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Module 3 · L10 of 12 35 min ⚡ +50 XP in Learn · +25 to complete

Galvanic Cells — Inert Electrodes & Predicting Reactions

The Trans-Alaska Pipeline (constructed 1974–1977) runs 1,300 km through permafrost. Every 300 metres, zinc sacrificial anodes are bolted to the steel exterior. The zinc corrodes first — because Zn is higher in the activity series than Fe — protecting the pipeline from electrochemical corrosion. The same standard reduction potentials used to predict galvanic cells predict exactly which metal will sacrifice itself.

Today’s hook — The Trans-Alaska Pipeline (constructed 1974–1977) runs 1,300 km through permafrost. Every 300 metres, zinc sacrificial anodes are bolted to the steel exterior. The zinc corrodes first — because Zn is higher in the activity series than Fe — protecting the pipeline from electrochemical corrosion. The same standard reduction potentials used to predict galvanic cells predict exactly which metal will sacrifice itself.

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Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Recall — your gut answer first

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An offshore oil pipeline runs along the seafloor for hundreds of kilometres. The steel pipe is in constant contact with seawater — one of the most corrosive environments on Earth. Engineers attach blocks of zinc metal at regular intervals along the outside of the pipe.

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What you'll master

Know

Key facts

What an inert electrode is and why it is needed — conducts electrons without participating in the reaction; required when both half-cell species are aqueous or gaseous
Common inert electrode materials: platinum (Pt) and graphite (C)
The principles of cathodic protection: attaching a more reactive metal (sacrificial anode) forces the structure to be the cathode, preventing its corrosion

Understand

Concepts

Why the Fe²⁺/Fe³⁺ half-reaction requires an inert electrode — both species are aqueous; there is no solid Fe that forms a usable electrode for this couple
How to use E° values to predict spontaneity of a reaction and identify which electrode is anode/cathode in cells with inert electrodes
Why cathodic protection works electrochemically — the more negative E° of zinc forces it to be oxidised (anode), leaving iron as the cathode where reduction (not corrosion) occurs

Can do

Skills

Identify when an inert electrode is needed and specify the correct material (Pt or graphite)
Perform E°cell calculations for cells involving inert electrodes; describe electrode changes (or lack thereof)
Predict whether a metal dissolves in dilute acid using E° values; explain cathodic protection using E° values and anode/cathode roles

Key terms

Inert electrode

An electrode that does not participate in the redox reaction; used when gases or non-solid species are involved (e.g., Pt, graphite).

Standard reduction potential (E°)

The voltage of a half-cell measured under standard conditions (25°C, 1 mol L⁻¹, 1 atm) relative to the standard hydrogen electrode.

Standard hydrogen electrode (SHE)

The reference half-cell with E° = 0.00 V; consists of H₂(g) at 1 atm over a Pt electrode in 1 mol L⁻¹ H⁺.

Predicting spontaneity

A galvanic cell reaction is spontaneous if E°cell > 0; the more negative half-cell is the anode.

Daniell cell

A classic galvanic cell with Zn(s)/Zn²⁺(aq) at the anode and Cu²⁺(aq)/Cu(s) at the cathode; E° = +1.10 V.

Table of standard reduction potentials

Lists half-reactions in order of E°; a species higher in the table is a stronger oxidant (more easily reduced).

Inert Electrodes — When the Electrode Doesn’t React

core concept

In the Hofmann voltameter, two platinum wires sit in dilute sulfuric acid. When current is applied, bubbles form at both wires — hydrogen at one end, oxygen at the other. The platinum rods themselves do not dissolve, grow, or change in any measurable way. The electrode is inert: it conducts electrons but plays no chemical role in the reaction.

For example, the Fe²⁺/Fe³⁺ half-reaction: Fe²⁺(aq) → Fe³⁺(aq) + e⁻. There is no solid iron metal involved — both species are in solution. An inert electrode is used in these cases — it conducts electrons to or from the solution without itself being oxidised or reduced.

Platinum (Pt) is the most common inert electrode because it is highly unreactive and an excellent electrical conductor. Graphite (C) is also used. The inert electrode does not change in mass during cell operation — it is neither dissolved nor deposited upon.

