9701 Chemistry · Topics 6 & 24 · AS + A Level
Redox and Electrochemistry Cheat Sheet — A Level Chemistry 9701
Electrochemistry splits neatly into the AS half — oxidation numbers, balancing half-equations and electrolysis — and the A2 half built around the electrochemical series. These sheets cover both, including the Faraday calculation, how to combine two standard electrode potentials without a sign error, and how the Nernst equation shifts E when concentrations are not standard.
✅ 27 worked sections
🎓 CAIE 9701 syllabus-mapped
⭐ 4.8 rated tutors
What’s on this cheat sheet
Redox and Electrolysis — AS
01 · Oxidation and reduction
Oxidation — loss of electrons, oxidation number rises. Reduction — gain of electrons, oxidation number falls. OIL RIG.
The oxidising agent is itself reduced; the reducing agent is itself oxidised. Name the agent, not the process, when asked which is which.
02 · Oxidation number rules
| Species | O.N. |
|---|---|
| uncombined element | 0 |
| simple ion | the charge |
| Group 1 / 2 | +1 / +2 |
| H | +1 (−1 in hydrides) |
| O | −2 (−1 in peroxides, +2 in OF₂) |
| F | always −1 |
Sum = 0 for a neutral compound, = charge for an ion. Roman numerals in a name give the oxidation number: manganate(VII), iron(III).
03 · Building redox equations
1 · balance the main element
2 · balance O with H₂O
3 · balance H with H⁺
4 · balance charge with e⁻
5 · scale so electrons cancel, then add
Fe²⁺ → Fe³⁺ + e⁻ (× 5)
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
04 · Disproportionation
One element is simultaneously oxidised and reduced in the same reaction.
Cl₂ + 2NaOH → NaCl + NaClO + H₂O — chlorine goes from 0 to −1 and to +1. Also 2H₂O₂ → 2H₂O + O₂ and Cu⁺ in acid.
05 · Electrolysis
Cathode (−) attracts cations, which are reduced. Anode (+) attracts anions, which are oxidised. Current flows only when ions are mobile — molten or in solution.
In aqueous solution the water can discharge instead: H₂ at the cathode for reactive metals, O₂ at the anode unless a halide is concentrated.
06 · Faraday calculations
F = 96 500 C mol⁻¹
Q = 2.00 × 1800 = 3600 C
n(e⁻) = 3600 ÷ 96 500 = 0.0373 mol
Cu²⁺ + 2e⁻ → Cu, so n(Cu) = 0.0187 mol
m = 0.0187 × 63.5 = 1.19 g
07 · Common oxidising and reducing agents
| Reagent | Becomes | Colour change |
|---|---|---|
| MnO₄⁻ / H⁺ | Mn²⁺ | purple → colourless |
| Cr₂O₇²⁻ / H⁺ | Cr³⁺ | orange → green |
| I₂ | I⁻ | brown → colourless |
| Fe²⁺ | Fe³⁺ | pale green → yellow |
| S₂O₃²⁻ | S₄O₆²⁻ | used in iodine titrations |
Manganate(VII) titrations are self‑indicating — the first permanent pink is the end point. Use dilute H₂SO₄, never HCl (chloride is oxidised) and never HNO₃ (itself an oxidising agent).
08 · Electrolysis products in solution
| Electrolyte | Cathode | Anode |
|---|---|---|
| conc. NaCl(aq) | H₂ | Cl₂ |
| dilute NaCl(aq) | H₂ | O₂ |
| CuSO₄(aq), Pt | Cu | O₂ |
| molten NaCl | Na | Cl₂ |
A metal below hydrogen in reactivity is deposited; a more reactive one leaves H₂ instead. With copper electrodes in CuSO₄ the anode dissolves — the basis of electroplating and purification.
09 · Electrode half‑equations
Cu²⁺ + 2e⁻ → Cu
2H₂O + 2e⁻ → H₂ + 2OH⁻
Anode (oxidation)
2Cl⁻ → Cl₂ + 2e⁻
2H₂O → O₂ + 4H⁺ + 4e⁻
The same quantity of charge passes through both electrodes, so the mole ratio of products follows the electrons in each half‑equation.
