Batteries, Fuel Cells and Corrosion

How do batteries and fuel cells work, why does iron rust, and how can rusting be stopped?

IntermediateRedox & ElectrochemistryLast reviewed 6 October 2026

What is it?

A battery is one or more galvanic cells packaged so that a spontaneous redox reaction pushes electrons through a circuit. A fuel cell is a galvanic cell that is fed its reactants continuously. Corrosion is a galvanic cell nobody wanted: metal slowly oxidized by its surroundings.

DeviceAnode (oxidation, −)Cathode (reduction, +)Voltage per cell
Alkaline cellzincmanganese(IV) oxideabout 1.5 V
Lead–acid (car) batteryleadlead(IV) oxide, in sulfuric acidabout 2.0 V (six cells: 12 V)
Lithium-ionlithium in graphitelithium metal oxideabout 3.7 V
Hydrogen fuel cellhydrogenoxygen1.23 V (standard)

Key idea

Every battery, fuel cell and rusting nail is the same thing: an anode where something is oxidized, a cathode where something is reduced, an electrolyte that lets ions move, and a path for electrons. Change the chemicals and you change the voltage.

Why does it matter?

  • Energy storage. Phones, laptops and electric cars run on lithium-ion batteries; grid storage helps renewable power.
  • Clean energy. Hydrogen fuel cells produce only water as their exhaust.
  • Cost of corrosion. Rust weakens bridges, pipelines and ships. Preventing it is a huge part of engineering.

How does it work?

1. Primary and rechargeable batteries

A primary battery (such as an alkaline cell) is used once: its reaction cannot easily be reversed. A secondary (rechargeable) battery is recharged by forcing current through it backwards, which drives the reaction in reverse: electrolysis. In a lithium-ion cell, LiX+\ce{Li+} ions shuttle between the graphite anode and the metal-oxide cathode, in one direction during discharge and in the other during charging.

2. The hydrogen fuel cell

anode: 2 HX2→4 HX++4 eX−cathode: OX2+4 HX++4 eX−→2 HX2Ooverall: 2 HX2+OX2→2 HX2O\begin{aligned} &\text{anode: } \ce{2H2 -> 4H+ + 4e-} \\[4pt] &\text{cathode: } \ce{O2 + 4H+ + 4e- -> 2H2O} \\[4pt] &\text{overall: } \ce{2H2 + O2 -> 2H2O} \end{aligned}

E°cell=1.23 V−0.00 V=1.23 VE°_\text{cell} = 1.23\ \text{V} - 0.00\ \text{V} = 1.23\ \text{V}. Unlike a battery, a fuel cell does not run down as long as hydrogen and oxygen are supplied.

3. How iron rusts

Rusting needs both oxygen and water. A drop of water on iron acts as a tiny galvanic cell:

  • Anode (where the iron is pitted): Fe→FeX2++2 eX−\ce{Fe -> Fe^2+ + 2e-}
  • Cathode (at the edge of the drop, where oxygen dissolves): OX2+4 HX++4 eX−→2 HX2O\ce{O2 + 4H+ + 4e- -> 2H2O} (or, in neutral water, OX2+2 HX2O+4 eX−→4 OHX−\ce{O2 + 2H2O + 4e- -> 4OH-})

The FeX2+\ce{Fe^2+} is then oxidized further by air to hydrated iron(III) oxide, FeX2OX3 ⋅ x HX2O\ce{Fe2O3.xH2O}: rust. Salt speeds rusting because it makes the water a better electrolyte.

4. Preventing corrosion

  • Barriers: paint, oil, grease or plastic coating keep out water and oxygen.
  • Galvanizing: coating iron with zinc. Zinc (E°=−0.76E° = -0.76 V) is more easily oxidized than iron (−0.44-0.44 V), so even if the coating is scratched, zinc corrodes instead of the iron.
  • Sacrificial anodes: blocks of magnesium or zinc are bolted to ship hulls and buried pipelines and slowly corrode in place of the steel.
  • Alloys: stainless steel contains chromium, which forms a protective oxide layer.

Think of it like this

A sacrificial anode is a bodyguard for the iron. Because magnesium or zinc gives up electrons more readily, the corrosion “attacks” the bodyguard first, and the iron is left untouched until the bodyguard is used up and replaced.