Feature	Active Electrode	Inert Electrode
Participates in reaction?	Yes — dissolves or gains mass	No — conducts only
Mass changes?	Yes	No
Common examples	Zn, Cu, Fe, Ag	Pt, graphite (C)
Used when?	Solid metal / metal ion reaction	Ion-to-ion or gas half-reaction

Must Know: When a question describes a galvanic cell involving Fe²⁺/Fe³⁺, Cl₂/Cl⁻, MnO₄⁻/Mn²⁺, or any half-reaction with no solid metal product, specify that a platinum (inert) electrode is required. Writing “iron electrode” for a Fe²⁺/Fe³⁺ cell is wrong — the iron electrode would dissolve as Fe(s) → Fe²⁺ + 2e⁻, not participate in the Fe²⁺/Fe³⁺ couple.

Common Error: Students assume that any cell involving iron must use an iron electrode. The Fe²⁺/Fe³⁺ couple requires a platinum electrode because both species are aqueous. The Fe(s)/Fe²⁺(aq) couple uses an iron electrode. Always check which species are solid and which are aqueous before assigning electrode material.

Inert electrodes (Pt, graphite) are used when both half-cell species are aqueous or gaseous — e.g. Fe²⁺/Fe³⁺, Cl₂/Cl⁻. They conduct electrons but are neither oxidised nor reduced, and their mass does not change. Required whenever there is no solid metal form of the electrode species.

Pause — copy the highlighted definition into your book before moving on.

Explain it: A student sets up a galvanic cell with one half-cell containing Fe²⁺(aq) and Fe³⁺(aq) and uses an iron electrode. Explain why this is incorrect and what electrode material should be used instead. (2–3 sentences)

Predicting Reactions Using the Standard Reduction Potential Table

core concept

We just saw that inert electrodes are used when half-cell species are in aqueous form. That raises a question: how do we use the standard reduction potential (SRP) table to systematically predict which reactions are spontaneous — not just for simple Zn/Cu cells, but for any combination? This card answers it → assign cathode (higher E°) and anode (lower E°); calculate E°cell = E°cathode − E°anode; if positive, the reaction is spontaneous.

The standard reduction potential table lists half-reactions in order of decreasing E° — from strongest oxidant at the top (most positive E°) to strongest reductant at the bottom (most negative E°).

Two systematic rules govern predictions:

A spontaneous reaction occurs when a species with more positive E° (higher in the table) oxidises a species with more negative E° (lower in the table).
E°cell = E°cathode − E°anode > 0 confirms spontaneity quantitatively.

Prediction procedure: (1) Identify the two half-reactions. (2) Assign cathode (more positive E°) and anode (more negative E°). (3) Calculate E°cell. (4) If E°cell > 0 → spontaneous as written; if E°cell < 0 → reverse reaction is spontaneous.

Must Know: To predict whether a metal dissolves in dilute acid: use the H⁺/H₂ couple (E° = 0.00 V) as cathode. If the metal has E° < 0 → E°cell > 0 → metal dissolves in dilute acid. If E° > 0 (e.g. Cu, Ag, Au) → E°cell < 0 → no reaction with dilute acids.

Insight: The standard reduction potential table and the activity series encode exactly the same reactivity ranking. The activity series is a qualitative ordering; the E° table gives quantitative values. Any prediction you can make from the activity series can also be made from the E° table — with the added benefit of knowing the exact cell voltage.

SRP prediction procedure: (1) identify the two half-reactions; (2) assign cathode (more positive E°) and anode; (3) calculate E°cell = E°cathode − E°anode; (4) E°cell > 0 → spontaneous. The SRP table and activity series encode identical reactivity ranking.

Add the highlighted procedure to your notes before the check below.

Mini-task: Use the standard reduction potential table to predict whether the following reaction is spontaneous as written: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s). Use E°(Cu²⁺/Cu) = +0.34 V and E°(Ag⁺/Ag) = +0.80 V. Identify the anode and cathode, calculate E°cell, and state your conclusion. Show full working. (3–4 sentences)

Comparing Cells with Active vs Inert Electrodes

core concept

We just saw the SRP prediction procedure for spontaneity. That raises a question: how does the analysis differ when comparing a standard Zn/Cu cell (active electrodes) with a cell using inert Pt electrodes and ion couples in solution? This card answers it → the E°cell calculation is identical; only the physical changes differ — active electrodes change mass while inert electrodes do not.

