogy

Two water reservoirs at different heights can drive flow even if the pipes and turbines are identical. The energy comes from a difference between the reservoirs, not different turbine materials. A concentration cell similarly draws on chemical-potential difference, although ion and electron pathways are separate.

Real-world example

Two copper electrodes immersed in copper-ion solutions of different activities can form a concentration cell. A sensitive high-impedance voltmeter detects a small open-circuit EMF. Interpreting the reading accurately requires control of solution junction effects and temperature.

Why?

Why does metal dissolve on the dilute side? That side has a lower reduction potential for M²⁺/M under the simple Nernst relation. Oxidation there supplies electrons, while reducing ions on the concentrated side tends to equalize activities.

Common misconception

“Identical electrodes must give zero voltage under all conditions.” They give zero standard potential difference under identical activities. Unequal activities change each electrode's actual potential and can generate a nonzero cell EMF.

Worked example

For Zn²⁺/Zn, n=2 at 298 K. If a high=0.100 and a low=0.00100, their ratio is 100. The ideal Ecell=(0.05916/2)log₁₀(100)=0.02958×2=0.05916 V. Zn dissolves at the lower-activity side and deposits at the higher-activity side. A real meter may include a junction contribution.

Quick check

1. What is E°cell for a simple concentration cell using identical half-reactions? Answer: Zero under the same standard states. 2. Which side is the cathode for M²⁺/M with a high>a low? Answer: The high-activity side, where M²⁺ is reduced.

Exam focus

Use activity ratios and the correct n; show electrode directions through the Nernst equation rather than guessing. At equal activities, ideal E approaches zero. State that liquid-junction potential can affect measured EMF.

Advanced insight

The net reaction may be written as transfer of ionic species from a high- to low-activity compartment, but electroneutrality requires counterion transport through the junction. A rigorous treatment therefore includes transport and junction contributions, not merely metal deposition and dissolution at separated electrodes.

Summary

Concentration cells obtain reversible voltage from unequal chemical potentials despite identical electrode materials and E°cell=0. The high-activity metal-ion side is the reduction cathode in a simple cell, and operation tends to reduce the activity difference. Measured voltage can include junction effects.

Practice questions

1. Estimate ideal E at 298 K for a divalent-metal concentration cell with a high/a low=10. Answer: (0.05916/2)log₁₀10≈0.0296 V. 2. Why does the voltage tend toward zero as the cell operates reversibly? Answer: Ion activities move toward equality, removing the chemical-potential difference. 3. Why might measured EMF differ from a concentration-only Nernst prediction? Answer: Activity coefficients and liquid-junction potentials can make the simple concentration approximation inaccurate.