More precisely

Coating iron with a less reactive metal, such as tin (tin cans) or copper, protects it only while the coating is unbroken; once scratched, the iron becomes the anode and corrodes faster. Aluminium does not rust away because it forms a thin, tough layer of AlX2OX3\ce{Al2O3} that seals the surface (passivation). Lithium-ion batteries can fail dangerously if overcharged or damaged, because the electrolyte is flammable.

Visualise it

A water droplet on an iron surface. In the centre of the drop, where oxygen is scarce, iron atoms lose electrons and become Fe2+ ions: this is the anode. The electrons travel through the iron to the edge of the drop, where oxygen from the air is reduced: this is the cathode. Fe2+ ions and hydroxide ions meet in the water and form rust.
Rusting is a tiny galvanic cell: iron is oxidized under the drop; oxygen is reduced at its edge.

Worked example

Worked example: The energy of a fuel cell

Question: Calculate ΔG° for 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O} in a fuel cell (E°=1.23E° = 1.23 V, n=4n = 4), and per mole of hydrogen.

ΔG°=−nFE°=−4×96 485 Cmol×1.23 V=−4.75×105 J=−475 kJ\begin{aligned} &\Delta G° = -nFE° \\[4pt] &= -4 \times 96\,485\ \tfrac{\text{C}}{\text{mol}} \times 1.23\ \text{V} \\[4pt] &= -4.75 \times 10^{5}\ \text{J} = -475\ \text{kJ} \end{aligned}

That is for 2 mol of HX2\ce{H2}, so −237-237 kJ per mole of hydrogen.

Worked example: The cell behind rusting

Question: Calculate E°E° for the rusting cell 2 Fe+OX2+4 HX+→2 FeX2++2 HX2O\ce{2Fe + O2 + 4H+ -> 2Fe^2+ + 2H2O}.

E°cell=E°cathode−E°anode=1.23 V−(−0.44 V)=1.67 V\small\begin{aligned} &E°_\text{cell} \\[4pt] &= E°_\text{cathode} - E°_\text{anode} \\[4pt] &= 1.23\ \text{V} - (-0.44\ \text{V}) \\[4pt] &= 1.67\ \text{V} \end{aligned}

A large positive value: rusting is strongly spontaneous whenever oxygen and water are present.

Worked example: Which metal protects iron?

Question: Which of zinc, magnesium and copper could act as a sacrificial anode for iron?

Only a metal more easily oxidized than iron (more negative E°E° than −0.44 V) protects it: zinc (−0.76 V) and magnesium (−2.37 V). Copper (+0.34 V) would make the iron corrode faster.

Worked example: Battery capacity

Question: A phone battery is rated at 2000 mAh and about 3.7 V. Calculate (a) the charge it delivers (b) the moles of electrons (c) the mass of lithium that must move, if each LiX+\ce{Li+} carries one electron’s worth of charge (Li: 6.94 g/mol) (d) the energy stored.

  1. (a) 2.000 A h×3600 s/h=7200 C2.000\ \text{A h} \times 3600\ \text{s/h} = 7200\ \text{C} (1 A s = 1 C)
  2. (b) 7200 C96 485 C/mol=0.07462 mol e−\dfrac{7200\ \text{C}}{96\,485\ \text{C/mol}} = 0.07462\ \text{mol e}^-
  3. (c) 0.07462 mol×6.94 g/mol=0.518 g Li0.07462\ \text{mol} \times 6.94\ \text{g/mol} = 0.518\ \text{g Li}
  4. (d) E=QV=7200 C×3.7 V=2.7×104 JE = QV = 7200\ \text{C} \times 3.7\ \text{V} = 2.7 \times 10^{4}\ \text{J}, about 27 kJ.

Common mistake

Common mistake: Thinking iron rusts in dry air or in pure, air-free water

Rusting needs both oxygen and water. Iron stays bright in dry air and in boiled water sealed from air.

Common mistake: Coating iron with any metal

Only a more reactive metal (zinc, magnesium) gives sacrificial protection. A less reactive coating (tin, copper) speeds corrosion once it is scratched.

Common mistake: Confusing a fuel cell with a battery

A battery stores its reactants inside and runs down. A fuel cell is supplied with fuel continuously and keeps working while fuel flows.

Notation note

  • In a battery the anode is the negative terminal and the cathode the positive terminal (the opposite of electrolysis).
  • mAh (milliampere-hours) measures charge: 1 mAh = 3.6 C.