The procedure for analysing a galvanic cell is identical whether the electrodes are active or inert — identify the half-reactions, assign anode and cathode by E° values, write equations, and calculate E°cell.

Cell A (active electrodes): Zn|Zn²⁺||Cu²⁺|Cu
Anode: Zn(s) → Zn²⁺(aq) + 2e⁻ (E° = −0.76 V). Cathode: Cu²⁺(aq) + 2e⁻ → Cu(s) (E° = +0.34 V).
E°cell = +0.34 − (−0.76) = +1.10 V. Zinc electrode loses mass; copper electrode gains mass.

Cell B (inert electrodes): Pt|Fe²⁺(aq),Fe³⁺(aq)||MnO₄⁻(aq),Mn²⁺(aq),H⁺(aq)|Pt
Anode (less positive E°): Fe²⁺(aq) → Fe³⁺(aq) + e⁻ (E° = +0.77 V for Fe³⁺/Fe²⁺). Cathode (more positive E°): MnO₄⁻(aq) + 8H⁺(aq) + 5e⁻ → Mn²⁺(aq) + 4H₂O(l) (E° = +1.51 V).
E°cell = +1.51 − (+0.77) = +0.74 V. Neither platinum electrode changes in mass.

Feature	Cell A (Active)	Cell B (Inert)
Anode material	Zinc metal	Platinum
Cathode material	Copper metal	Platinum
Anode change	Mass decreases	No change
Cathode change	Mass increases	No change
E°cell	+1.10 V	+0.74 V

Must Know: Cell notation convention: anode half-cell is always on the LEFT; cathode half-cell is always on the RIGHT. Single vertical line | = phase boundary (between solid electrode and solution). Double vertical line || = salt bridge.

Cell notation: anode | anode solution || cathode solution | cathode. Active cells (Zn/Cu): both electrode masses change. Inert electrode cells (Pt): electrode masses unchanged — only solution concentrations change. E°cell calculation is identical in both cases.

Pause — write the highlighted convention into your book.

Quick check: A galvanic cell uses a Pt electrode in a solution containing Fe²⁺(aq) and Fe³⁺(aq). What change occurs at the platinum electrode during operation when it acts as the anode?

Cathodic Protection — Applied Galvanic Cell Engineering

core concept

We just saw that E°cell values predict which metal corrodes when two metals are in electrical contact. That raises a question: how is this deliberately exploited in engineering to protect steel structures? This card answers it → by attaching a more reactive metal (more negative E°), engineers ensure steel is always the cathode — where reduction, not corrosion, occurs.

Cathodic protection prevents steel corrosion by ensuring the steel structure always acts as a cathode (where reduction occurs) rather than an anode (where oxidation/corrosion occurs).

Method 1 — Sacrificial anode protection: A block of more reactive metal (zinc or magnesium) is attached directly to the steel structure. Zinc (E° = −0.76 V) has a more negative reduction potential than iron (E° = −0.44 V), so zinc is more readily oxidised. Zinc becomes the anode and corrodes preferentially; iron is the cathode and is protected. Applications: ship hulls, offshore pipelines, underground gas mains, jetty pilings. Zinc blocks must be replaced periodically.

Method 2 — Impressed current cathodic protection (ICCP): An external DC power supply forces current through the steel structure, making it the cathode. Used for large structures where sacrificial anodes would be impractical.

Must Know: In HSC questions on cathodic protection, always explain the mechanism using E° values: “Zinc has a more negative standard reduction potential than iron, so zinc is preferentially oxidised (acts as the anode) while iron acts as the cathode and is protected from oxidation.”

Common Error: Students say zinc “prevents oxygen from reaching the iron.” This describes barrier protection (paint, galvanising as a physical coat) — not cathodic protection. Cathodic protection works electrochemically even when the iron surface is exposed to the corrosive environment, because the zinc forces the iron to be a cathode regardless of whether oxygen reaches it.

Cathodic protection: attach Zn (E° = −0.76 V) or Mg to steel (E° = −0.44 V) — the more negative E° forces zinc to be the anode (corrodes sacrificially) while steel is the cathode (protected). ICCP alternative: external DC supply makes steel the cathode directly.

