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.
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
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.
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.
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)
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.
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])
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
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.
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 galvanic cell with identical electrodes and electrolytes differing only in concentration.
E° = 0 because the two half-cells are the same couple.
The dilute half-cell, where oxidation raises the concentration.
E = -(0.0592/n) log([dilute]/[concentrated]).
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.
E = -(0.0592 / n) log([dilute] / [concentrated])
ΔE = 0.0592 / n volts per tenfold ratio
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.
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.
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.
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.
| Mistake | Why it's wrong | Check 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. |
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After the lesson · Question bank
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28 topic questions across 3 difficulty levels.
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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.
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