Chemistry • Year 11 • Module 3 • Lesson 10

Galvanic Cells — Inert Electrodes & Predicting Reactions

Apply E° values and the prediction rules to real cell setups, quantitative data, corrosion scenarios and a diagram critique.

Apply • Data & Reasoning

1. Interpret experimental cell voltage data

A class experiment measured the actual cell voltage (E_{cell, measured}) for five galvanic cells constructed under near-standard conditions. The table includes the relevant standard reduction potentials. 8 marks

Cell	Anode half-reaction	E°(anode) / V	Cathode half-reaction	E°(cathode) / V	E_{cell,measured} / V
P	Zn(s) → Zn²⁺(aq) + 2e⁻	−0.76	Cu²⁺(aq) + 2e⁻ → Cu(s)	+0.34	1.07
Q	Fe²⁺(aq) → Fe³⁺(aq) + e⁻	+0.77	MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O	+1.51	0.71
R	Fe(s) → Fe²⁺(aq) + 2e⁻	−0.44	Fe³⁺(aq) + e⁻ → Fe²⁺(aq)	+0.77	1.18
S	Zn(s) → Zn²⁺(aq) + 2e⁻	−0.76	2H⁺(aq) + 2e⁻ → H₂(g)	0.00	0.72
T	Mg(s) → Mg²⁺(aq) + 2e⁻	−2.37	Cu²⁺(aq) + 2e⁻ → Cu(s)	+0.34	2.68

1.1 Calculate E°cell for each cell using E°cell = E°cathode − E°anode. Compare each calculated value to E_{cell,measured} and describe the general relationship. 3 marks

1.2 For Cell Q, identify which electrode must be inert. Justify your choice by referring to the species involved in both half-reactions. 2 marks

1.3 Cell R uses an iron anode with a half-reaction involving Fe(s) and an iron cathode compartment containing Fe²⁺ and Fe³⁺ ions. For the cathode half-cell, what electrode material must be used, and why? 2 marks

1.4 The measured voltage for each cell is slightly lower than the calculated E°cell. Suggest one reason for this difference under classroom conditions. 1 mark

Stuck? Use E°cell = E°cathode − E°anode. Standard conditions are 25°C, 1 mol L⁻¹ concentrations, 1 atm — which classroom cells rarely achieve precisely.

2. Interpret the standard reduction potential bar chart

The chart below shows E° values for selected half-reactions. Use it to answer the questions. 7 marks

Figure 2. Standard reduction potentials for selected half-reactions under standard conditions (25°C, 1 mol L⁻¹, 1 atm). Values from NESA reference data.

2.1 Identify the species that would act as the cathode if Zn²⁺/Zn were paired with Fe²⁺/Fe. Calculate E°cell and state whether the reaction is spontaneous. 2 marks

2.2 A student claims that Cl₂/Cl⁻ could oxidise Fe²⁺ to Fe³⁺ spontaneously. Use E° values from the chart to confirm or refute this claim, showing your calculation. 2 marks

2.3 Using the chart, predict and justify which of the metals shown (Fe, Zn, Al, Mg) would dissolve most rapidly in dilute acid, and explain why the rate relates to E° value. 3 marks

Stuck? For 2.2: think about which species is the cathode (more positive E° = Cl₂) and which is the anode (Fe²⁺ → Fe³⁺). For 2.3: E°cell = 0.00 − E°(M). The most negative E°(M) gives the largest positive E°cell.

3. Compare active and inert electrodes across five criteria

Complete the comparison table. Write a brief phrase or sentence for each cell. 8 marks

Feature	Active electrode	Inert electrode
Definition
Mass change during operation
Typical materials used
Example half-reaction requiring this type
Why this electrode type is chosen

Stuck? Revisit lesson Card 02 and the data table comparing active vs inert electrodes.

4. Apply to a real-world scenario — Australian chlor-alkali industry

Orica Limited operates one of Australia’s largest chlor-alkali plants at Botany, NSW, which electrolytically produces chlorine (Cl₂) and sodium hydroxide from brine (NaCl solution). The industrial process is related to electrochemical principles: in the cell, chloride ions are oxidised at an electrode and water is reduced at another electrode.

