Chemistry • Year 11 • Module 3 • Lesson 10
Galvanic Cells — Inert Electrodes & Predicting Reactions
Build HSC Band 5–6 extended-response technique: synthesise electrode theory, quantitative E° reasoning and engineering trade-offs in extended written form.
1. Extended response — protecting an Australian offshore pipeline (Band 5–6)
8 marks Band 5–6
Scenario. The North West Shelf gas pipeline network runs approximately 1500 km along the Western Australian seabed. The pipelines are constructed from carbon steel. A materials engineer is assessing sacrificial anode options for a new 200 km section, with data collected from three trial sites after 18 months of deployment in seawater.
| Trial site | Anode metal | E°(anode) / V | E°cell vs Fe / V | Mass of anode consumed in 18 months / kg | Pipeline corrosion rate (mm year−1) |
|---|---|---|---|---|---|
| Site 1 | Tin | −0.14 | −0.30 | 0.2 | 0.81 |
| Site 2 | Zinc | −0.76 | +0.32 | 3.1 | 0.04 |
| Site 3 | Magnesium | −2.37 | +1.93 | 14.8 | 0.03 |
Table 1. Trial anode performance data. E°(Fe2+/Fe) = −0.44 V. Pipeline corrosion rate measured independently by ultrasonic wall-thickness testing. Source: illustrative data adapted from NACE International pipeline corrosion standards.
Q1. Analyse the data in Table 1 and evaluate which anode material best balances effective cathodic protection of the pipeline against practical maintenance requirements. In your response you must:
- Define cathodic protection and explain the electrochemical mechanism by which a sacrificial anode prevents pipeline corrosion — referring to standard reduction potentials and the roles of anode and cathode.
- Explain why tin (Site 1) fails to protect the pipeline. Include an E°cell calculation and an explanation of what happens electrochemically when tin is in contact with the iron pipeline.
- Compare zinc (Site 2) and magnesium (Site 3) on at least three criteria drawn from the data: effectiveness of protection (corrosion rate), anode consumption rate, and E°cell value.
- Reach a justified recommendation for which metal the engineer should choose for the 200 km pipeline section. Include a consideration of the long-term cost of anode replacement.
2. Multi-step calculation and interpretation — fuel cell with inert electrodes (Band 5–6)
7 marks Band 5–6
Scenario. The CSIRO has been involved in Australian hydrogen economy research, including development of hydrogen fuel cell technology for stationary power and transport applications. A hydrogen–oxygen fuel cell operates using the following half-reactions under standard conditions:
Anode (oxidation): H2(g) → 2H+(aq) + 2e− E° = 0.00 V
Cathode (reduction): O2(g) + 4H+(aq) + 4e− → 2H2O(l) E° = +1.23 V
Both electrodes use porous platinum electrodes.
(a) Calculate E°cell for the hydrogen–oxygen fuel cell. 1 mark
(b) Write the balanced overall cell equation. Confirm electron balance. 2 marks
(c) Explain why platinum electrodes are used at both the anode and cathode. Your explanation must refer to the species in each half-reaction and what “inert” means in this context. 2 marks
(d) A critic claims: “The E°cell of the hydrogen fuel cell (+1.23 V) is lower than the Daniell cell (+1.10 V) — the Daniell cell must be a better energy source.” Evaluate this claim. What does E°cell measure, and what does it not tell us about the practical energy output of a cell? 2 marks
Q1 — Sample Band 6 response (8 marks), annotated
Define cathodic protection + mechanism. Cathodic protection is an electrochemical technique that prevents the corrosion (oxidation) of a metal structure by ensuring the structure always acts as the cathode in any galvanic cell formed in the surrounding electrolyte. A more reactive sacrificial metal (the anode) is placed in electrical contact with the structure. The anode has a more negative standard reduction potential than iron: E°(Zn2+/Zn) = −0.76 V and E°(Mg2+/Mg) = −2.37 V, both more negative than E°(Fe2+/Fe) = −0.44 V. Because the anode is more readily oxidised, it undergoes the oxidation half-reaction (e.g. Zn(s) → Zn2+(aq) + 2e−), while the iron pipeline is forced to be the cathode (reduction occurs). Iron at the cathode cannot be oxidised, so it cannot corrode. [2 marks: 1 for defining cathodic protection with electrochemical mechanism; 1 for correctly assigning anode/cathode roles using E° values.]
Tin fails — calculation and explanation. For tin as the sacrificial anode: E°cell = E°cathode(Fe) − E°anode(Sn) = −0.44 − (−0.14) = −0.30 V. E°cell < 0 → the proposed cell (tin anode, iron cathode) is non-spontaneous. The spontaneous cell has iron as the anode (−0.44 V is more negative than tin’s −0.14 V when iron and tin are in contact in seawater, the iron will corrode preferentially, not the tin). Table 1 confirms this: Site 1 shows a pipeline corrosion rate of 0.81 mm year−1 — far higher than with zinc or magnesium — meaning tin is actively accelerating corrosion. [2 marks: 1 for correct E°cell calculation; 1 for explaining that tin makes iron the anode and accelerates corrosion, supported by data.]