10 · Worked example — oxidation numbers
S in H₂SO₄: 2(+1) + x + 4(−2) = 0 → x = +6
N in NH₄⁺: x + 4(+1) = +1 → x = −3
Cl in ClO₃⁻: x + 3(−2) = −1 → x = +5
11 · Uses of electrolysis
Extraction of aluminium — molten Al₂O₃ in cryolite, which lowers the melting point and cuts the energy cost. Carbon anodes burn away as CO₂ and must be replaced.
Purification of copper — impure Cu anode dissolves, pure Cu deposits on the cathode; the impurities drop as anode sludge.
Electroplating — the object is the cathode and the plating metal the anode, in a solution of that metal’s salt.
12 · Worked example — two cells in series
Ag⁺ + e⁻ → Ag: n(Ag) = 1.08 ÷ 107.9 = 0.0100 mol
n(e⁻) = 0.0100 mol — the same in both cells
Cu²⁺ + 2e⁻ → Cu: n(Cu) = 0.00500 mol
m(Cu) = 0.00500 × 63.5 = 0.318 g
13 · Worked example — redox titration
MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
n(MnO₄⁻) = 0.0200 × 0.0215 = 4.30 × 10⁻⁴ mol
n(Fe²⁺) = 5 × 4.30 × 10⁻⁴ = 2.15 × 10⁻³ mol
c(Fe²⁺) = 2.15 × 10⁻³ ÷ 0.0250 = 0.0860 mol dm⁻³
End point: the first permanent pale pink.
Marks lost here
— Naming the oxidising agent as the species that is oxidised.
— Leaving time in minutes inside Q = It.
— Forgetting the electron ratio from the half‑equation before converting to mass.
Electrode Potentials — A2
14 · Standard conditions
All solutions 1 mol dm⁻³, gases at 100 kPa, temperature 298 K, measured against the standard hydrogen electrode which is defined as 0.00 V.
A platinum electrode is used where no solid metal is involved — it is inert and provides a surface for electron transfer. The salt bridge (KNO₃) completes the circuit and balances charge without mixing the solutions.
15 · Reading E° values
Data‑booklet half‑equations are written as reductions. A more positive E° means a greater tendency to be reduced, so that species is the better oxidising agent.
A more negative E° means the reduced form is the better reducing agent. The more negative half‑cell runs backwards and forms the negative electrode.
16 · Cell e.m.f.
E°cell = +0.34 − (−0.76) = +1.10 V
Zn is oxidised (negative electrode), Cu²⁺ reduced.
Zn + Cu²⁺ → Zn²⁺ + Cu
Cell diagram: Zn | Zn²⁺ ⋮⋮ Cu²⁺ | Cu
Positive E°cell → the reaction is feasible. Feasible does not mean fast: a large Ea can still make it immeasurably slow.
17 · Effect of concentration
Apply Le Chatelier to the half‑equation. Raising the concentration of the oxidised species drives the reduction forward and makes E° more positive; raising the reduced species makes it more negative.
For MnO₄⁻/Mn²⁺, raising [H⁺] makes E more positive — which is why acidified manganate(VII) is the stronger oxidising agent.
18 · ΔG and E°
n = moles of electrons transferred, F = 96 500 C mol⁻¹. Answer in J mol⁻¹ — divide by 1000 for kJ mol⁻¹. A positive E°cell gives a negative ΔG°, the same feasibility test seen twice.
19 · Cells in use
Fuel cell (H₂/O₂, alkaline): anode 2H₂ + 4OH⁻ → 4H₂O + 4e⁻, cathode O₂ + 2H₂O + 4e⁻ → 4OH⁻. Only product is water; efficiency is high, but storing and transporting H₂ is the difficulty.
Rechargeable cells reverse the electrode reactions on charging. Redox potentials also explain corrosion: a metal with a more negative E° protects one with a less negative E° (sacrificial protection of iron by zinc).
