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SCIENCE · INVESTIGATE AND EXPLAIN

Keep voltage intensive while balancing electron transfer

Relate E°, ΔG° and K with correct electron balance and equation scaling.

Learning outcome: Relate E°, ΔG° and K with correct electron balance and equation scaling.

Starting recap

A volt is a joule per coulomb. Reaction energy scales with transferred charge; voltage does not.

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Try

A fictional balanced two-electron reaction has E°=+0.500 V.

Open the experiment

Predict

If you double its written equation, should E°, ΔG° and ln K all double?

Experiment

Keep electron count 2 and compare equation multipliers 1, 2 and 3. Then explain the result using energy per charge.

Scale reactions without scaling voltage

Worked teaching model · synthetic values · Hypothetical balanced base reaction has E°=+0.500 V. Multiplying its equation scales n and ΔG°, not E°. Electron counts are assigned complete balanced-reaction values, not independently adjustable half-reaction coefficients.

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Conditional calculations, not measured observations
QuantityValueUnit
Scaled electrons n2mol e−/mol written reaction
E°0.5V
ΔG°-96.48533kJ/mol written reaction
ln K38.92174dimensionless

Voltage stays +0.500 V while reaction free energy and ln K scale.

Reduction potentials are intensive. Balance electrons, then subtract anode reduction potential from cathode reduction potential. Reversing the written reaction reverses E° and ΔG° and reciprocates K.

Predict before changing a control. Save two contrasting captures and explain what is held fixed. Displayed digits are calculation precision, not experimental uncertainty.

Voltage stays +0.500 V while reaction free energy and ln K scale.

Reduction potentials are intensive. Balance electrons, then subtract anode reduction potential from cathode reduction potential. Reversing the written reaction reverses E° and ΔG° and reciprocates K.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Derive the units

For a reversible electrochemical reaction, ΔrG°=−nFE°. Here n is moles of electrons per mole of reaction as written, F is charge per mole of electrons, and E° is joules per coulomb. Their product is joules per mole of reaction. Convert to kJ only after constructing the charge-energy ledger.

With n=2, F=96485.33212 C/mol and E°=0.500 V, ΔrG°≈−96.485 kJ/mol reaction. The negative sign corresponds to the favorable written direction. Reversing the complete reaction changes E° to −0.500 V and ΔrG° to +96.485 kJ/mol reverse reaction.

Balance electrons without multiplying a potential

Tabulated half-cell values are reduction potentials. If a selected cathodic reduction is +0.34 V and the anodic couple’s reduction value is −0.76 V, E°cell = 0.34−(−0.76)=1.10 V. Reverse the anode reaction when writing the chemical sum, but do not multiply its tabulated potential when multiplying its coefficients.

For a hypothetical one-electron cathodic half-reaction paired with a two-electron anodic half-reaction, multiply the cathodic chemical equation by two. The combined electron count is two, not three, and the cathodic potential stays its original value. It is energy and charge that double together.

Equation scaling and equilibrium

Doubling the complete equation doubles n and ΔrG°, while E° stays fixed. The equilibrium constant is squared because every quotient exponent doubles. Equivalently ln K = nFE°/(RT) doubles. Use the lab to verify this distinction for three equation multipliers, keeping temperature at the reference value.

K is always positive. E°>0 corresponds to K>1 and ΔrG° < 0 for the same written reaction. E°=0 corresponds to K=1 under this standard-state relation. A nonstandard cell can nevertheless have E=0 with K far from one because its actual Q equals K; the next lesson separates those states.

Transfer

A three-electron reaction has E°=0.200 V. Calculate ΔrG°≈−57.891 kJ/mol. Triple the equation: n=9 and ΔrG°≈−173.674 kJ/mol written reaction, but E° remains 0.200 V. State why the calculation does not predict battery runtime: capacity requires available reactant quantities, and delivered energy also depends on changing composition and operating losses.

Source scope and continuation

Original teaching explanation, calculations and practice. Framework alignment: AP Chemistry topic 9.9 in the College Board course framework, effective Fall 2024, checked 30 October 2026. These pages do not reproduce restricted examination questions or establish AP authorization. The investigation is a bounded computational model, not a conducted laboratory.

Return to the preceding Unit 9 lesson when its reasoning is unfamiliar.

Continue to Explain potential changes with the reaction quotient.

Apply

Doubling a balanced cell reaction at fixed conditions changes:

Public practice checks a specific claim. It does not establish independent mastery.

Show your calculation or representation, explain the changed case, and separate the stated model from measured evidence. Relate E°, ΔG° and K with correct electron balance and equation scaling.

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