Electrochemistry

Learn and apply Electrochemistry in the published Chemistry course sequence.

GCE A-Level H2 Chemistry 9476-2027

Learning goals

Standard Electrode Potentials and the SHE
Cell Potentials and Spontaneity
Writing Redox Equations from Half-equations
ΔG = −nFE
Electrolysis Predictions and Faraday’s Law
Batteries and Fuel Cells

What You’ll Learn

Explain redox using electron transfer and oxidation-number change.
Describe and measure standard electrode potentials using the standard hydrogen electrode.
Calculate E⦵_cell, deduce electron flow and judge feasibility with its limitations.
Construct redox equations and apply Δ G⦵ = -nFE⦵_cell.
Predict electrolysis products, calculate amounts using charge and explain the two specified industrial processes.
Compare batteries and hydrogen–oxygen fuel cells using evidence relevant to their application.

Electrochemistry is prescribed in H2 9476. H1 8873 retains foundation redox reasoning elsewhere but does not require this electrode-potential, cell, Faraday and industrial-electrolysis sequence.

Galvanic cell vs electrolysis cell

Two-panel comparison of electrode signs, electron direction, ion direction, and oxidation-reduction labels.

Galvanic (voltaic): spontaneous

Anode: negative, oxidation

Cathode: positive, reduction

e- flow: anode -> cathode (external circuit)

Ion flow: cations -> cathode, anions -> anode

Electrolysis: non-spontaneous

Anode: positive, oxidation

Cathode: negative, reduction

e- flow: anode -> cathode in the cell path

Power supply drives the reaction direction

Exam rule: never swap the definitions - anode = oxidation, cathode = reduction.

Oxidation is always at the anode and reduction is always at the cathode. The electrode signs change between a spontaneous galvanic cell and electrolysis.

Lessons (Recommended Order)

Standard Electrode Potentials and SHE — define, construct and measure a half-cell.
Cell Potentials and Spontaneity — calculate voltage, direction and qualified feasibility.
Writing Redox Equations from Half-equations — turn the chosen direction into a balanced equation.
Δ G⦵ = -nFE⦵ — connect voltage, electron count and maximum electrical work.
Electrolysis Predictions and Faraday’s Law — predict products, calculate yield and explain industrial cells.
Batteries and Fuel Cells — transfer the core model to operating devices.

Follow the order on a first pass. During revision, enter at the lesson matching the evidence in the question: a data table, a full equation, energy data, current–time data or a device comparison.

Quick Reference

Question evidence	First move	Essential check
one unknown half-cell	compare it with the SHE	298 K, 1 bar, 1.0 mol dm⁻³
two reduction potentials	more positive value is reduced	E⦵_cell = E⦵_cathode-E⦵_anode
overall redox equation	reverse one half-equation and equalise electrons	never multiply E⦵
feasibility claim	inspect the sign of E⦵_cell	say “as written under standard conditions”; rate is separate
Gibbs-energy data	find n from the balanced reaction	use J mol⁻¹ in Δ G⦵ = -nFE⦵
electrolysis current and time	Q = It, then n(e⁻) = Q/F	time in s; use the half-equation ratio
aqueous electrolyte	list water and solute competitors	use potential/concentration evidence given
industrial cell	identify both electrode reactions	link them to oxide growth or copper transfer

F = Le connects the charge per mole of electrons, the Avogadro constant and the charge on one electron.

Hub Quiz

Use the Electrochemistry hub quiz after all six lessons. Then use structured Electrochemistry practice to rehearse equations, calculations and explanations with visible working.

Practise

Work through questions with marking and feedback as you learn.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only and doesn't count towards mastery.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

This check needs JavaScript to record answers. The reviewed questions are listed below for study.

A copper half-cell is connected to a standard hydrogen electrode to measure E⦵(Cu²⁺/Cu). Which arrangement and result is correct?

A copper rod in 1.00 mol dm⁻³ Cu²⁺(aq) at 298 K, joined to the SHE by a wire through a high-resistance voltmeter and by a KNO₃(aq) salt bridge; the reading is 0.34 V with copper as the positive electrode.
The two solutions are joined by a copper wire instead of a salt bridge.
A low-resistance ammeter is used, so that the maximum current can be measured.
The reading is 0.34 V with copper as the negative electrode.