Remember this

Remember this

  • Batteries and fuel cells are galvanic cells: oxidation at the anode (−), reduction at the cathode (+).
  • Rechargeable batteries are recharged by electrolysis (reaction driven backwards).
  • Hydrogen fuel cell: 2 HX2+OX2→2 HX2O\ce{2H2 + O2 -> 2H2O}, E°=1.23E° = 1.23 V, only water as exhaust.
  • Rusting needs oxygen and water; salt speeds it up. E°cell=1.67E°_\text{cell} = 1.67 V.
  • Protection: barriers, galvanizing and sacrificial anodes (metals with more negative E° than iron).

Test yourself

Check your understanding before moving on.

Flashcards

Batteries, Fuel Cells and Corrosion: Flashcards

10 cards

  1. Question
    What is a battery, in electrochemical terms?
    Answer

    One or more galvanic cells: a spontaneous redox reaction drives electrons through a circuit.

  2. Question
    What is the difference between a primary and a secondary battery?
    Answer

    Primary: used once. Secondary: rechargeable, by driving the reaction backwards with an external current (electrolysis).

  3. Question
    In a battery, which electrode is negative?
    Answer

    The anode (oxidation); the cathode is positive.

  4. Question
    What moves between the electrodes in a lithium-ion battery?
    Answer

    Li⁺ ions, between the graphite anode and the metal-oxide cathode.

  5. Question
    Give the overall reaction and E° of a hydrogen fuel cell.
    Answer

    2H₂ + O₂ → 2H₂O; E° = 1.23 V.

  6. Question
    What two substances are needed for iron to rust?
    Answer

    Oxygen and water.

  7. Question
    Give the anode half-equation for rusting.
    Answer

    Fe→FeX2++2 eX−\ce{Fe -> Fe^2+ + 2e-}

  8. Question
    Why does salt speed up rusting?
    Answer

    It makes the water a better electrolyte, so ions move more easily.

  9. Question
    Why does zinc protect iron even when the coating is scratched?
    Answer

    Zinc (E° −0.76 V) is more easily oxidized than iron (−0.44 V), so it corrodes instead (sacrificial protection).

  10. Question
    How many coulombs is 2000 mAh?
    Answer

    2.000 A × 3600 s = 7200 C

Quiz

Batteries, Fuel Cells and Corrosion: Quiz

7 questions

  1. Question 1EasyWhat happens when a rechargeable battery is charged?
    Show answer

    Answer: An external current drives the cell reaction in reverse

    Charging is electrolysis: electrical energy forces the non-spontaneous reverse reaction, restoring the reactants.

  2. Question 2EasyWhat is the only product of a hydrogen–oxygen fuel cell?
    Show answer

    Answer: Water

    Overall reaction: 2H₂ + O₂ → 2H₂O.

  3. Question 3EasyIn which conditions does an iron nail rust fastest?
    Show answer

    Answer: Salt water open to air

    Rusting needs oxygen and water; dissolved salt makes the water a better electrolyte and speeds the reaction.

  4. Question 4MediumWhich metal could be used as a sacrificial anode to protect a steel pipeline?
    Show answer

    Answer: Magnesium

    Magnesium (E° −2.37 V) is oxidized more easily than iron (−0.44 V), so it corrodes instead. Copper, silver and tin are less reactive than iron.

  5. Question 5MediumWhat is E°(cell) for rusting, 2Fe + O₂ + 4H⁺ → 2Fe²⁺ + 2H₂O? (O₂/H₂O +1.23 V; Fe²⁺/Fe −0.44 V)
    Show answer

    Answer: 1.67 V

    E°cell = E°cathode − E°anode = 1.23 V − (−0.44 V) = 1.67 V. 0.79 V comes from adding the values.

  6. Question 6HardWhat is ΔG° per mole of H₂ in a hydrogen fuel cell? (n = 4 for 2H₂ + O₂; E° = 1.23 V)
    Show answer

    Answer: −237 kJ

    ΔG° = −4 × 96 485 C/mol × 1.23 V = −475 kJ for 2 mol H₂, which is −237 kJ per mole of H₂.

  7. Question 7HardWhy does a scratched tin coating make iron rust faster?
    Show answer

    Answer: Iron becomes the anode, because iron is more reactive than tin

    Where both metals touch the electrolyte, the more reactive iron is oxidized and the tin acts as the cathode, speeding corrosion of the iron.

Notes and downloads

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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