Chemistry • Year 11 • Module 3 • Lesson 9
Galvanic Cells
Apply standard reduction potentials, E°cell calculations and cell-operation reasoning to real data, scenario questions and a compare-and-contrast task.
1. Interpret graph — cell voltage over time in a Zn/Cu galvanic cell
The graph below shows the voltage of a Zn/Cu Daniell cell recorded every 5 minutes as the cell discharged through a fixed resistor at 25°C under standard initial concentrations (1.00 mol/L). 8 marks
Figure 1.1. Voltage-time trace for a Zn/Cu Daniell cell (1.00 mol/L initial concentrations, 25°C, fixed external resistance). Adapted from standard electrochemistry discharge curves.
1.1 Using the graph, describe the trend in cell voltage over the 60-minute period. Include figures in your description. 2 marks
1.2 Explain, using the lesson’s content on changes during cell operation, why the cell voltage decreases over time. Refer to changes at both the anode and the cathode solution. 3 marks
1.3 Predict what the cell voltage would approach if the cell were allowed to run for several hours without replacement of solutions. Justify your prediction with reference to equilibrium. 2 marks
1.4 A student claims: “The voltage drop occurs because the electrons in the wire are used up.” Identify the error in this claim and provide the correct explanation. 1 mark
2. Interpret data table — comparing galvanic cell pairs
The table below shows four galvanic cell combinations and their measured E°cell values at 25°C. Use the data to answer the questions. 7 marks
| Cell | Anode metal | Cathode metal | E° anode (V) | E° cathode (V) | E°cell (V) |
|---|---|---|---|---|---|
| 1 | Zinc | Copper | −0.76 | +0.34 | +1.10 |
| 2 | Iron | Silver | −0.44 | +0.80 | +1.24 |
| 3 | Magnesium | Lead | −2.37 | −0.13 | +2.24 |
| 4 | Copper | Gold | +0.34 | +1.50 | +1.16 |
2.1 Which cell produces the highest voltage? Identify one factor from the table that explains why. 2 marks
2.2 For Cell 3, write the balanced half-equation at each electrode and the overall cell equation. Show electron balance. 3 marks
2.3 Use the data to verify E°cell for Cell 2 using the formula E°cell = E°cathode − E°anode. Show full working. 1 mark
2.4 A student proposes Cell 5 using copper as the anode in a copper/zinc arrangement (i.e. reversing Cell 1). Predict E°cell and state whether this cell is spontaneous. 1 mark
3. Cause-and-effect — what happens when the salt bridge is removed?
A working Zn/Cu galvanic cell has its salt bridge carefully removed while a voltmeter monitors the cell. Trace the cause-and-effect consequences in the boxes below. 5 marks
Cause: salt bridge removed from the working Zn/Cu galvanic cell.
Effect 1: What happens to the charge balance in the anode half-cell?
Effect 2: What happens to the charge balance in the cathode half-cell?
Effect 3: What does the voltmeter reading do?
Overall outcome: What does this demonstrate about the role of the salt bridge?
4. Compare and contrast — anode vs cathode in a galvanic cell
Complete the table comparing the anode and cathode in a galvanic cell. 8 marks (1 per cell)
| Feature | Anode | Cathode |
|---|---|---|
| Type of reaction occurring | ||
| Charge (polarity) of electrode | ||
| Direction electrons travel (in/out of electrode) | ||
| What happens to electrode mass over time | ||
| What happens to metal ion concentration in solution | ||
| Which ion from salt bridge migrates here | ||
| Half-equation type (write “M → Mⁿ⁺ + ne⁻” or “Mⁿ⁺ + ne⁻ → M”) | ||
| E° value relative to the other electrode |
Q1.1 — Trend description (2 marks)
The cell voltage starts at approximately 1.10 V at t = 0 and remains relatively stable (above 1.00 V) for approximately the first 25–30 minutes [1 mark]. After this initial plateau the voltage declines steadily, reaching approximately 0.85 V by 60 minutes [1 mark]. Accept similar numerical estimates ±0.05 V.