20 · Predicting reactions from E°
Br₂ + 2e⁻ → 2Br⁻ E° = +1.07 V
Fe³⁺ + e⁻ → Fe²⁺ E° = +0.77 V
Br₂ has the more positive E°, so it is reduced and Fe²⁺ oxidised.
E°cell = 1.07 − 0.77 = +0.30 V → feasible.
I₂ (E° = +0.54 V) would not oxidise Fe²⁺: E°cell = −0.23 V.
21 · Half‑cell types
| Type | Example |
|---|---|
| metal / metal ion | Zn(s) | Zn²⁺(aq) |
| gas / ion, Pt | Pt | H₂(g) | H⁺(aq) |
| two ions, Pt | Pt | Fe²⁺, Fe³⁺ |
| ion / solid, Pt | Pt | MnO₄⁻, Mn²⁺, H⁺ |
In a cell diagram, | is a phase boundary and ⋮⋮ the salt bridge; the oxidised form is written next to the bridge on each side.
22 · Worked example — ΔG° from E°
ΔG° = −nFE°
ΔG° = −2 × 96 500 × 1.10
ΔG° = −212 300 J mol⁻¹
ΔG° = −212 kJ mol⁻¹
Strongly negative, so the reaction is thermodynamically feasible.
23 · Disproportionation from E° values
A species disproportionates when its own reduction has a more positive E° than its own oxidation — the two half‑reactions combine to give a positive E°cell.
Cu²⁺ + e⁻ → Cu⁺ E° = +0.15 V
E°cell = 0.52 − 0.15 = +0.37 V
2Cu⁺ → Cu + Cu²⁺ is feasible, which is why Cu⁺ is unstable in aqueous solution.
24 · E° values worth remembering
| Half‑cell | E° / V |
|---|---|
| F₂ / F⁻ | +2.87 |
| MnO₄⁻, H⁺ / Mn²⁺ | +1.52 |
| Cr₂O₇²⁻, H⁺ / Cr³⁺ | +1.33 |
| Fe³⁺ / Fe²⁺ | +0.77 |
| Cu²⁺ / Cu | +0.34 |
| 2H⁺ / H₂ | 0.00 |
| Zn²⁺ / Zn | −0.76 |
The list is the reactivity series in another form: the most negative metals are the strongest reducing agents, the most positive non‑metals the strongest oxidising agents.
25 · Corrosion and protection
Rusting is electrochemical: Fe → Fe²⁺ + 2e⁻ at the anodic region, while oxygen is reduced at the cathodic region. Both water and oxygen are needed, and salt speeds it up by carrying the current.
Sacrificial protection — attach a metal with a more negative E° (Zn or Mg) so it corrodes instead. Galvanising does both: the zinc coats the surface and protects sacrificially where it is scratched.
Marks lost here
— Multiplying E° when scaling a half‑equation; E° is intensive and never multiplied.
— Subtracting the wrong way round and reporting a negative E°cell for a feasible cell.
— Treating “feasible” as “will actually happen quickly”.
— Forgetting that the quoted values only hold under standard conditions.
Get all 25 topic cheat sheets for 9701 Chemistry
One free PDF pack, AS + A Level, straight to WhatsApp.
Redox and Electrochemistry — Frequently Asked Questions
How do you calculate the cell EMF from two electrode potentials?
Eᴼᶜᵉˡˡ = Eᴼ(reduction half, the more positive) − Eᴼ(oxidation half, the more negative). A positive cell EMF means the reaction as written is feasible.
What does a more positive Eᴼ value tell you?
The more positive the standard electrode potential, the greater the tendency of that species to be reduced — so it is the stronger oxidising agent. The more negative the value, the stronger the reducing agent on the right-hand side of the half-equation.
How does the Nernst equation change E?
Increasing the concentration of the oxidised species makes E more positive; increasing the reduced species makes it more negative. At 298 K, E = Eᴼ + (0.059/z) log([oxidised]/[reduced]).