A student studying galvanic cells notes that Cl₂(g)/Cl⁻(aq) has E° = +1.36 V, and asks: “If I constructed a galvanic cell with a Cl₂/Cl⁻ half-cell paired against Zn²⁺/Zn (E° = −0.76 V), what electrode material would the Cl₂/Cl⁻ half-cell need, and why?” 5 marks

4.1 Answer the student’s question: what electrode material is needed for the Cl₂/Cl⁻ half-cell, and why? 2 marks

4.2 Calculate E°cell for this Zn|Zn²⁺||Cl⁻|Cl₂|Pt cell. Is the reaction as written (Zn oxidised, Cl₂ reduced) spontaneous? 2 marks

4.3 Write the overall cell equation and identify which species acts as the oxidising agent. 1 mark

Stuck? Cl₂ is a gas and Cl⁻ is aqueous — neither is a solid metal. What type of electrode does this require? For E°cell, use E°cell = E°cathode − E°anode.

5. Predict and justify — sacrificial anode selection

A submerged steel (iron) water main in Sydney requires cathodic protection. An engineer has three candidate metals for sacrificial anodes: tin (E° = −0.14 V), zinc (E° = −0.76 V) and aluminium (E° = −1.66 V). E°(Fe²⁺/Fe) = −0.44 V. 7 marks

5.1 For each candidate metal, calculate E°cell when paired with steel. Determine whether each would actually provide cathodic protection. Show working. 3 marks

5.2 Explain why tin would not protect the steel, even though it is a metal. Refer to E° values. 2 marks

5.3 The engineer chooses zinc over aluminium for the Sydney water main. Using E°cell values and the concept of rate of anode consumption, justify this choice. 2 marks

Stuck? For a metal to act as a sacrificial anode for iron, its E° must be more negative than iron’s E° = −0.44 V. The more negative E°, the more strongly the metal is oxidised — but also the faster it corrodes.

Answers — Do not peek before attempting

Q1.1 — E°cell calculations

P: +0.34 − (−0.76) = +1.10 V (measured 1.07 V). Q: +1.51 − (+0.77) = +0.74 V (measured 0.71 V). R: +0.77 − (−0.44) = +1.21 V (measured 1.18 V). S: 0.00 − (−0.76) = +0.76 V (measured 0.72 V). T: +0.34 − (−2.37) = +2.71 V (measured 2.68 V). The measured values are consistently 0.02–0.04 V lower than calculated; this is a systematic positive relationship with a small negative offset due to non-standard conditions.

Q1.2 — Inert electrode in Cell Q

The anode half-cell of Cell Q (Fe²⁺ → Fe³⁺ + e⁻) requires a platinum (inert) electrode because both Fe²⁺ and Fe³⁺ are aqueous ions — no solid metal forms or dissolves at the electrode surface. A platinum rod conducts electrons without itself reacting. The MnO₄⁻ cathode also requires platinum for the same reason (MnO₄⁻ and Mn²⁺ are both aqueous).

Q1.3 — Electrode for Cell R cathode

The cathode half-cell of Cell R (Fe³⁺ + e⁻ → Fe²⁺) requires an inert platinum electrode. Both Fe³⁺ and Fe²⁺ are in aqueous solution; no solid iron would deposit at the cathode (unlike the anode where Fe(s) dissolves). Using an iron electrode for the cathode would introduce an Fe(s)/Fe²⁺ couple that interferes with the measurement.

Q1.4 — Why measured voltage is lower

Accept any one of: non-standard ion concentrations (not exactly 1 mol L⁻¹); temperature not exactly 25°C; internal resistance of the salt bridge and solutions; electrode surface impurities that reduce efficiency of electron transfer.

Q2.1 — Zn/Fe cell

Fe²⁺/Fe (E° = −0.44 V) is more positive, so iron is the cathode. E°cell = −0.44 − (−0.76) = +0.32 V > 0 → spontaneous. Zn is oxidised; Fe²⁺ is reduced.

