Zinc blocks bolted to the hulls of ships and offshore pipelines corrode slowly in seawater so that the steel structure around them doesn’t — this is cathodic protection, applied galvanic cell chemistry saving billions in infrastructure every year.
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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.
Why would attaching a different metal to the pipe make the pipe corrode less, not more? You might expect that adding a reactive metal like zinc would make corrosion worse overall. Write your prediction before reading on — use what you know from the activity series and galvanic cells.
Type your prediction below — you will revisit it at the end of the lesson.
Write your prediction in your book. You will revisit it at the end.
📚 Content
In L09, both half-cells involved a metal electrode dissolving or being deposited — an active electrode. However, some half-reactions involve species that are all in aqueous solution or gaseous form and cannot form a solid electrode themselves.
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 |
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:
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.
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 |
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.
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.
📐 Worked Examples
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).
Anode (oxidation, inert Pt): Fe²⁺(aq) → Fe³⁺(aq) + e⁻
Cathode (reduction, Ag): Ag⁺(aq) + e⁻ → Ag(s)
Electrons equal (1e⁻) — combine directly:
Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)
Charge check: left = +2+1 = +3. Right = +3+0 = +3. ✓
E°cell = E°cathode − E°anode = +0.80 − (+0.77) = +0.03 V
E°cell > 0 → spontaneous (just barely).
Platinum anode: no change in mass (inert; Fe²⁺ is oxidised to Fe³⁺ in solution — no solid deposited or dissolved at the electrode itself).
Silver cathode: increases in mass as Ag⁺ ions are reduced and deposit as solid silver.
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).
E°cell = E°cathode − E°anode = −0.44 − (−0.76) = −0.44 + 0.76 = +0.32 V
E°cell > 0 → the zinc-iron galvanic cell is spontaneous in seawater.
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 functioning as a cathode. The zinc corrodes instead — it is “sacrificed” to maintain the iron as the cathode.
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. Zinc corrodes preferentially, protecting the steel structure.
✏️ Activities
For each galvanic cell below, (i) state whether the described electrode should be active or inert, and (ii) name an appropriate electrode material.
Type your active/inert classification and electrode material for each.
Write your answers in your workbook.
An underground gas pipeline is made of steel (iron, E° = −0.44 V). Engineers are considering attaching sacrificial anodes from either zinc (E° = −0.76 V) or magnesium (E° = −2.37 V).
Question A: Calculate E°cell for both the zinc-iron and magnesium-iron galvanic cells. Show working.
Question B: Both cells are spontaneous and will protect the pipeline. Explain why engineers might prefer zinc over magnesium as the sacrificial anode, despite both being effective. Consider the E°cell values and the rate of corrosion of the anode.
Type your calculations and explanation.
Show calculations and write your explanation in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: In a galvanic cell, electrons flow through the salt bridge.
Right: 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.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
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)
Type your explanation and examples.
Write your response in your book.
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)
Type your E°cell calculations and predictions.
Show calculations in your book.
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)
Type your full extended response.
Write your full response in your book.
Earlier you were asked: Can you explain precisely why attaching zinc reduces iron corrosion — using E° values and the concept of anode/cathode?
The key insight: zinc has a more negative standard reduction potential (E° = −0.76 V) than iron (E° = −0.44 V). This means zinc is more easily oxidised. When Zn and Fe are in electrical contact in the presence of an electrolyte, zinc becomes the anode (oxidises: Zn → Zn²¹ + 2e−) and iron is forced to become the cathode. Because iron is the cathode, it cannot be oxidised — its reduction reaction occurs instead. The zinc sacrifices itself to protect the iron, which is why it is called a sacrificial anode.
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.
1. Active. Electrode material: zinc (the solid zinc participates directly — Zn(s) → Zn²⁺ + 2e⁻).
2. Inert. Electrode material: platinum. Both Fe²⁺ and Fe³⁺ are in aqueous solution; no solid metal forms or dissolves at the electrode.
3. Inert. Electrode material: platinum. Cl₂ is a gas and Cl⁻ is aqueous; no solid electrode material is involved in the half-reaction.
4. Active. Electrode material: silver. The Ag⁺/Ag couple involves solid Ag depositing or dissolving at the electrode surface.
A. Zn-Fe: E°cell = −0.44 − (−0.76) = +0.32 V (spontaneous). Mg-Fe: E°cell = −0.44 − (−2.37) = +1.93 V (spontaneous). Both are effective — both have E°cell > 0.
B. Zinc is often preferred over magnesium despite magnesium providing a higher cell voltage because: (i) the higher E°cell for Mg-Fe means the galvanic cell operates with a larger driving force — magnesium corrodes much more rapidly; (ii) zinc anodes last longer between replacements, reducing maintenance costs and intervals; (iii) zinc is cheaper and more widely available for marine applications; (iv) in seawater conditions, the rate of zinc corrosion is well-characterised and predictable, making maintenance scheduling easier. A higher E°cell means faster corrosion of the sacrificial anode, not necessarily better long-term protection.