Q1.2 — Explanation of voltage decrease (3 marks)
As the cell operates, zinc at the anode dissolves (Zn → Zn²⁺ + 2e⁻), so the concentration of Zn²⁺ in the anode solution increases [1]. At the cathode, Cu²⁺ ions are removed from solution by reduction (Cu²⁺ + 2e⁻ → Cu), so the concentration of Cu²⁺ in the cathode solution decreases [1]. These concentration changes reduce the driving force (chemical potential difference) of the overall redox reaction, which lowers the EMF of the cell over time [1].
Q1.3 — Voltage approaching zero (2 marks)
The cell voltage would approach zero [1]. At this point the system has reached equilibrium — the forward and reverse reactions occur at the same rate, there is no net driving force for the redox reaction, and the cell no longer produces a net current (“flat battery”) [1].
Q1.4 — Error correction (1 mark)
The error is that electrons are not “used up”. Electrons are produced continuously at the anode (by oxidation of Zn) and consumed continuously at the cathode (by reduction of Cu²⁺). The voltage drops because the concentrations in the two half-cells change, reducing the chemical driving force, not because electrons are exhausted.
Q2.1 — Highest voltage (2 marks)
Cell 3 (Mg/Pb) produces the highest voltage at +2.24 V [1]. The large E°cell arises because magnesium has a very negative standard reduction potential (−2.37 V), giving a very large difference between E°cathode (−0.13 V) and E°anode (−2.37 V) [1].
Q2.2 — Cell 3 half-equations and overall (3 marks)
Anode (oxidation): Mg(s) → Mg²⁺(aq) + 2e⁻ [1]
Cathode (reduction): Pb²⁺(aq) + 2e⁻ → Pb(s) [1]
Overall (electrons balance at 2e⁻ — add directly): Mg(s) + Pb²⁺(aq) → Mg²⁺(aq) + Pb(s) [1]. Charge check: left = 0 + 2 = +2; right = +2 + 0 = +2. ✓
Q2.3 — Cell 2 verification (1 mark)
E°cell = E°cathode − E°anode = +0.80 − (−0.44) = +0.80 + 0.44 = +1.24 V. ✓ Matches the table.
Q2.4 — Reversed Cell (copper anode) (1 mark)
E°cell = E°cathode − E°anode = −0.76 − (+0.34) = −1.10 V. This is non-spontaneous — copper cannot displace Zn²⁺ ions from solution because copper has a more positive reduction potential (is less reactive) than zinc.
Q3 — Salt bridge cause-and-effect (5 marks)
Effect 1: The anode half-cell accumulates excess positive charge because Zn²⁺ ions keep forming but there are no anions entering to balance them [1].
Effect 2: The cathode half-cell accumulates a negative charge deficit because Cu²⁺ ions are removed from solution without cations entering to replace them [1].
Effect 3: The voltmeter reading drops rapidly to zero — the current stops almost immediately [1].
Overall outcome: This demonstrates that the salt bridge is essential for maintaining electrical neutrality in both half-cells; without it, the charge imbalance opposes further electron flow and the cell stops producing current [2].
Q4 — Anode vs cathode comparison (8 marks)
| Feature | Anode | Cathode |
|---|---|---|
| Type of reaction | Oxidation | Reduction |
| Polarity | Negative (−) | Positive (+) |
| Electron direction | Electrons flow OUT | Electrons flow IN |
| Electrode mass | Decreases (metal dissolves) | Increases (metal deposits) |
| Metal ion concentration | Increases | Decreases |
| Salt bridge ions migrating here | Anions (e.g. NO₃⁻) | Cations (e.g. K⁺) |
| Half-equation type | M → Mⁿ⁺ + ne⁻ | Mⁿ⁺ + ne⁻ → M |
| E° value | More negative (less positive) | More positive (less negative) |