MCAT content library / Chemistry & physics / Galvanic cells: connect half-reactions to cell potential
The same redox account must explain the electrodes, electron direction, ion movement, and voltage. Concentration can change that voltage without changing the metals.
About 23 minutes, including the worked examples and checks.
Useful foundations
On this page
- Use one reaction account across both half-cells
- A salt bridge completes an ionic pathway
- Worked application: read a reduction-potential table
- Balance electrons without multiplying potentials
- Separate standard state from the current solution composition
- Changed case: an initially symmetric cell is diluted
- Check your reasoning
- Continue learning
- Sources
Use one reaction account across both half-cells
A galvanic cell uses a spontaneous redox reaction to drive charge through an external circuit. Oxidation occurs at the anode and reduction at the cathode. Electrons released at the anode travel through the conducting connection to the cathode, where a species accepts them. AAMC category 4C connects these half-reactions, electrode potentials, and concentration cells; category 5E connects the driving force to reaction thermodynamics.12
“Anode” means the oxidation electrode in either a galvanic or an electrolytic cell. For a discharging galvanic cell, the anode is negative and the cathode positive. The sign convention changes in an externally driven electrolytic setup, so identify oxidation and reduction before relying on an electrode’s sign.
A salt bridge completes an ionic pathway
Separating oxidation and reduction into two compartments prevents their electrons from simply transferring locally between adjacent reactants. The external wire provides an electronic pathway; the salt bridge or suitable separator provides an ionic pathway that maintains bulk charge balance. Sustained current requires both parts of this circuit.3
For the metal-ion example below, oxidation releases positive metal ions into the anode solution. Salt-bridge anions can enter that compartment to compensate. At the cathode, positive metal ions are removed from solution as metal forms, so salt-bridge cations can enter. Electrons do not travel through the aqueous bridge in place of these ions. The direction of conventional current in the external wire is opposite the electron-flow direction.
Worked application: read a reduction-potential table
The following M and X labels, values, and diagram form an original hypothetical case. They are not measured potentials for particular named elements. Both entries are written as reductions at 25°C, relative to the same reference electrode.
| Reduction half-reaction | E°red |
|---|---|
| M²⁺(aq) + 2e⁻ → M(s) | −0.40 V |
| X³⁺(aq) + 3e⁻ → X(s) | +0.20 V |
Under the stated standard conditions, the more positive reduction potential belongs to the cathode reaction: X³⁺ accepts electrons. Reverse the M half-reaction to supply them: M(s) → M²⁺ + 2e⁻. This identifies M as the anode before any numerical subtraction is performed.4
Multiply the M oxidation by three and the X reduction by two to match six transferred electrons. The complete reaction is 3M(s) + 2X³⁺(aq) → 3M²⁺(aq) + 2X(s). Both sides have charge +6, and the six electrons have canceled. Electrode mass decreases at M and increases at X for these specified deposition/dissolution reactions.
Balance electrons without multiplying potentials
When both tabulated numbers are reduction potentials, E°cell = E°cathode,red − E°anode,red. Here E°cell = +0.20 − (−0.40) = +0.60 V. Equivalently, add +0.20 V for the reduction and +0.40 V for the reversed oxidation. Do not both reverse the sign and subtract it again.4
Multiplying half-reaction coefficients does not multiply E°. A volt is a joule per coulomb: doubling the reaction doubles the energy change and transferred charge together. The same ratio remains. The factors three and two therefore balance electrons but do not turn the calculation into 2(+0.20) − 3(−0.40).
The relationship ΔG° = −nFE° makes the difference clear. For the balanced equation as written, n = 6 mol electrons per mole of reaction. With F ≈ 96,485 C/mol, ΔG° ≈ −347 kJ per mole of that reaction. Doubling the equation doubles n and ΔG°, while the potential remains +0.60 V.5
A positive standard potential indicates favorable reaction direction under the specified standard conditions, not an unlimited current or a reaction rate. Real terminal voltage under load can be reduced by resistance and other departures from reversible operation. Changing concentrations can also change the thermodynamic cell potential.
Separate standard state from the current solution composition
Standard potentials refer to specified standard states and temperature. For solutes, unit activity is the relevant condition; treating it as 1 M is an ideal-solution approximation. Pure solids have unit activity in this convention. Temperature is stated here because standard state does not itself mean that every table must be at 25°C.
A concentration cell makes the composition effect especially clear. Place the same M(s)/M²⁺ couple in both compartments at the same temperature. If their ion activities are equal, the electrode potentials cancel and E = 0. Their intrinsic standard reduction potentials remain identical even when one solution is subsequently diluted.
Changed case: an initially symmetric cell is diluted
For an idealized M²⁺ concentration cell at 25°C, neglect liquid-junction potentials and approximate activities by concentrations. If chigh > clow, the supplied Nernst relationship is E = (0.0592 V/2)log10(chigh/clow). The factor two comes from the M²⁺ + 2e⁻ → M half-reaction, not from having two compartments.5
| Low-side concentration | High-side concentration | Ratio | Predicted E |
|---|---|---|---|
| 1.00 M | 1.00 M | 1 | 0 V |
| 0.010 M | 1.00 M | 100 | +0.0592 V |
| 0.100 M | 1.00 M | 10 | +0.0296 V |
After diluting one side from 1.00 M to 0.010 M, reduction is favored at the concentrated side, making it the cathode. Oxidation at the dilute-side anode creates additional M²⁺ there. Discharge consequently reduces the concentration difference. The direction is not selected by a different metal, because both electrodes have the same metal.