E⦵ values: Fe³⁺/Fe²⁺ +0.77 V; I₂/I⁻ +0.54 V; Br₂/Br⁻ +1.07 V. Which statement about aqueous Fe³⁺ under standard conditions is correct?

Fe³⁺ oxidises I⁻ to I₂ (E⦵cell = +0.23 V) but does not oxidise Br⁻ (E⦵cell = −0.30 V).
Fe³⁺ oxidises both I⁻ and Br⁻.
Fe³⁺ oxidises I⁻ to I₂, with E⦵cell = +1.31 V.
Fe³⁺ oxidises Br⁻ but not I⁻, because Br₂/Br⁻ has the more positive E⦵.

Use MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and O₂ + 2H⁺ + 2e⁻ → H₂O₂. Which is the balanced equation for the oxidation of H₂O₂ by acidified manganate(VII)?

2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
MnO₄⁻ + 8H⁺ + H₂O₂ → Mn²⁺ + 4H₂O + O₂
2MnO₄⁻ + 16H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
2MnO₄⁻ + 10H⁺ + 3H₂O₂ → 2Mn²⁺ + 8H₂O + 3O₂

For Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, E⦵ values are Cl₂/Cl⁻ +1.36 V and Br₂/Br⁻ +1.07 V. What is ΔG⦵ for the reaction? (F = 96 500 C mol⁻¹)

−56.0 kJ mol⁻¹
−28.0 kJ mol⁻¹
−469 kJ mol⁻¹
+56.0 kJ mol⁻¹

Aqueous copper(II) sulfate is electrolysed using copper electrodes. Which statement is correct?

Copper is deposited on the cathode and the copper anode dissolves as Cu²⁺, so the Cu²⁺ concentration stays constant.
Copper is deposited on the cathode, oxygen forms at the anode and the solution becomes acidic.
Hydrogen forms at the cathode and the copper anode dissolves.
Copper dissolves from the cathode and is deposited on the anode.

Dilute sulfuric acid is electrolysed with a current of 2.00 A for 965 s. What volume of hydrogen forms at the cathode, measured at r.t.p.? (F = 96 500 C mol⁻¹; molar gas volume 24.0 dm³ mol⁻¹)

0.240 dm³
0.480 dm³
0.120 dm³
0.360 dm³

Lithium (Ar 6.9; E⦵ Li⁺/Li = −3.04 V) is used as the negative electrode material in improved batteries for electric vehicles. Which explanation is correct?

Its very negative E⦵ gives a high cell voltage, and its low Ar means a low mass of electrode material for the charge it supplies.
Its E⦵ is the most positive of the metals, so Li⁺ is easily reduced at the negative electrode.
Lithium is a very dense metal, so the cells can be made small.
Its very negative E⦵ gives a low cell voltage, which makes the battery safer.

Practise after feedback

After a check, practise the skills it showed you need to work on.

About 10 minutes

Questions are picked at random each time you start. You'll see the answer after each question. It's for practice only and doesn't count towards mastery.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

This check needs JavaScript to record answers. The reviewed questions are listed below for study.

A copper half-cell is connected to a standard hydrogen electrode to measure E⦵(Cu²⁺/Cu). Which arrangement and result is correct?

A copper rod in 1.00 mol dm⁻³ Cu²⁺(aq) at 298 K, joined to the SHE by a wire through a high-resistance voltmeter and by a KNO₃(aq) salt bridge; the reading is 0.34 V with copper as the positive electrode.
The two solutions are joined by a copper wire instead of a salt bridge.
A low-resistance ammeter is used, so that the maximum current can be measured.
The reading is 0.34 V with copper as the negative electrode.

E⦵ values: Fe³⁺/Fe²⁺ +0.77 V; I₂/I⁻ +0.54 V; Br₂/Br⁻ +1.07 V. Which statement about aqueous Fe³⁺ under standard conditions is correct?

Fe³⁺ oxidises I⁻ to I₂ (E⦵cell = +0.23 V) but does not oxidise Br⁻ (E⦵cell = −0.30 V).
Fe³⁺ oxidises both I⁻ and Br⁻.
Fe³⁺ oxidises I⁻ to I₂, with E⦵cell = +1.31 V.
Fe³⁺ oxidises Br⁻ but not I⁻, because Br₂/Br⁻ has the more positive E⦵.

Use MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O and O₂ + 2H⁺ + 2e⁻ → H₂O₂. Which is the balanced equation for the oxidation of H₂O₂ by acidified manganate(VII)?

