Thermodynamics and Electrochemistry
Unit 9 of AP Chemistry, worth 7–9% of the exam. 14 questions below, each with the working. Every answer was checked by a second pass before it was published.
Entropy, Gibbs free energy and thermodynamic favourability, coupled reactions, galvanic and electrolytic cells, cell potential, Nernst equation, electrolysis and Faraday's laws.
How this unit is tested
Start by treating thermodynamic favorability as a single question with three interchangeable answers: the sign of ΔG°, the size of K, and the sign of E°cell. Learn to convert freely between them using ΔG° = -RT ln K and ΔG° = -nFE°cell, because AP problems love to give you one and ask for another. Before touching a formula, decide qualitatively whether ΔS should be positive or negative (more gas moles, dissolution, mixing all increase entropy) and whether ΔH is exothermic or endothermic — this lets you predict at a glance whether a reaction is favorable at all temperatures, no temperatures, or only above/below some crossover T.
For electrochemistry, always identify oxidation and reduction half-reactions first and assign anode and cathode before calculating anything (remember AN OX, RED CAT: anode=oxidation, cathode=reduction). In a galvanic cell electrons flow spontaneously from anode to cathode through the external wire and E°cell is positive; in an electrolytic cell you are forcing a nonspontaneous reaction by supplying electrical energy, so E°cell is negative and current must be driven by an outside source. Compute E°cell as E°cathode minus E°anode using standard reduction potentials directly from a table — never flip a sign just because a half-reaction is written as oxidation.
When concentrations are not standard (1 M, 1 atm, 25°C), switch to the Nernst equation. Memorize the 25°C shortcut E = E° - (0.0592/n) log Q but know it only applies at 25°C; otherwise use the RT/nF form. Finally, treat electrolysis stoichiometry problems as a chain: current × time gives charge, charge ÷ Faraday's constant gives moles of electrons, and the balanced half-reaction converts moles of electrons into moles (then mass or volume) of product.
What you have to know
Gibbs free energy equation
ΔG° = ΔH° − TΔS°. A reaction is thermodynamically favorable (spontaneous) when ΔG° < 0.
ΔG° and the equilibrium constant
ΔG° = −RT ln K, where R = 8.314 J/(mol·K) and T is in kelvin. A large K corresponds to a very negative ΔG°.
ΔG° and cell potential
ΔG° = −nFE°cell, where n is moles of electrons transferred and F = 96,485 C/mol e⁻. A positive E°cell corresponds to a negative ΔG° (spontaneous).
Standard cell potential
E°cell = E°cathode − E°anode, using standard reduction potentials for both half-reactions as written for reduction.
Nernst equation
E = E° − (RT/nF) ln Q; at 25°C this simplifies to E = E° − (0.0592/n) log Q. It gives cell potential under nonstandard concentrations/pressures.
Faraday's law of electrolysis
Moles of electrons passed = (current × time)/F = It/F. This converts directly to moles of product via the balanced half-reaction stoichiometry.
14 practice questions
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Predict the sign of ΔS° for the reaction 2 NO2(g) → N2O4(g).
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Answer. Negative
Two moles of gas combine to form one mole of gas, decreasing the number of independent gas particles and thus the dispersal of matter, so entropy decreases. -
A reaction has ΔH° = −92.0 kJ/mol and ΔS° = −198 J/(mol·K). Is it thermodynamically favorable at 298 K?
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Answer. Yes, ΔG° ≈ −33.0 kJ/mol
ΔG° = ΔH° − TΔS° = −92,000 J − (298 K)(−198 J/K) = −92,000 + 59,004 = −32,996 J/mol ≈ −33.0 kJ/mol, which is negative, so the reaction is favorable at this temperature despite the unfavorable entropy term. -
Which combination of ΔH° and ΔS° guarantees a reaction is thermodynamically favorable at every temperature?
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Answer. ΔH° negative and ΔS° positive
Since ΔG° = ΔH° − TΔS°, if ΔH° is negative and −TΔS° is also negative (ΔS° positive), every term is negative for any positive T, so ΔG° is always negative. -
A reaction has ΔH° = +120 kJ/mol and ΔS° = +150 J/(mol·K). At what minimum temperature does it become thermodynamically favorable?
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Answer. 800 K
Favorability begins where ΔG° = 0, so T = ΔH°/ΔS° = 120,000 J/(150 J/K) = 800 K. Above this temperature the −TΔS° term outweighs the positive ΔH°, making ΔG° negative. -
A reaction at 298 K has ΔG° = −10.0 kJ/mol. Estimate the equilibrium constant K.
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Answer. K ≈ 57
Using ΔG° = −RT ln K, ln K = −ΔG°/RT = 10,000/(8.314 × 298) ≈ 4.04, so K = e^4.04 ≈ 57. A negative ΔG° correctly predicts K greater than 1. -
In cellular respiration, ATP hydrolysis (ΔG° very negative) is coupled to an otherwise nonspontaneous biosynthetic reaction (ΔG° positive). Why does the coupled process proceed?
