Practice a College Board-style short-answer question on Applications of Thermodynamics. Write your response, then reveal the model answer to see exactly how each part earns its point.
Short Answer Question · Unit 9 · Free Energy & Galvanic Cells
A galvanic cell is constructed using the half-reactions Zn²⁺(aq) + 2e⁻ → Zn(s), E° = −0.76 V and Cu²⁺(aq) + 2e⁻ → Cu(s), E° = +0.34 V. The cell operates under standard conditions at 298 K.
A
Identify which half-reaction occurs at the anode and which occurs at the cathode in this galvanic cell, and calculate E°cell.
✓ Model answer (earns the point)
For a galvanic cell, the half-reaction with the more positive (or less negative) E° occurs as reduction at the cathode, while the other occurs as oxidation at the anode. Here, Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V) is the cathode (reduction), and Zn²⁺ + 2e⁻ → Zn (E° = −0.76 V) is reversed to Zn → Zn²⁺ + 2e⁻ at the anode (oxidation). E°cell = E°cathode − E°anode = 0.34 V − (−0.76 V) = +1.10 V.
Why it scores: Correctly identifies the anode and cathode half-reactions AND correctly calculates E°cell using E°cathode − E°anode.
B
Using the value of E°cell from Part A, calculate ΔG° for this reaction (n = 2 mol electrons, F = 96,485 C/mol). Explain what the sign of your answer indicates about the reaction.
✓ Model answer (earns the point)
ΔG° = −nFE° = −(2 mol)(96,485 C/mol)(1.10 V) = −2.12 × 10⁵ J, or −212 kJ. The negative sign indicates that this redox reaction is thermodynamically favorable (spontaneous) under standard conditions — consistent with the fact that this is a galvanic cell, which by definition relies on a spontaneous reaction to generate current.
Why it scores: Correctly applies ΔG° = −nFE° with the right sign AND correctly interprets the negative ΔG° as indicating a thermodynamically favorable reaction.
C
As the cell operates over time, predict what happens to E°cell, and explain your reasoning in terms of Q and K.
✓ Model answer (earns the point)
The cell potential will gradually decrease over time. As the cell operates, Zn is oxidized (consumed) and Cu²⁺ is reduced (consumed), so the reaction quotient Q increases as the reaction proceeds toward equilibrium. According to the Nernst equation, as Q approaches K (the equilibrium constant), the cell potential E approaches zero — the cell potential decreases until the cell is "dead" once equilibrium (Q = K) is reached.
Why it scores: Correctly predicts decreasing cell potential AND explains the mechanism using Q approaching K (via the Nernst equation), rather than just stating the result without reasoning.
How to score points on SAQs
Show your work for every calculation. Writing out ΔG° = −nFE° (or ΔG° = −RT ln K) before plugging in numbers often earns credit even with a small arithmetic slip.
Always connect a numerical answer back to its physical meaning. A negative ΔG° isn't just a number — say explicitly that it means the reaction is thermodynamically favorable.
Know which half-reaction goes where. The half-reaction with the higher (more positive) E° is always the reduction/cathode in a galvanic cell; the other is reversed and become the oxidation/anode.
Use the Q vs. K framework to explain changing cell potential over time. "The cell potential decreases" alone scores less than connecting it to Q approaching K via the Nernst equation.
Keep it tight. 2–4 sentences per part is plenty. Long answers don't score higher; they just waste exam time.