Increasing the dilute side from 0.010 M to 0.100 M reduces the concentration ratio from 100 to 10 and halves this model’s voltage. Increasing it until the activities match makes the concentration-cell driving force zero. Keeping electrode identity, reaction direction, and solution composition together produces a consistent account from the half-reactions to the measured voltage.
ACTIVE PRACTICE · ELECTROCHEMISTRY
Identify the Cathode Before Calculating Voltage
Use the original hypothetical metal half-cells at 25°C: M²⁺ + 2e⁻ → M, E° = −0.40 V; X³⁺ + 3e⁻ → X, E° = +0.20 V. These are not potentials of named real metals.
Electron balance: 3M(s) + 2X³⁺ → 3M²⁺ + 2X(s), transferring 6 e⁻. Standard E°cell = +0.60 V.
Concentration-cell transfer: Two M/M²⁺ electrodes have identical standard potentials. Hold high-side concentration at 1.00 M and vary the low side. Treat concentrations as activities and neglect junction potentials. Use E = (0.0592/2)log₁₀(chigh/clow) V.
Predicted concentration-cell E = V.
More concentrated M²⁺ side is cathode; the dilute side is anode.
Standard potentials are intensive quantities and are not multiplied by stoichiometric coefficients. The concentration-cell E is reversible ideal EMF, not guaranteed terminal voltage under a load.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
Two hypothetical standard reduction potentials are P²⁺ + 2e⁻ → P, E° = −0.25 V, and Q⁺ + e⁻ → Q, E° = +0.35 V. Under standard conditions, which electrode assignment and external electron direction give a discharging galvanic cell?
A. P is the anode; electrons flow from P to Q.
Correct. Q has the more positive reduction potential and is reduced at the cathode. P is oxidized at the anode, releasing electrons toward Q.
B. P is the cathode; electrons flow from P to Q.
Review this choice. A cathode accepts the electrons used in reduction. The stated direction and cathode assignment are inconsistent.
C. Q is the anode; electrons flow from P to Q.
Review this choice. Electrons leave the oxidation electrode. If Q were the anode, they would leave Q, and the standard reaction direction would be unfavorable.
D. Q is the cathode; electrons flow from Q to P.
Review this choice. The cathode assignment is correct, but electrons arrive at Q rather than leave it in the external wire.
The principle to carry forward
Choose the favored reduction, identify the complementary oxidation, and then follow the released electrons.
Check 2
The supplied reductions are R³⁺ + 3e⁻ → R at −0.10 V and S²⁺ + 2e⁻ → S at +0.50 V. For 2R + 3S²⁺ → 2R³⁺ + 3S under standard conditions, what is E°cell?
A. +0.20 V.
Review this choice. This does not equal the difference between the given cathode and anode reduction potentials.
B. +0.40 V.
Review this choice. Adding the tabulated reduction values without reversing the anode loses the required sign change.
C. +0.60 V.
Correct. E°cell = +0.50 − (−0.10) = +0.60 V. The factors two and three balance six electrons but do not scale electrode potentials.
D. +1.70 V.
Review this choice. This results from 3(+0.50) − 2(−0.10). Reaction coefficients do not multiply a potential, which is energy per charge.
The principle to carry forward
Reaction coefficients scale energy and charge together; they leave the electrode-potential ratio unchanged.
Check 3
An ideal two-electron concentration cell at 25°C has ion concentrations 0.020 M and 0.200 M. Use E = (0.0592 V/2)log₁₀(c_high/c_low), neglecting junction effects. The lower side is changed to 0.0020 M while the higher side remains 0.200 M. How does the model voltage change?
A. It falls from 0.0296 V to 0.00296 V.
Review this choice. The concentration ratio grows rather than shrinks. Voltage depends logarithmically on that ratio.
B. It remains 0.0296 V because the electrodes are the same.
Review this choice. Identical electrodes cancel their standard potentials, but unequal solution activities can still produce a voltage.
C. It rises from 0.0296 V to 0.0592 V.
Correct. The ratio changes from 10 to 100. Its base-ten logarithm changes from 1 to 2, so the predicted voltage doubles.
D. It rises from 0.0296 V to 0.296 V.
Review this choice. The concentration ratio rises tenfold, but its logarithm rises from 1 to 2. The voltage is not proportional to the raw ratio.
The principle to carry forward
Use the concentration ratio inside the logarithm; do not scale voltage directly with concentration.
Continue learning
Related lessons
- Redox bookkeeping: balance atoms, charge, and electrons
- Electric fields and potential: keep the charge sign
- Resistor circuits: follow branches and place meters correctly
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Sources
1. AAMC: C/P content category 4C
2. AAMC: C/P content category 5E
3. OpenStax Chemistry 2e: Galvanic Cells
4. OpenStax Chemistry 2e: Electrode and Cell Potentials
5. OpenStax Chemistry 2e: Potential, Free Energy, and Equilibrium