Compare zinc and magnesium on three criteria.
- Effectiveness (corrosion rate): Both zinc (0.04 mm year−1) and magnesium (0.03 mm year−1) provide near-equivalent pipeline protection — a 95–96% reduction in corrosion rate compared to tin.
- E°cell: Magnesium has a much larger E°cell (+1.93 V vs +0.32 V for zinc), meaning the thermodynamic driving force for magnesium oxidation is far greater. This translates to a higher rate of electron delivery and hence faster galvanic action.
- Anode consumption rate: The greater driving force causes magnesium to be consumed nearly five times faster than zinc (14.8 kg vs 3.1 kg over 18 months). Magnesium anodes would require far more frequent replacement, increasing maintenance costs and downtime for a 200 km pipeline.
[2 marks: 1 for correctly comparing at least two criteria using data values; 1 for discussing both E°cell and mass consumption rate.]
Recommendation. Zinc is the recommended anode material for the 200 km offshore section. While magnesium provides slightly superior corrosion protection (0.03 vs 0.04 mm year−1), the difference is negligible for pipeline integrity standards. The dramatically higher consumption rate of magnesium (4.8 times faster than zinc) means maintenance dives or pipeline access events would be required far more frequently, at significantly greater cost for a remote offshore installation. Zinc provides adequate protection (corrosion rate well below the typical industry threshold of <0.1 mm year−1) at a sustainable replacement interval, making it the more practical and economical choice. Magnesium may be preferred for short spans or highly aggressive microenvironments where the higher driving force is essential. [2 marks: 1 for a justified recommendation referencing data; 1 for incorporating a cost/maintenance argument linked to anode consumption rate.]
Marking criteria summary.
- 1 mark — Defines cathodic protection and links it to the electrochemical mechanism of forcing the structure to be the cathode.
- 1 mark — Correctly assigns anode and cathode roles using E° values for both effective metals (E° of anode must be more negative than E° of Fe).
- 1 mark — Calculates E°cell for tin vs iron correctly (−0.30 V) and concludes the cell is non-spontaneous (tin cannot act as the sacrificial anode).
- 1 mark — Explains electrochemically what happens with tin: iron becomes the anode and corrodes, supported by the corrosion-rate data from Site 1.
- 1 mark — Compares zinc and magnesium on corrosion effectiveness, citing data values from Table 1.
- 1 mark — Compares zinc and magnesium on E°cell and mass consumption rate, correctly linking larger E°cell to faster oxidation and higher mass loss.
- 1 mark — Reaches an explicit, justified recommendation that balances protection effectiveness against anode consumption and maintenance practicality.
- 1 mark — Response uses precise lesson terminology throughout: standard reduction potential, anode, cathode, oxidation, spontaneous, E°cell.
Q2(a) — E°cell for hydrogen–oxygen fuel cell
E°cell = E°cathode − E°anode = +1.23 − 0.00 = +1.23 V. Spontaneous (E°cell > 0).
Q2(b) — Balanced overall equation
Anode (scaled ×2): 2H2(g) → 4H+(aq) + 4e−. Cathode: O2(g) + 4H+(aq) + 4e− → 2H2O(l). Cancel 4H+ and 4e−. Overall: 2H2(g) + O2(g) → 2H2O(l). Electron check: 4e− produced at anode, 4e− consumed at cathode. ✓
Q2(c) — Why platinum at both electrodes
Both half-reactions involve only gases (H2, O2) and aqueous ions (H+). There is no solid metal that forms as a product or dissolves as a reactant at either electrode. A platinum (inert) electrode is therefore required at both anode and cathode: it provides a conducting surface through which electrons flow between the gas phase and the solution, without platinum itself participating in or being consumed by either half-reaction. Platinum also catalyses the slow O2 and H2 electrode reactions efficiently at low temperatures.
Q2(d) — Evaluate the critic’s claim
The critic has made a factual error: the hydrogen–oxygen fuel cell has E°cell = +1.23 V, which is greater than the Daniell cell (+1.10 V), not lower. The claim is wrong on its own arithmetic. More importantly, E°cell measures the electromotive force (voltage) under standard conditions; it does not determine total energy output, which also depends on the amount of reactant available (capacity) and the current delivered. A larger E°cell does indicate a greater thermodynamic driving force per electron transferred, but practical energy output is E × charge, not E alone. The hydrogen fuel cell produces water as its only product and can operate continuously with a hydrogen supply, offering advantages in energy density and emissions that a Daniell cell cannot match.
Marking: 1 mark for identifying the factual error in the E°cell comparison (+1.23 V > +1.10 V); 1 mark for explaining what E°cell does and does not measure (voltage ≠ total energy output; energy also depends on charge/capacity).