Add the highlighted explanation to your notes before the check below.

Explain it: A zinc block is bolted to a steel pipeline submerged in seawater. Using E° values and the concepts of anode/cathode, explain why this arrangement prevents the steel from corroding. (3–4 sentences)

Predicting Whether Metals Will Dissolve in Acid

core concept

We just saw how cathodic protection exploits the spontaneity rule: more negative E° → forced to be anode. That raises a question: how do we use the same rule to predict whether a specific metal will dissolve when added to dilute acid? This card answers it → use the H⁺/H₂ couple (E° = 0.00 V) as the cathode; if E°(metal) is negative, E°cell > 0 and the metal dissolves.

Using the standard reduction potential table, you can predict whether any metal will dissolve in dilute acid without needing the activity series.

The relevant reduction half-reaction for dilute acid: 2H⁺(aq) + 2e⁻ → H₂(g), E° = 0.00 V.

For any metal M: E°cell = E°(H⁺/H₂) − E°(Mⁿ⁺/M) = 0.00 − E°(M).
If E°(M) is negative → E°cell > 0 → metal dissolves in dilute acid.
If E°(M) is positive → E°cell < 0 → metal does NOT dissolve in dilute acid.

This is why Zn (E° = −0.76 V), Fe (E° = −0.44 V), and Mg (E° = −2.37 V) all dissolve in dilute HCl, while Cu (+0.34 V), Ag (+0.80 V), and Au (+1.50 V) do not. The activity series rule “metals above hydrogen dissolve in dilute acid” is now fully quantified: “above hydrogen” simply means E° < 0.00 V.

Insight: Gold is famously unreactive with most acids — even concentrated HCl and HNO₃ individually. The only acid mixture that dissolves gold is aqua regia (3:1 HCl:HNO₃), which works through a different mechanism involving chloride complexes that lower the effective E° for the Au³⁺/Au couple. (Aqua regia chemistry is covered in Year 12 Module 7 — noted here for interest only.) This is one reason gold has been used as currency for millennia.

To predict whether a metal dissolves in dilute acid: E°cell = 0.00 − E°(metal). If E°(metal) < 0 → E°cell > 0 → dissolves (Zn, Fe, Mg). If E°(metal) > 0 → E°cell < 0 → does not dissolve (Cu, Ag, Au). "Metals above hydrogen dissolve in dilute acid" means E° < 0.00 V.

Pause — write the highlighted rule into your book.

Mini-task: A student drops three metals — zinc (E° = −0.76 V), copper (E° = +0.34 V), and aluminium (E° = −1.66 V) — separately into dilute hydrochloric acid. Use E°cell = E°cathode − E°anode to predict for each metal whether it dissolves. Show a calculation for each and state your conclusion. [E°(2H⁺/H₂) = 0.00 V]

Cross-lesson links: In L09 you built galvanic cells with active (dissolving/depositing) electrodes. In this lesson the electrodes are inert (platinum or graphite) — the reaction happens in solution, not at the metal. The standard reduction potentials from L09 are used here to predict cell potential and spontaneity.

Worked examples · reveal as you go

Worked example +5 XP on full reveal

A galvanic cell uses one half-cell containing Fe²⁺(aq) and Fe³⁺(aq) with a platinum electrode, and another containing Ag⁺(aq) with a silver electrode. Use E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Ag⁺/Ag) = +0.80 V. (a) Identify anode and cathode. (b) Write half-equations and overall equation. (c) Calculate E°cell. (d) Describe the change at each electrode.

Ag⁺/Ag: E° = +0.80 V (more positive) → silver is the cathode (reduction). Fe³⁺/Fe²⁺: E° = +0.77 V (less positive) → Fe²⁺ is oxidised → platinum electrode is the anode (inert).

The species with the more positive E° value acts as the cathode (reduction); the less positive acts as the anode (oxidation).

Anode (oxidation, inert Pt): Fe²⁺(aq) → Fe³⁺(aq) + e⁻
Cathode (reduction, Ag): Ag⁺(aq) + e⁻ → Ag(s)
Overall: Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)

Electrons balance (1e⁻), so combine the half-equations directly. Check charge: left +2+1=+3; right +3+0=+3.