Q2.2 — Cl₂ oxidising Fe²⁺

If Cl₂/Cl⁻ is cathode (E° = +1.36 V) and Fe³⁺/Fe²⁺ is anode (E° = +0.77 V): E°cell = +1.36 − (+0.77) = +0.59 V > 0. Confirmed: Cl₂ can spontaneously oxidise Fe²⁺ to Fe³⁺ (Cl₂ is a stronger oxidant than Fe³⁺, so it drives Fe²⁺ to Fe³⁺).

Q2.3 — Most rapid dissolver in acid

Mg dissolves most rapidly. For each metal, E°cell = 0.00 − E°(M): Fe = +0.44 V; Zn = +0.76 V; Al = +1.66 V; Mg = +2.37 V. Mg has the largest E°cell, meaning the thermodynamic driving force for its oxidation (dissolution) is greatest, corresponding to the fastest rate of reaction. [Award 1 mark: identify Mg; 1 mark: calculate E°cell; 1 mark: link larger positive E°cell to greater thermodynamic driving force / faster rate.]

Q3 — Compare/contrast table

Active electrode: Participates directly in the redox reaction (dissolves at anode or has metal deposited at cathode); mass changes (increases at cathode, decreases at anode); materials include Zn, Cu, Fe, Ag; example: Zn(s) → Zn²⁺(aq) + 2e⁻; chosen when the half-reaction involves a solid metal and its ion.

Inert electrode: Only conducts electrons; does not participate in the half-reaction; mass does not change; materials include Pt, graphite (C); example: Fe²⁺(aq) → Fe³⁺(aq) + e⁻; chosen when both species in the half-cell are aqueous or gaseous — no solid electrode material is produced or consumed.

Q4.1 — Electrode for Cl₂/Cl⁻ half-cell

An inert electrode such as platinum is required. Both Cl₂(g) and Cl⁻(aq) are non-metallic species that cannot form a solid conducting electrode — there is no solid metal to dissolve or be deposited. Platinum conducts electrons to/from the solution without reacting with Cl₂ or Cl⁻.

Q4.2 — E°cell and spontaneity

E°cell = E°cathode − E°anode = +1.36 − (−0.76) = +2.12 V. E°cell > 0 → the reaction (Zn oxidised, Cl₂ reduced) is spontaneous.

Q4.3 — Overall equation and oxidising agent

Anode: Zn(s) → Zn²⁺(aq) + 2e⁻. Cathode: Cl₂(g) + 2e⁻ → 2Cl⁻(aq). Overall: Zn(s) + Cl₂(g) → Zn²⁺(aq) + 2Cl⁻(aq). Oxidising agent: Cl₂ (it gains electrons and is reduced).

Q5.1 — E°cell for each candidate

Tin: E°cell = −0.44 − (−0.14) = −0.30 V < 0 → NOT spontaneous; tin will not protect steel. Zinc: E°cell = −0.44 − (−0.76) = +0.32 V > 0 → spontaneous; zinc will protect steel. Aluminium: E°cell = −0.44 − (−1.66) = +1.22 V > 0 → spontaneous; aluminium will protect steel.

Q5.2 — Why tin fails

Tin (E° = −0.14 V) has a more positive E° than iron (E° = −0.44 V). This means iron would be preferentially oxidised (act as the anode) in a tin–iron galvanic cell, not the tin. The E°cell = −0.30 V confirms the proposed cell (Sn anode, Fe cathode) is non-spontaneous; the spontaneous cell has iron as the anode — exactly the corrosion we are trying to prevent. Tin would accelerate corrosion of the iron pipeline, not protect it.

Q5.3 — Zinc vs aluminium choice

Both protect steel, but aluminium has a much larger E°cell (+1.22 V vs +0.32 V), meaning the aluminium-iron cell has a far greater driving force. The higher voltage drives a much faster oxidation rate for the aluminium anode — it corrodes very rapidly and must be replaced far more frequently than zinc. Zinc’s lower E°cell means it corrodes at a slower, more economical rate, making it more practical and cost-effective for long-term protection of the Sydney water main. Zinc is also less prone to passivation issues than aluminium under alkaline soil conditions.