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

Redox Reactions and Galvanic Cells

Ever wondered how your phone battery actually generates electricity? Redox reactions are chemical processes where electrons transfer between substances, and they power everything from phone batteries to your body's energy production. You'll learn to write half-equations (which show exactly which atoms lose or gain electrons) and use standard electrode potentials to predict which direction electricity flows in a galvanic cell—essential skills for VCE chemistry exams and understanding real-world applications like batteries and corrosion.

Learning objective: Write half-equations and overall redox equations, and predict the direction of current flow in a galvanic cell using standard electrode potentials.

By the end of this lesson you can:

• I identify species oxidised and reduced and their oxidation numbers.
• I use the electrochemical series to predict cell voltage.
• I label the anode (oxidation) and cathode (reduction) correctly.

A galvanic cell is a waterfall of electrons. Electrons fall from the stronger reductant to the stronger oxidant through a wire, doing useful work on the way.

Redox and galvanic cells

Oxidation is loss of electrons; reduction is gain. The species losing electrons is the reducing agent; the species gaining electrons is the oxidising agent. Track changes using oxidation numbers — up = oxidised, down = reduced.

A half-equation shows only one half-reaction (e.g. Cu²⁺ + 2e⁻ → Cu). Two half-equations combine to form the overall redox equation. Balance atoms and charges on each side.

A galvanic cell uses a spontaneous redox reaction to generate electricity. It has two half-cells connected by a salt bridge (maintains neutrality by ion flow) and an external wire (carries electrons). The anode is where oxidation happens; the cathode is where reduction happens. Electrons flow through the wire from anode to cathode.

Use the electrochemical series (standard electrode potentials E°) to predict which half-cell oxidises and which reduces. The species with the HIGHER E° is reduced (cathode); the LOWER E° is oxidised (anode). Cell voltage = E°(cathode) − E°(anode). If positive, the reaction is spontaneous.

Redox rules

• OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons).
• Reducing agent: Is itself oxidised.
• Oxidising agent: Is itself reduced.
• Galvanic cell direction: Electrons: anode → wire → cathode.
• Cell voltage: E°(cathode) − E°(anode). Positive = spontaneous.

Quick reference

| Feature | Galvanic cell |
|---|---|
| Anode | Oxidation, negative |
| Cathode | Reduction, positive |
| Electron flow | Anode → cathode |
| Salt bridge | Maintains neutrality |
| Prediction | Electrochemical series |

Worked example (Hard)

A galvanic cell uses Zn/Zn²⁺ and Cu/Cu²⁺ half-cells. E°(Zn²⁺/Zn) = −0.76 V; E°(Cu²⁺/Cu) = +0.34 V. Identify the anode, cathode, and calculate the cell voltage.

1. Higher E° = reduction (cathode). Cu has the higher E°. Cathode: Cu²⁺ + 2e⁻ → Cu
2. Lower E° = oxidation (anode). Zn is oxidised. Anode: Zn → Zn²⁺ + 2e⁻
3. Cell voltage. Subtract. E°cell = 0.34 − (−0.76) = +1.10 V
4. Spontaneity. Positive → spontaneous.

Answer: Anode = Zn (oxidised); cathode = Cu (reduced). Cell voltage = +1.10 V. Reaction is spontaneous.

Worked example (Hard)

A galvanic cell uses Zn(s)/Zn²⁺(aq) and Cu(s)/Cu²⁺(aq). Write the cell reaction and calculate E°cell. (E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = −0.76 V)

1. Identify electrodes. Zn is more reactive (lower E°) → anode (oxidation). Cu²⁺ is reduced at cathode.
2. Half-equations. Anode: Zn(s) → Zn²⁺(aq) + 2e⁻. Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s).
3. Cell voltage. E°cell = E°cathode − E°anode. E°cell = +0.34 − (−0.76) = +1.10 V

Answer: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). E°cell = +1.10 V (spontaneous).

Common mistakes

• Subtracting E° values in the wrong order (must be cathode minus anode) This is a commonly penalised error in VCAA exams. Fix: Review the underlying concept and practise similar problems.
• Multiplying E° by the number of electrons transferred (E° is intensive, not extensive) Incorrect application leads to wrong answers. Fix: Always return to first principles and verify your working.
• Confusing which electrode is anode and cathode in a galvanic cell A frequent source of confusion in VCE Chemistry. Fix: Distinguish between related concepts through targeted practice.

Electrochemistry in everyday life

Electrochemistry governs batteries, corrosion, and electroplating. All batteries are galvanic cells converting chemical to electrical energy. The voltage depends on the electrode potentials of the half-cells. Corrosion (rusting) is an electrochemical process where iron is oxidised at anodic sites and oxygen is reduced at cathodic sites. Galvanic corrosion occurs when two dissimilar metals are in contact — the more reactive metal corrodes preferentially. This principle is used in cathodic protection: sacrificial anodes (zinc blocks) are attached to steel structures so the zinc corrodes instead of the iron.

Practice questions

1. At the anode of a galvanic cell:

   1. Oxidation occurs
   2. Reduction occurs
   3. Current enters
   4. Cations are reduced

   (2 marks)

2. In the half-equation Fe²⁺ → Fe³⁺ + e⁻, iron is:

   1. Oxidised from +2 to +3
   2. Reduced from +3 to +2
   3. Losing a proton
   4. Gaining a neutron

   (3 marks)

3. Using E°(Ag⁺/Ag) = +0.80 V and E°(Fe²⁺/Fe) = −0.44 V, calculate the cell voltage and state whether the reaction is spontaneous.

   (3 marks)

Summary

• Identify species oxidised and reduced and their oxidation numbers.
• Use the electrochemical series to predict cell voltage.
• Label the anode (oxidation) and cathode (reduction) correctly.