E°cell = E°cathode − E°anode = +0.80 − (+0.77) = +0.03 V

E°cell > 0 means spontaneous (just barely). For inert electrodes, only the solution species change; electrode mass remains constant.

Platinum anode: no change in mass (inert; Fe²⁺ is oxidised to Fe³⁺ in solution). Silver cathode: increases in mass as Ag⁺ ions are reduced and deposit as solid silver.

At an inert electrode, the electrode itself does not participate. At the active (Ag) cathode, solid Ag accumulates, so mass increases.

Worked example +5 XP on full reveal

A zinc block is bolted to a steel (iron) structure submerged in seawater. Use E°(Zn²⁺/Zn) = −0.76 V and E°(Fe²⁺/Fe) = −0.44 V. (a) Identify anode and cathode. (b) Calculate E°cell. (c) Explain why this arrangement protects the iron.

Zn: E° = −0.76 V (more negative) → more readily oxidised → zinc is the anode (sacrificial). Fe: E° = −0.44 V (more positive) → more readily reduced → iron is the cathode (protected).

The more negative E° value is oxidised at the anode. The less negative (more positive) value is reduced at the cathode. This difference drives the spontaneous reaction.

E°cell = E°cathode − E°anode = −0.44 − (−0.76) = −0.44 + 0.76 = +0.32 V

E°cell > 0 confirms the zinc-iron galvanic cell is spontaneous in seawater. This spontaneity drives zinc to corrode preferentially.

Because zinc is the anode, it undergoes oxidation: Zn(s) → Zn²⁺(aq) + 2e⁻. Iron is the cathode — it undergoes reduction, not oxidation. Iron cannot corrode (be oxidised) while it is the cathode.

Corrosion is oxidation. At the cathode, reduction occurs, not oxidation. Therefore iron is protected. Zinc corrodes instead — it is "sacrificed" to protect iron.

Summary: Zinc = anode (sacrificial). Iron = cathode (protected). E°cell = +0.32 V (spontaneous). The spontaneous zinc-iron galvanic cell forces iron to be the cathode, preventing its oxidation while zinc corrodes preferentially.

This is cathodic protection — protecting a structure by making it act as a cathode. The sacrificial anode bears the cost (corrosion), so the main structure is spared.

Formula reference · this lesson

core formula

📐

Key Formulas — This Lesson

E°cell = E°cathode − E°anode (same as L09 — applied to new contexts)

Spontaneous: E°cell > 0 | Non-spontaneous: E°cell < 0

Inert electrode: conductor that does not participate in the cell reaction (e.g. Pt, graphite)

Required when both species in a half-cell are in solution or gas phase (no solid metal electrode)

Metal dissolves in dilute acid if E°(Mⁿ⁺/M) < 0.00 V (i.e. metal is above H in activity series)

E°cell = 0.00 − E°(Mⁿ⁺/M) > 0 if E°(M) is negative

Common errors · the 3 traps that cost marks

Common misconception

In a galvanic cell, electrons flow through the salt bridge.

Fix: Electrons flow through the external wire from anode to cathode. The salt bridge allows ions to flow internally to maintain charge balance — it does not conduct electrons. Confusing electron flow with ion migration is a common source of lost marks.

An inert electrode loses mass at the anode

A student states that the platinum anode in a Fe²⁺/Fe³⁺ cell loses mass because oxidation occurs there.

Fix: An inert electrode (Pt, graphite) undergoes no change in mass — it conducts electrons but does not participate in the reaction. Oxidation occurs in the half-cell (Fe²⁺ → Fe³⁺ + e⁻ in solution), not at the electrode itself. Always state clearly: "The platinum electrode undergoes no change in mass; only the [ion] concentration in solution changes."

Cathodic protection physically coats the iron

A student writes that zinc blocks prevent iron corrosion by "coating the iron surface so oxygen cannot reach it."

Fix: This describes barrier protection (paint, galvanising as a physical coat) — not cathodic protection. Cathodic protection works electrochemically: the more negative E° of zinc forces zinc to be the anode and iron to be the cathode. Iron as a cathode undergoes reduction, not oxidation — so it cannot corrode regardless of whether oxygen reaches the surface. The protection is electrical, not physical.

