Electrochemistry (Redox)

Physical Chemistry Weightage: AS · Papers 1, 2 and 3 (assumed in Papers 4 and 5) Topic 6

“Welcome! Redox looks like a lot of rules, but it comes down to one idea: keep track of the electrons. Once you can write oxidation numbers quickly and build a half-equation in four steps, you can balance almost anything, and the same skills carry into the titration calculations that appear in both the theory and the practical papers.”

— SCORECHEM ACADEMIC TEAM

0. What Examiners Want

How marks are earned in redox

Oxidation numbers: the sign counts. Write +3, not 3, and give the number for the element, not for the whole ion.
Explaining redox: the mark is for the change: name the element, give both oxidation numbers (or the electrons lost or gained) and say whether that is oxidation or reduction.
Equations: a balanced equation needs matching atoms and matching charge. A half-equation has electrons in it.
Calculations (Papers 2 and 3): mole ratio from the balanced equation, the concentration, and the correct final unit; ecf is allowed if the ratio is stated.
Practical (Paper 3): the end-point colour, why the acid is used, how to read the burette, and choosing the concordant titres for the mean.

1. Electron Transfer and Redox Agents

Oxidation: loss of electrons, and an increase in oxidation number.
Reduction: gain of electrons, and a decrease in oxidation number.
Redox reaction: a reaction in which oxidation and reduction happen together. The electrons lost by one species are exactly those gained by another.
Oxidising agent: takes electrons from another species, so it is itself reduced.
Reducing agent: gives electrons to another species, so it is itself oxidised.

Zn + 2Ag+ → Zn2+ + 2Ag. The reducing agent loses electrons and is oxidised (its oxidation number rises); the oxidising agent gains electrons and is reduced (its oxidation number falls). The electrons lost equal the electrons gained.

Metals tend to be reducing agents because they lose electrons to form positive ions. Non-metals such as the halogens and oxygen tend to be oxidising agents because they gain electrons to form negative ions.

⚠️ Examiner Trap: Agent versus process. The oxidising agent is the species that gets reduced. If a question asks “which species is the oxidising agent”, look for the one whose oxidation number falls, and name it (or its formula) rather than writing “the one that is oxidised”.

2. Oxidation Numbers

The oxidation number of an element in a species is the charge that its atom would have if all its bonding electrons were assigned to the more electronegative atom. Use these rules in order.

Rule	Oxidation number
Uncombined element (Fe, O₂, S₈, P₄)	0
Simple monatomic ion	the charge on the ion (Al³⁺ is +3, S²⁻ is −2)
Sum for a neutral compound	0
Sum for a polyatomic ion	the charge on the ion
Group 1 and Group 2 in compounds	+1 and +2
Fluorine in compounds	−1
Oxygen in most compounds	−2 (but −1 in peroxides such as H₂O₂ and Na₂O₂, and +2 in OF₂)
Hydrogen in most compounds	+1 (but −1 in metal hydrides such as NaH and CaH₂)
Cl, Br and I	−1 in halides, but positive when combined with oxygen or fluorine

Worked example 1. Find the oxidation number of sulfur in the tetrathionate ion, S₄O₆²⁻, and of chromium in CrO₄²⁻.

S₄O₆²⁻: 4x + 6(−2) = −2, so 4x = +10 and x = +2.5. This is an average value, because the four sulfur atoms are not all in the same environment.
CrO₄²⁻: x + 4(−2) = −2, so x = +6.

Now use the ladder to see what the numbers say about a reaction. The same element can be found in many oxidation states, and any step up is oxidation and any step down is reduction.

Nitrogen shows almost every oxidation number from +5 to −3. Moving up the ladder is oxidation and moving down is reduction. For example, nitrate(V) in dilute nitric acid is reduced to NO when it oxidises a metal: +5 → +2, a gain of 3 electrons per nitrogen atom.

⚠️ Examiner Trap: The exceptions. Oxygen is −1 in peroxides (H₂O₂) and +2 in OF₂; hydrogen is −1 in CaH₂. Halogens are only −1 when they are the more electronegative atom, so chlorine in ClO₃⁻ is not −1. Always check that your numbers add up to the overall charge.

3. Naming with Oxidation Numbers

A Roman numeral in a name gives the oxidation number of the element it follows. It is written in brackets, with no sign and no space.

