Redox processes
Section: 6 Electrochemistry | Syllabus: Cambridge AS Level Chemistry (9701)
Oxidation and Reduction - Electron Transfer
Redox reactions involve the simultaneous transfer of electrons between species. A useful memory aid is OIL RIG:
OIL RIGOxidation Is Loss (of electrons) | Reduction Is Gain (of electrons)
Oxidation Loss of electrons by a species, or an increase in oxidation number. Reduction Gain of electrons by a species, or a decrease in oxidation number. Redox Reaction A reaction in which both oxidation and reduction occur simultaneously - electrons are transferred from one species to another. Example: Zinc displacing copperZn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Zn → Zn²⁺ + 2e⁻ (oxidation - loses electrons)
Cu²⁺ + 2e⁻ → Cu (reduction - gains electrons)
Oxidising Agents and Reducing Agents
Oxidising Agent A substance that oxidises another species by accepting electrons from it. The oxidising agent is itself reduced in the process. Reducing Agent A substance that reduces another species by donating electrons to it. The reducing agent is itself oxidised in the process.| Species | What it does to electrons | What happens to it | Change in oxidation number |
|---|---|---|---|
| Oxidising agent | Accepts electrons | Is reduced | Decreases |
| Reducing agent | Donates electrons | Is oxidised | Increases |
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
Zn is the reducing agent - it donates electrons and is oxidised (0 → +2).
Cu²⁺ is the oxidising agent - it accepts electrons and is reduced (+2 → 0).
Oxidation Numbers (Oxidation States)
Oxidation numbers are a bookkeeping tool to track electron transfer in redox reactions. They are the charge an atom would have if the compound were fully ionic.
Rules for Assigning Oxidation Numbers
- Uncombined elements have oxidation number 0 (e.g. Na, O₂, Cl₂, Fe).
- Simple monatomic ions have oxidation number equal to their charge (e.g. Na⁺ = +1, Cl⁻ = −1, Fe³⁺ = +3).
- In compounds, oxygen is usually −2 (exception: in peroxides, e.g. H₂O₂, O = −1; in OF₂, O = +2).
- In compounds, hydrogen is usually +1 (exception: in metal hydrides, e.g. NaH, H = −1).
- Fluorine is always −1 (most electronegative element).
- The sum of oxidation numbers in a neutral compound = 0.
- The sum of oxidation numbers in a polyatomic ion = the charge on the ion.
H₂SO₄ is neutral: sum = 0
2(+1) + S + 4(−2) = 0
2 + S − 8 = 0 → S = +6
Oxidation number of Cr in Cr₂O₇²⁻Sum of oxidation numbers = −2 (charge on ion)
2(Cr) + 7(−2) = −2
2Cr − 14 = −2 → 2Cr = 12 → Cr = +6
Oxidation number of Mn in MnO₄⁻Mn + 4(−2) = −1
Mn − 8 = −1 → Mn = +7
Figure 6.1: Oxidation Number Rules - Summary Card
(Table/card format listing all seven rules with examples: element = 0; monatomic ion = charge; O = −2 (−1 in peroxides); H = +1 (−1 in hydrides); F = −1; compound sum = 0; ion sum = charge. Include one example for each rule.)
Roman Numerals for Oxidation Numbers
When a metal can have more than one oxidation state, a Roman numeral in brackets after the element name specifies the oxidation state in that particular compound.
| Name | Roman numeral | Ion | Example compound |
|---|---|---|---|
| Iron(II) | Fe(II) | Fe²⁺ | FeCl₂ (iron(II) chloride) |
| Iron(III) | Fe(III) | Fe³⁺ | FeCl₃ (iron(III) chloride) |
| Copper(I) | Cu(I) | Cu⁺ | Cu₂O (copper(I) oxide) |
| Copper(II) | Cu(II) | Cu²⁺ | CuO (copper(II) oxide) |
| Manganese(VII) | Mn(VII) | Mn in MnO₄⁻ | KMnO₄ (potassium manganate(VII)) |
| Chromium(VI) | Cr(VI) | Cr in Cr₂O₇²⁻ | K₂Cr₂O₇ (potassium dichromate(VI)) |
The Roman numeral gives the oxidation number of the metal only - not the overall charge of the compound. Cu₂O: each Cu is +1 (copper(I)), even though the compound is neutral overall (2 × +1 balances 1 × −2 from O).
Identifying Redox Reactions Using Oxidation Numbers
A reaction is a redox reaction if one or more elements change oxidation number. Non-redox reactions show no change in oxidation numbers.
Is this a redox reaction? 2Mg + O₂ → 2MgOMg: 0 → +2 (increase → oxidation) | O: 0 → −2 (decrease → reduction)
Yes - redox reaction. Mg is oxidised; O₂ is reduced.
Is this a redox reaction? NaOH + HCl → NaCl + H₂ONa: +1 → +1, O: −2 → −2, H: +1 → +1, Cl: −1 → −1
No change in any oxidation numbers → not a redox reaction (acid–base neutralisation).
Figure 6.2: Tracking Oxidation Number Changes in a Redox Equation
(Balanced equation: 2Mg + O₂ → 2MgO. Above the equation, show oxidation numbers for each atom: Mg(0), O(0) on left; Mg(+2), O(−2) on right. Draw curved arrows above Mg showing "+2" increase (oxidation) and below O showing "−2" decrease (reduction). Label "oxidised" and "reduced" with brackets.)
Using Oxidation Numbers to Balance Equations
Changes in oxidation numbers must balance - total increase in oxidation number must equal total decrease. This principle helps balance redox equations.
- Step 1: Assign oxidation numbers to all atoms.
- Step 2: Identify which atoms are oxidised and which are reduced.
- Step 3: Calculate the change in oxidation number for each.
- Step 4: Balance the changes - multiply species so that total increase = total decrease.
- Step 5: Balance remaining atoms (O, H, charge) using H₂O and H⁺ (in acid solution) or OH⁻ (in alkaline solution).
Fe²⁺ → Fe³⁺ (change = +1 per Fe - oxidation)
Mn⁷⁺ → Mn²⁺ (change = −5 per Mn - reduction)
To balance: need 5 Fe²⁺ per MnO₄⁻
5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O
Check: charge left = 5(2+) + (1−) + 8(1+) = 10 − 1 + 8 = 17+ | charge right = 5(3+) + 2+ = 17+ ✓
Disproportionation
Disproportionation A redox reaction in which the same element is simultaneously oxidised and reduced - the element exists in a single oxidation state in the reactant but is converted to two different oxidation states in the products. Example 1: Disproportionation of chlorine in waterCl₂(aq) + H₂O(l) ⇌ HCl(aq) + HClO(aq)
Cl in Cl₂ = 0 | Cl in HCl = −1 (reduced) | Cl in HClO = +1 (oxidised)
The same Cl₂ is both oxidised (+1) and reduced (−1) → disproportionation.
Example 2: Disproportionation of copper(I)2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
Cu⁺ = +1 | Cu²⁺ = +2 (oxidised) | Cu = 0 (reduced)
Cu⁺ is simultaneously oxidised and reduced → disproportionation.
Figure 6.3: Disproportionation of Cl₂
(Equation: Cl₂ → HCl + HClO. Above Cl₂ label oxidation state = 0. Arrow branching to two products: left arrow to HCl labelled "Cl = −1, reduced"; right arrow to HClO labelled "Cl = +1, oxidised". Curved arrows show the simultaneous decrease and increase from the same Cl₂ starting material.)
Interactive revision notes, videos and practice questions load below.