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3.1 Physical Chemistry · Year 12

3.1.7 Redox

Track oxidation states, electron transfer and half-equations accurately.

Revise the specification Practise redox equations

What you need to know

Open a line for a quick recap. If it feels obvious, move straight to the linked practice.

3.1.7 Define oxidation by electron loss and recognise an oxidising agent as an electron acceptor. Quick revision

Oxidation is electron loss. An oxidising agent accepts those electrons and is therefore reduced itself. I find it safest to identify the electron transfer before attaching the word “agent”.

Oxidation state gives the same diagnosis: the species being oxidised increases in oxidation state, while the oxidising agent contains an element whose oxidation state decreases.

Watch forThe oxidising agent is reduced; the reducing agent is oxidised. The “agent” name describes its effect on the other reactant.
Practise redox equations →
3.1.7 Define reduction by electron gain and recognise a reducing agent as an electron donor. Quick revision

For reduction, keep electron gain at the centre of the definition. A reducing agent supplies those electrons and is oxidised itself, so its name tells you what it does to the other species.

A decrease in oxidation state identifies reduction. In any complete redox process, electrons lost in oxidation must equal electrons gained in reduction.

Practise redox equations →
3.1.7 Apply oxidation-state rules consistently to elements, compounds and ions. Quick revision

Apply oxidation-state rules consistently: an element in its standard elemental form is 0, a simple ion has the value of its charge, O is usually −2 and H usually +1. The total must equal the overall species charge.

Write the sum as a short equation when the value is unfamiliar. That makes oxidation-state work reliable bookkeeping and avoids guesswork.

Watch forThe sum of oxidation states must equal the overall charge of the species, not always zero.
3.1.7 Find the oxidation state of an element from a species formula and overall charge. Quick revision

Assign oxidation states systematically: free elements are 0, simple ions equal their charge, and the total must add up to the overall charge. Use the usual O = −2 and H = +1 rules unless you have a known exception.

For MnO₄⁻, for example, Mn + 4(−2) = −1, so Mn = +7. Write the algebra if the species is unfamiliar; it is quicker than guessing.

Worked example

Find the oxidation state of Mn in MnO₄⁻.

  1. Let Mn = x.
  2. x + 4(−2) = −1
  3. x − 8 = −1

Answer Mn has oxidation state +7.

Practise redox equations →
3.1.7 Write oxidation and reduction half-equations with electrons balanced correctly. Quick revision

Oxidation is an increase in oxidation state and reduction is a decrease. You can also track electron loss and gain: oxidation loses electrons, reduction gains them.

Balance atoms and charge, using electrons to balance charge. In acidic aqueous systems H₂O and H⁺ may be needed; the final half-equation must conserve both atoms and total charge.

Zn(s) → Zn²⁺(aq) + 2e⁻
Cu²⁺(aq) + 2e⁻ → Cu(s)
Watch forA half-equation must balance both atoms and charge. Use electrons to balance charge.
Practise redox equations →
3.1.7 Combine compatible half-equations into an overall redox equation. Quick revision

Write the two half-equations first, then multiply them so the number of electrons lost equals the number gained. Add the equations and cancel the electrons and any identical species on both sides.

The final redox equation must conserve atoms and total charge. If either fails, go back to the half-equations and rebalance them before combining them again.

Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Worked example

Combine Fe²⁺ → Fe³⁺ + e⁻ with Cl₂ + 2e⁻ → 2Cl⁻.

  1. multiply the iron half-equation by 2
  2. 2Fe²⁺ → 2Fe³⁺ + 2e⁻
  3. add and cancel 2e⁻

Answer 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻.

Watch forMultiply whole half-equations so the electrons cancel; do not alter species formulae to force cancellation.
Practise redox equations →