Work mode · how are you completing this lesson?

Quick-fire practice · 5 reps +2 XP per reveal

Q8 (4 marks): Explain the difference between an active electrode and an inert electrode in a galvanic cell. Give one example of a half-reaction that requires each type, and explain your reasoning.

Q9 (4 marks): Use E° values to predict whether each of the following metals will dissolve in dilute sulfuric acid (H₂SO₄). Show your E°cell calculation for each. E°(2H⁺/H₂) = 0.00 V; E°(Zn²⁺/Zn) = −0.76 V; E°(Cu²⁺/Cu) = +0.34 V; E°(Al³⁺/Al) = −1.66 V. (a) Zinc. (b) Copper. (c) Aluminium.

Q10 (5 marks): An offshore drilling platform has a steel hull. Engineers are choosing between zinc (E° = −0.76 V) and magnesium (E° = −2.37 V) sacrificial anodes. (a) For each metal, calculate E°cell when used as a sacrificial anode with the steel hull (E°(Fe) = −0.44 V) and confirm that both are effective. (b) Explain the electrochemical mechanism by which the sacrificial anode protects the steel from corrosion. (c) Evaluate which metal (Zn or Mg) would be a more practical choice for this marine application, considering both the level of protection and the rate of consumption of the anode.

Predict, using E° values, whether iron dissolves in dilute HCl. [E°(Fe²⁺/Fe) = −0.44 V; E°(2H⁺/H₂) = 0.00 V]. Show working.

Explain, using an E°cell calculation, why gold (E° = +1.50 V) does not dissolve in dilute HCl.

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Revisit your thinking

The Trans-Alaska Pipeline's zinc anodes work because zinc has a lower standard reduction potential than iron: E°(Zn²⁺/Zn) = −0.76 V vs E°(Fe²⁺/Fe) = −0.44 V. In a galvanic couple, the more negative E° metal oxidises preferentially — Zn → Zn²⁺ + 2e⁻ occurs before Fe → Fe²⁺ + 2e⁻. The pipeline steel is the cathode (protected); the zinc is the anode (sacrificed). The cell potential for the Zn–Fe couple: E°cell = −0.44 − (−0.76) = +0.32 V — positive, confirming the corrosion is spontaneous. A positive E°cell always means the forward reaction is spontaneous.

Now revisit your initial response. What did you get right? What has changed in your thinking?

Look back at your initial response in your book. Annotate it with what you now understand differently.

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Interactive Tool — Redox Reactions Lab Open fullscreen ↗

The Redox Reactions Lab tool shows that when a zinc block is attached to a steel structure in seawater, the zinc acts as the anode because…

Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next.

Short answer

UnderstandBand 3

Q1. 8. (4 marks) Explain the difference between an active electrode and an inert electrode in a galvanic cell. Give one example of a half-reaction that requires each type, and explain your reasoning. (2 + 2 marks)

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ApplyBand 4

Q2. 9. (4 marks) Use E° values to predict whether each of the following metals will dissolve in dilute sulfuric acid (H₂SO₄). Show your E°cell calculation for each. E°(2H⁺/H₂) = 0.00 V; E°(Zn²⁺/Zn) = −0.76 V; E°(Cu²⁺/Cu) = +0.34 V; E°(Al³⁺/Al) = −1.66 V. (a) Zinc. (b) Copper. (c) Aluminium. (1 + 1 + 2 marks)

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EvaluateBand 5

Q3. 10. (5 marks) An offshore drilling platform has a steel hull. Engineers are choosing between zinc (E° = −0.76 V) and magnesium (E° = −2.37 V) sacrificial anodes. (a) For each metal, calculate E°cell when used as a sacrificial anode with the steel hull (E°(Fe) = −0.44 V) and confirm that both are effective. (b) Explain the electrochemical mechanism by which the sacrificial anode protects the steel from corrosion. (c) Evaluate which metal (Zn or Mg) would be a more practical choice for this marine application, considering both the level of protection and the rate of consumption of the anode. (2 + 2 + 1 marks)

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Science Jump · Galvanic Cells — Inert Electrodes & Predicting Reactions

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