Q1 C: Platinum is inert — it provides a conducting surface but does not participate in the Cl₂/Cl⁻ half-reaction. Since both Cl₂ and Cl⁻ are gas/aqueous, no solid electrode material is consumed or produced.
Q2 C: Mg = anode (more negative E°); Fe = cathode (more positive E°). E°cell = −0.44 − (−2.37) = +1.93 V. Options A and B use wrong arithmetic. Option D incorrectly adds the absolute values.
Q3 C: Zinc has E° = −0.76 V < E°(Fe) = −0.44 V. Zinc is the anode (preferentially oxidised). Iron is the cathode (protected). Option A describes barrier protection, not cathodic protection. Option B incorrectly states zinc has more positive E°.
Q4 B: For a metal to dissolve in dilute acid, E°cell = 0.00 − E°(M) > 0, i.e. E°(M) must be negative. Cu has E° = +0.34 V > 0 → E°cell = 0.00 − 0.34 = −0.34 V < 0 → non-spontaneous → Cu does NOT dissolve.
Q5 A: Fe³⁺/Fe²⁺ (E° = +0.77 V, less positive) is the anode (inert Pt). Ag⁺/Ag (E° = +0.80 V, more positive) is the cathode. E°cell = +0.80 − +0.77 = +0.03 V. Pt does not change mass; Ag electrode gains mass.
Q6 C: For cathodic protection, the sacrificial anode must have a MORE NEGATIVE E° than the metal being protected (iron, −0.44 V). Tin E° = −0.14 V, which is MORE POSITIVE than iron. Therefore iron would be the anode and would corrode. Tin is unsuitable as a sacrificial anode for iron.
Q7 B: E°cell = E°cathode − E°anode. With Cu as cathode (+0.34 V): +0.46 = +0.34 − E°(M) → E°(M) = +0.34 − 0.46 = −0.12 V. Since E°(M) = −0.12 V < 0.00 V, M would dissolve in dilute acid.
Q8: Active electrode: the electrode material participates directly in the half-reaction, either dissolving (anode) or having metal deposited onto it (cathode); its mass changes during operation. Example: Cu(s)/Cu²⁺ couple — Cu(s) → Cu²⁺(aq) + 2e⁻ (copper electrode dissolves). Inert electrode: the electrode material does not participate in the half-reaction; it only provides a conducting surface for electron transfer; its mass does not change. Example: Fe²⁺/Fe³⁺ couple — Fe²⁺(aq) → Fe³⁺(aq) + e⁻ (both species in solution; platinum electrode required).
Q9: (a) Zn: E°cell = 0.00 − (−0.76) = +0.76 V > 0 → spontaneous → zinc dissolves in H₂SO₄. (b) Cu: E°cell = 0.00 − (+0.34) = −0.34 V < 0 → non-spontaneous → copper does NOT dissolve in dilute H₂SO₄. (c) Al: E°cell = 0.00 − (−1.66) = +1.66 V > 0 → spontaneous based on E° → thermodynamically, aluminium should dissolve in dilute H₂SO₄. However, in practice, Al forms a protective Al₂O₃ passivation layer that resists further reaction at moderate acid concentrations. With concentrated acid or if the oxide layer is disrupted, reaction can occur. This is an important practical caveat to the E° prediction.
Q10: (a) Zn-Fe: E°cell = −0.44 − (−0.76) = +0.32 V > 0 → effective. Mg-Fe: E°cell = −0.44 − (−2.37) = +1.93 V > 0 → effective. Both confirm spontaneous galvanic cells that protect the steel. (b) Mechanism: when the sacrificial anode (e.g. Zn) is electrically connected to the steel hull (Fe) in seawater (electrolyte), a galvanic cell is formed. Zinc has a more negative standard reduction potential than iron, so zinc is preferentially oxidised at the anode: Zn(s) → Zn²⁺(aq) + 2e⁻. Iron functions as the cathode — where reduction occurs. Since iron is at the cathode (gaining electrons), it cannot be oxidised (corroded). The zinc corrodes instead, protecting the steel. (c) Zinc is generally more practical for marine applications. While magnesium (E°cell = +1.93 V) provides stronger cathodic protection, the much higher driving force means magnesium corrodes very rapidly, requiring more frequent replacement. Zinc (E°cell = +0.32 V) provides adequate protection while corroding at a slower, more economical rate. Zinc is also denser and mechanically robust, suitable for bolting onto a ship hull. Magnesium might be preferred in highly aggressive environments or for smaller structures where its higher driving force is needed to protect a large steel surface area from a small anode block.
Put your knowledge of Galvanic Cells — Inert Electrodes & Predicting Reactions to the test. Answer correctly to deal damage — get it wrong and the boss hits back. Pool: lessons 1–10.