VCE Chemistry (Year 12) · Unit 3 — Supplying Energy: Fuels, Cells and Electrolysis · 14m read

Fuel Cells and Electrolysis

What if you could turn electricity into hydrogen, then burn that hydrogen to make electricity again? Fuel cells do exactly this, offering a cleaner alternative to traditional engines by converting chemical energy directly into power. You'll explore how the hydrogen-oxygen fuel cell works, compare it with combustion engines, and use Faraday's laws — simple rules linking electrical charge to chemical change — to predict how much product forms during electrolysis, skills essential for chemistry assessments and understanding tomorrow's green energy systems.

Learning objective: Describe the hydrogen–oxygen fuel cell, compare it with combustion engines, and apply Faraday's laws to electrolysis calculations.

By the end of this lesson you can:

• I write anode/cathode half-equations for the H₂–O₂ fuel cell.
• I apply Q = I·t and n(e⁻) = Q/F (F = 96 500 C mol⁻¹) to electrolysis problems.
• I describe why fuel cells are more efficient than combustion engines.

Electrolysis is pushing the waterfall uphill. A power supply forces a non-spontaneous reaction; fuel cells run continuously on supplied fuel.

Fuel cells and electrolysis

A fuel cell is a galvanic cell where fuel and oxygen are supplied continuously. The hydrogen–oxygen fuel cell (with KOH electrolyte):
Anode: 2 H₂ + 4 OH⁻ → 4 H₂O + 4 e⁻
Cathode: O₂ + 2 H₂O + 4 e⁻ → 4 OH⁻
Overall: 2 H₂ + O₂ → 2 H₂O
No moving parts, no heat loss to mechanical conversion — fuel cells are 40–60% efficient vs combustion engines at 25–35%. Only product is water.

Electrolysis is the opposite: electrical energy drives a non-spontaneous redox reaction. In a molten salt, the cation reduces at the cathode and the anion oxidises at the anode. In aqueous solution, water can be oxidised or reduced instead, depending on E° values and concentration.

Faraday's laws quantify electrolysis:
Charge: Q = I·t (coulombs = amperes × seconds)
Moles of electrons: n(e⁻) = Q / F, where F = 96 500 C/mol
Then use the half-equation stoichiometry to find moles of product, and convert to mass or volume.

Electrolysis formulas

• Charge: Q = I × t (A × s = C).
• Moles of electrons: n(e⁻) = Q / F; F = 96 500 C/mol.
• Stoichiometric ratio: Use half-equation ratio to convert electrons → product.
• H₂ fuel cell: Overall: 2 H₂ + O₂ → 2 H₂O. No CO₂.
• Efficiency: Fuel cells more efficient than combustion engines.

Quick reference

| Cell | Energy conversion | Anode sign |
|---|---|---|
| Galvanic | Chemical → electrical | Negative |
| Electrolytic | Electrical → chemical | Positive |
| Fuel cell | Chemical → electrical (continuous) | Negative |
| Faraday | n(e⁻) = Q/F; Q = It | |

Worked example (Hard)

Electrolysis of molten NaCl with a current of 5.00 A for 30 minutes. Calculate the mass of Na produced.

1. Charge. Q = I × t. 30 min = 1800 s. Q = 5.00 × 1800 = 9000 C
2. Moles of electrons. Q / F. n(e⁻) = 9000 / 96 500 = 0.0933 mol
3. Moles of Na. Na⁺ + e⁻ → Na (1:1 ratio). n(Na) = 0.0933 mol
4. Mass of Na. m = n × M. M(Na) = 23.0 g/mol. m = 0.0933 × 23.0 = 2.15 g

Answer: Mass of Na produced ≈ 2.15 g.

Worked example (Hard)

Calculate the mass of aluminium deposited when 5.00 A of current is passed through molten Al₂O₃ for 1.00 hour. (M(Al) = 26.98 g/mol, F = 96485 C/mol)

1. Charge. Q = It = 5.00 × 3600 = 18000 C.
2. Moles of electrons. n(e⁻) = Q/F = 18000/96485 = 0.1866 mol.
3. Moles of Al. Al³⁺ + 3e⁻ → Al. Ratio 1:3. n(Al) = 0.1866/3 = 0.06219 mol.
4. Mass. m = nM = 0.06219 × 26.98 = 1.68 g.

Answer: 1.68 g of aluminium. The 3e⁻ per Al³⁺ is critical — Al requires 3 electrons for reduction, not 1 or 2.

Common mistakes

• Using the wrong number of electrons per metal ion (Al³⁺ needs 3e⁻, Cu²⁺ needs 2e⁻) This is a commonly penalised error in VCAA exams. Fix: Review the underlying concept and practise similar problems.
• Not converting time to seconds for the Q = It calculation Incorrect application leads to wrong answers. Fix: Always return to first principles and verify your working.
• Confusing anode and cathode roles in electrolysis vs galvanic cells A frequent source of confusion in VCE Chemistry. Fix: Distinguish between related concepts through targeted practice.

Faraday's laws and quantitative electrolysis

Faraday’s laws provide the quantitative link between charge passed and mass deposited in electrolysis. The formula m = (Q × M)/(n × F) relates mass deposited (m) to charge passed (Q = It), molar mass (M), number of electrons per ion (n), and Faraday’s constant (F = 96485 C/mol). Industrial applications include: aluminium extraction (Hall-Héroult process), copper refining, electroplating (chrome, gold, silver), and chlor-alkali production (Cl₂, NaOH). The key to VCE calculations is always writing the half-equation first to determine the electron ratio.

Practice questions

1. In a hydrogen–oxygen fuel cell with an alkaline electrolyte, the anode reaction is:

   1. 2 H₂ + 4 OH⁻ → 4 H₂O + 4 e⁻
   2. O₂ + 2 H₂O + 4 e⁻ → 4 OH⁻
   3. H₂ → 2 H⁺ + 2 e⁻
   4. 2 H₂O → O₂ + 4 H⁺ + 4 e⁻

   (2 marks)

2. How many coulombs are delivered by a 2.0 A current flowing for 10 minutes?

   1. 1200 C
   2. 200 C
   3. 20 C
   4. 12 000 C

   (3 marks)

3. Calculate the mass of Cu deposited from electrolysis of CuSO₄ solution with 3.0 A for 20 minutes. (Cu²⁺ + 2e⁻ → Cu, M(Cu) = 63.5)

   (3 marks)

Summary

• Write anode/cathode half-equations for the H₂–O₂ fuel cell.
• Apply Q = I·t and n(e⁻) = Q/F (F = 96 500 C mol⁻¹) to electrolysis problems.
• Describe why fuel cells are more efficient than combustion engines.