Metals with variable oxidation number: iron(II) sulfate is FeSO₄; iron(III) chloride is FeCl₃; copper(I) oxide is Cu₂O.
Oxoanions: the ending is -ate and the numeral is on the central atom. Manganate(VII) is MnO₄⁻; sulfate(IV) is SO₃²⁻; nitrate(III) is NO₂⁻; iodate(V) is IO₃⁻.
Name to formula: work out the charge of the anion from the numeral, then balance the charges. Vanadium(V) oxide: V is +5 and O is −2, so the formula is V₂O₅.
Formula to name: work out the oxidation number by the rules. In NaIO₃, I = +5, so it is sodium iodate(V).

⚠️ Examiner Trap: Charge or oxidation number? The ion Fe³⁺ has a charge of 3+; iron in it has an oxidation number of +3. In a name only the oxidation number is used: iron(III), not iron(3+) and not iron(+3).

4. Half-Equations

A half-equation shows one of the two processes in a redox reaction, with the electrons written in. For a reduction the electrons are on the left; for an oxidation they are on the right.

Simple: Zn → Zn²⁺ + 2e⁻ (oxidation); Br₂ + 2e⁻ → 2Br⁻ (reduction).
In acid: balance O with H₂O, then H with H⁺, then charge with e⁻.

Building the half-equation for manganate(VII) in acid: Mn goes from +7 to +2, so 5 electrons are gained. Electrons appear on the left of a reduction and on the right of an oxidation.

Other half-equations you should know:

Species	Half-equation	Change
Fe²⁺	Fe²⁺ → Fe³⁺ + e⁻	oxidation
I⁻	2I⁻ → I₂ + 2e⁻	oxidation
Ethanedioate	C₂O₄²⁻ → 2CO₂ + 2e⁻	oxidation
H₂O₂ (as a reducing agent)	H₂O₂ → O₂ + 2H⁺ + 2e⁻	oxidation
H₂O₂ (as an oxidising agent)	H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O	reduction

Combining half-equations: multiply each so that the electrons are equal, add them, and cancel the electrons and any species that appear on both sides.

⚠️ Examiner Trap: Check the charge. A half-equation can have the right atoms and still be wrong. Add up the charges on each side: they must be equal. For MnO₄⁻ + 8H⁺ + 5e⁻, the left is −1 + 8 − 5 = +2, matching Mn²⁺.

5. Balancing Redox Equations

Method A: by oxidation-number change (quick for full equations)

Write the unbalanced equation and find the elements whose oxidation numbers change.
Find the change per atom (a rise for oxidation, a fall for reduction).
Choose coefficients so that the total rise equals the total fall.
Balance the remaining atoms (O with H₂O, H with H⁺ in acid) and check the charge.

Balancing by oxidation-number change: (1) find the atoms whose oxidation number changes; (2) write the change per atom; (3) choose the coefficients so the total increase equals the total decrease; (4) balance O, H and charge with H₂O and H⁺ (acid) and check.

Worked example 2. Dilute nitric acid oxidises copper: Cu + NO₃⁻ + H⁺ → Cu²⁺ + NO + H₂O.

Cu: 0 → +2 (rise of 2). N: +5 → +2 (fall of 3).
The lowest common multiple is 6: 3 Cu (rise of 6) and 2 N (fall of 6).
3Cu + 2NO₃⁻ → 3Cu²⁺ + 2NO. Oxygen needs 4H₂O on the right, and that needs 8H⁺ on the left.
3Cu + 2NO₃⁻ + 8H⁺ → 3Cu²⁺ + 2NO + 4H₂O. Charge: +6 = +6.

Method B: by half-equations (best when the half-equations are known)

Worked example 3. Acidified manganate(VII) with tin(II) ions, Sn²⁺ → Sn⁴⁺.

Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
Oxidation: Sn²⁺ → Sn⁴⁺ + 2e⁻.
Equalise the electrons (10): multiply the first by 2 and the second by 5.
2MnO₄⁻ + 16H⁺ + 5Sn²⁺ → 2Mn²⁺ + 8H₂O + 5Sn⁴⁺. Charge: −2 + 16 + 10 = +24; 4 + 20 = +24.