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Answer. Because the sum of the two ΔG° values is negative overall
Coupling adds the free energy changes of the two reactions together. As long as the large negative ΔG° of ATP hydrolysis outweighs the positive ΔG° of the biosynthetic step, the net ΔG° for the combined process is negative, making it thermodynamically favorable even though one step alone is not. -
Given E°(Cu²⁺/Cu) = +0.34 V and E°(Zn²⁺/Zn) = −0.76 V, identify the cathode and calculate E°cell for the galvanic cell built from these half-reactions.
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Answer. Cu is the cathode; E°cell = 1.10 V
The half-reaction with the higher (more positive) reduction potential occurs as written (reduction) at the cathode, so Cu²⁺/Cu is the cathode and Zn²⁺/Zn is the anode. E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V. -
Which statement correctly describes an electrolytic cell?
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Answer. An external power source forces a nonspontaneous redox reaction to occur
Electrolytic cells use an outside voltage source to push electrons in the nonspontaneous direction, meaning E°cell for the forced reaction is negative; oxidation still occurs at the anode and reduction at the cathode, just driven by external current. -
For the Zn/Cu cell (E°cell = 1.10 V, n = 2), calculate E when [Zn²⁺] = 0.010 M and [Cu²⁺] = 1.0 M at 25°C.
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Answer. E ≈ 1.16 V
Using E = E° − (0.0592/n) log Q with Q = [Zn²⁺]/[Cu²⁺] = 0.010/1.0 = 0.010: E = 1.10 − (0.0592/2) log(0.010) = 1.10 − 0.0296(−2) = 1.10 + 0.0592 ≈ 1.16 V. Lower product-side (Zn²⁺) concentration relative to standard increases the driving force, raising E above E°. -
A concentration cell is built with two Ag/Ag⁺ half-cells, one with [Ag⁺] = 1.0 M and the other with [Ag⁺] = 0.0010 M. What is E°cell for this cell, and will it produce a nonzero voltage?
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Answer. E°cell = 0 V, but the actual cell voltage is nonzero
Because both electrodes involve the identical half-reaction, the standard reduction potentials cancel, giving E°cell = 0. However, the concentration difference makes Q ≠ 1, so the Nernst equation still predicts a nonzero measured potential that drives ion flow to equalize concentrations. -
A current of 2.00 A is passed through a CuSO4 solution for 1.00 hour. What mass of copper metal is deposited at the cathode?
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Answer. ≈ 2.37 g Cu
Charge = It = (2.00 A)(3600 s) = 7200 C. Moles e⁻ = 7200/96,485 ≈ 0.0746 mol. Since Cu²⁺ + 2e⁻ → Cu, moles Cu = 0.0746/2 ≈ 0.0373 mol, and mass = 0.0373 mol × 63.55 g/mol ≈ 2.37 g. -
How long must a 1.50 A current run to deposit 1.00 g of silver from an AgNO3 solution (Ag⁺ + e⁻ → Ag)?
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Answer. ≈ 596 s (about 9.9 minutes)
Moles Ag = 1.00 g/107.9 g/mol ≈ 0.00927 mol, requiring the same moles of electrons (n = 1). Charge = moles e⁻ × F = 0.00927 × 96,485 ≈ 894 C. Time = charge/current = 894/1.50 ≈ 596 s. -
A reaction has E°cell = +0.75 V. What can be concluded about ΔG° and K for this reaction?
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Answer. ΔG° is negative and K is greater than 1
A positive E°cell gives ΔG° = −nFE°cell < 0, so the reaction is thermodynamically favorable, which through ΔG° = −RT ln K also means ln K > 0 and therefore K > 1; the three quantities are three consistent descriptions of the same favorability. -
A reaction is found to have ΔG° < 0 but is observed to proceed extremely slowly at room temperature. What does this tell you, and why is it not a contradiction?
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Answer. The reaction is thermodynamically favorable but kinetically slow (high activation energy)
ΔG° describes only the relative stability of products versus reactants (thermodynamics), not how fast equilibrium is reached (kinetics). A reaction can be highly favorable yet have a large activation energy barrier, making it proceed slowly without any contradiction, e.g. diamond converting to graphite at room temperature.
What people get wrong
- Flipping a sign when a half-reaction is written as oxidation instead of reduction. Always look up the reduction potential as tabulated and use E°cell = E°cathode − E°anode without reversing any table values.
- Forgetting to convert ΔS° from J/(mol·K) to kJ/(mol·K) before combining it with ΔH° in kJ/mol in the Gibbs equation. Keep units consistent throughout, or convert ΔH° to joules instead.
- Using the 0.0592/n shortcut in the Nernst equation at a temperature other than 25°C. That constant already has T = 298 K built in; at any other temperature you must use the full RT/nF form.
- Multiplying E° values by the number of electrons when combining half-reactions. Cell potential is an intensive property and does not get multiplied even when you scale a half-reaction to balance electrons — only ΔG° scales with n.
- In electrolysis stoichiometry, forgetting to multiply current by time to get charge before dividing by Faraday's constant, or plugging in current alone as if it were charge.
- Assuming an exothermic reaction is always spontaneous. Check the sign of ΔS° too — a reaction can have ΔH° < 0 but still be nonspontaneous at high T if ΔS° is sufficiently negative.
Drill this unit until it sticks
These questions come back on a schedule built from what you get wrong, alongside the rest of AP Chemistry. Free, and no account needed to start.