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Leave an iron nail in damp air and it rusts. Leave a galvanised bucket in the same air and it lasts for years. The difference is a table of standard electrode potentials you already have in your textbook. This post explains rusting as a galvanic cell, works through two calculations, and lists the mistakes that cost marks in board and JEE answers.
Is corrosion in your syllabus?
Corrosion is section 2.8 of the NCERT Class 12 Chemistry Part I electrochemistry unit. The CBSE Class 12 Chemistry syllabus for 2025–26 ends its electrochemistry unit with "fuel cells, corrosion", and the JEE (Advanced) 2026 information brochure lists "Corrosion" by name. The JEE (Main) 2026 syllabus names cells and fuel cells but not corrosion, so treat corrosion there as an application of electrode potentials rather than a separate topic. Check the current year's documents for your exam.
It is also an expensive problem. The NACE IMPACT study estimated the global cost of corrosion at US$2.5 trillion, equivalent to 3.4% of global GDP in 2013, and said existing corrosion control practices could save between 15% and 35% of that.
Why is rusting a galvanic cell?
NCERT describes rusting as two reactions at two different places on the same piece of iron.
At one spot, iron is oxidised. This spot is the anode: 2Fe → 2Fe²⁺ + 4e⁻, with E°(Fe²⁺/Fe) = −0.44 V.
The electrons travel through the metal to another spot, where oxygen is reduced in the presence of H⁺. This spot is the cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O, with E° = 1.23 V.
Overall: 2Fe + O₂ + 4H⁺ → 2Fe²⁺ + 2H₂O, with E°cell = 1.67 V.
NCERT says the H⁺ is believed to come from carbonic acid formed when carbon dioxide from the air dissolves in water, or from other acidic oxides. The Fe²⁺ ions are then oxidised further by air to Fe³⁺ and appear as rust, hydrated ferric oxide, Fe₂O₃·xH₂O.
Two things follow. Rusting needs both water and oxygen. And the place where iron dissolves does not have to be the place where oxygen is reduced, because the metal itself is the wire.
Worked example 1: how spontaneous is rusting?
Use ΔG° = −nFE°cell for the overall reaction as written.
Four electrons are transferred (two Fe atoms each lose two), so n = 4.
ΔG° = −4 × 96,485 C/mol × 1.67 V = −644,500 J.
So ΔG° ≈ −645 kJ for every 2 mol of iron oxidised.
A large negative ΔG° means the reaction is strongly favoured. Rust does not appear overnight because spontaneity says nothing about rate. Our post on cell EMF and the Nernst equation shows how concentration changes the actual cell potential.
How do coatings and sacrificial anodes protect iron?
NCERT lists three approaches: keep air and water off the surface (paint or chemicals such as bisphenol), cover the surface with another metal such as tin or zinc, or attach a sacrificial electrode of a more reactive metal such as magnesium or zinc, which corrodes in place of the iron.
The electrode potentials in the NCERT table explain why zinc and tin behave differently once the coating is scratched:
Zn²⁺/Zn: −0.76 V. Zinc is more easily oxidised than iron (−0.44 V), so at a scratch zinc becomes the anode and the iron stays protected.
Sn²⁺/Sn: −0.14 V. Tin is less easily oxidised than iron, so at a scratch iron becomes the anode and corrodes.
Mg²⁺/Mg: −2.36 V. Magnesium is even more easily oxidised, which is why it works as a sacrificial anode.
Worked example 2: how much zinc does a sacrificial anode use?
Suppose a zinc block must supply a steady protective current of 0.50 A for one year (365.25 days). Assume all the current comes from zinc dissolving.
Time: 365.25 × 24 × 3600 = 3.156 × 10⁷ s.
Charge: Q = It = 0.50 × 3.156 × 10⁷ = 1.578 × 10⁷ C.
Moles of electrons: 1.578 × 10⁷ / 96,485 = 163.5 mol.
Zn → Zn²⁺ + 2e⁻, so moles of zinc = 163.5 / 2 = 81.8 mol.
Mass = 81.8 × 65.38 g/mol = 5,350 g, about 5.35 kg.
Repeat with magnesium (24.31 g/mol, also two electrons) and you get about 2.0 kg for the same charge. Real anodes are not perfectly efficient, so actual use is higher. For more practice with this kind of calculation, see our post on electrolysis and Faraday's laws.
What mistakes do students make?
In interviews with final-year school chemistry students, Garnett and Treagust (1992) found that the sign of the anode and cathode was a common source of misconceptions. Students who thought the anode was negatively charged expected cations to move towards it. In corrosion questions this shows up as confusion about where the iron dissolves.
Anode means "where oxidation happens". On rusting iron, the anode is the spot where Fe → Fe²⁺. Do not decide by sign.
Tin protects like zinc. Only while the coating is unbroken. Once scratched, tin speeds up rusting of the iron below.
Rust is Fe₂O₃. Write it as hydrated ferric oxide, Fe₂O₃·xH₂O.
Wrong n. For the overall equation with 2Fe, n = 4, not 2.
If electrode potentials still feel like a list to memorise, the Fermi Concept Library covers every concept in chemistry with its own lesson, so you can go back to the electrochemical series before returning to corrosion.
A practice routine
Write the anode, cathode and overall reactions for rusting from memory, with E° values.
For zinc, tin, magnesium and copper coatings, use the NCERT table to say what happens at a scratch, in one line each.
Do two sacrificial-anode calculations with different currents and metals.
Explain, in three sentences, why salt water and acid rain make rusting worse, using the cell picture.
Next, compare this with the cells built on purpose in our post on batteries and fuel cells.