⚠️ Examiner Trap: Balance the charge too. An equation with equal atoms but unequal charges will not score the balancing mark. Add the charges on each side as your last check, and do not multiply the electrons by a different number in the two half-equations.

6. Disproportionation

Disproportionation is a redox reaction in which the same element is both oxidised and reduced. The element in the reactant has an intermediate oxidation number and forms one product at a lower and one at a higher oxidation number.

Disproportionation is the oxidation and reduction of the same element in the same reaction. Chlorine goes from 0 to −1 (reduced) and to +1 (oxidised) in cold alkali. Other examples: 2Cu⁺ → Cu²⁺ + Cu, and 2H₂O₂ → 2H₂O + O₂.

Chlorine with cold, dilute sodium hydroxide: Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O. The products are NaCl and NaClO (chlorate(I)), a bleach.
Hydrogen peroxide: 2H₂O₂ → 2H₂O + O₂. Oxygen goes from −1 to −2 and to 0.
Copper(I): 2Cu⁺ → Cu²⁺ + Cu. Copper goes from +1 to +2 and to 0, so Cu⁺ is not stable in aqueous solution.

⚠️ Examiner Trap: Same element. To earn the mark you must say that one element is both oxidised and reduced, and give the numbers. “Chlorine is oxidised and reduced” scores; “chlorine is oxidised and hydroxide is reduced” does not, because that is an ordinary redox reaction.

7. Redox Titrations

Potassium manganate(VII) is the standard oxidising agent for AS titrations. It reacts with iron(II) ions, ethanedioic acid and its salts, and hydrogen peroxide, all in acid.

Redox titration with acidified potassium manganate(VII). It is self-indicating: MnO₄⁻ is purple and Mn²⁺ is almost colourless, so the first drop in excess gives a permanent pale pink. Use dilute sulfuric acid, not hydrochloric or nitric acid.

Acid: an excess of dilute sulfuric acid is added to the flask because H⁺ is used up in the reduction of MnO₄⁻. Without enough acid, brown MnO₂ forms instead of Mn²⁺. Do not use hydrochloric acid (the chloride ions are oxidised) or nitric acid (an oxidising agent itself).
End-point: the first permanent pale pink colour (one drop of excess MnO₄⁻).
Burette: the solution is deeply coloured, so read the top of the meniscus.
Results: carry out a rough titration, then repeat until two or more titres are within 0.10 cm³, and average those.

Key ratios: MnO₄⁻ : Fe²⁺ = 1 : 5; MnO₄⁻ : C₂O₄²⁻ = 2 : 5.

Worked example 4. 25.0 cm³ of 0.0450 mol dm⁻³ ethanedioic acid, H₂C₂O₄, is acidified and titrated with potassium manganate(VII). The mean titre is 22.5 cm³. Calculate the concentration of the manganate(VII).

n(H₂C₂O₄) = 0.0450 × 25.0 ÷ 1000 = 1.125 × 10⁻³ mol.
Ratio 2MnO₄⁻ : 5C₂O₄²⁻, so n(MnO₄⁻) = 1.125 × 10⁻³ × 2/5 = 4.50 × 10⁻⁴ mol.
Concentration = 4.50 × 10⁻⁴ ÷ (22.5 ÷ 1000) = 0.0200 mol dm⁻³.

⚠️ Examiner Trap: The ratio and the dilution. The equation ratio is not 1 : 1. Write the ratio, use it, and then scale up if only a portion of a larger volume was titrated (for example, 25.0 cm³ from 250.0 cm³ is a factor of 10). Missing either step costs the last two marks.

8. Quick Sheet and Checklist

Idea	Say or write
Oxidation	loss of electrons; oxidation number increases
Reduction	gain of electrons; oxidation number decreases
Oxidising agent	takes electrons and is reduced
Reducing agent	gives electrons and is oxidised
Disproportionation	one element is both oxidised and reduced
MnO₄⁻ / Fe²⁺	1 : 5
MnO₄⁻ / C₂O₄²⁻	2 : 5
End-point colour	first permanent pale pink

Before you leave the question, check:

Every oxidation number has a sign, and the numbers add up to the overall charge.
The exceptions (peroxide, hydride, OF₂) have been considered.
The equation is balanced for atoms and charge.
The oxidising and reducing agents are named as species, not as processes.
For titrations: ratio written, dilution factor applied, and the answer given with a unit and sensible significant figures.