2MnO₄⁻ + 6H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
MnO₄⁻ + 8H⁺ + H₂O₂ → Mn²⁺ + 4H₂O + O₂
2MnO₄⁻ + 16H⁺ + 5H₂O₂ → 2Mn²⁺ + 8H₂O + 5O₂
2MnO₄⁻ + 10H⁺ + 3H₂O₂ → 2Mn²⁺ + 8H₂O + 3O₂

For Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂, E⦵ values are Cl₂/Cl⁻ +1.36 V and Br₂/Br⁻ +1.07 V. What is ΔG⦵ for the reaction? (F = 96 500 C mol⁻¹)

−56.0 kJ mol⁻¹
−28.0 kJ mol⁻¹
−469 kJ mol⁻¹
+56.0 kJ mol⁻¹

Aqueous copper(II) sulfate is electrolysed using copper electrodes. Which statement is correct?

Copper is deposited on the cathode and the copper anode dissolves as Cu²⁺, so the Cu²⁺ concentration stays constant.
Copper is deposited on the cathode, oxygen forms at the anode and the solution becomes acidic.
Hydrogen forms at the cathode and the copper anode dissolves.
Copper dissolves from the cathode and is deposited on the anode.

Dilute sulfuric acid is electrolysed with a current of 2.00 A for 965 s. What volume of hydrogen forms at the cathode, measured at r.t.p.? (F = 96 500 C mol⁻¹; molar gas volume 24.0 dm³ mol⁻¹)

0.240 dm³
0.480 dm³
0.120 dm³
0.360 dm³

Lithium (Ar 6.9; E⦵ Li⁺/Li = −3.04 V) is used as the negative electrode material in improved batteries for electric vehicles. Which explanation is correct?

Its very negative E⦵ gives a high cell voltage, and its low Ar means a low mass of electrode material for the charge it supplies.
Its E⦵ is the most positive of the metals, so Li⁺ is easily reduced at the negative electrode.
Lithium is a very dense metal, so the cells can be made small.
Its very negative E⦵ gives a low cell voltage, which makes the battery safer.

Check what I know

Start here to see which parts you already know.

About 8 minutes

Answer 7 short questions. It shows what to work on next and doesn't count towards mastery.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

This check needs JavaScript to record answers. The reviewed questions are listed below for study.

Which description of the standard hydrogen electrode is correct?

Hydrogen gas at 1 bar bubbled over a platinum electrode in a solution with [H⁺] = 1.00 mol dm⁻³ at 298 K; its electrode potential is defined as 0.00 V.
Hydrogen gas at 1 bar over a platinum electrode in 1.00 mol dm⁻³ H₂SO₄(aq) at 298 K.
Hydrogen gas at 1 bar over a copper electrode in 1.00 mol dm⁻³ HCl(aq) at 298 K.
Hydrogen gas at 1 bar over a platinum electrode in 1.00 mol dm⁻³ HCl(aq) at 273 K.

A cell is made from Zn²⁺/Zn (E⦵ = −0.76 V) and Ag⁺/Ag (E⦵ = +0.80 V) half-cells under standard conditions. What is E⦵cell, and which electrode is negative?

+1.56 V; the zinc electrode is negative
+0.04 V; the zinc electrode is negative
+2.36 V; the zinc electrode is negative
+1.56 V; the silver electrode is negative

Use Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O and Fe³⁺ + e⁻ → Fe²⁺. Which is the balanced equation for the oxidation of Fe²⁺ by acidified dichromate(VI)?

Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺
Cr₂O₇²⁻ + 14H⁺ + Fe²⁺ → 2Cr³⁺ + 7H₂O + Fe³⁺
Cr₂O₇²⁻ + 14H⁺ + 6Fe³⁺ → 2Cr³⁺ + 7H₂O + 6Fe²⁺
Cr₂O₇²⁻ + 7H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺

For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E⦵cell = +1.10 V. What is ΔG⦵ for this reaction? (F = 96 500 C mol⁻¹)

−212 kJ mol⁻¹
−106 kJ mol⁻¹
+212 kJ mol⁻¹
−212 300 kJ mol⁻¹

Concentrated aqueous sodium chloride is electrolysed with inert electrodes. E⦵ values: Na⁺/Na −2.71 V; H₂O/H₂ −0.83 V; O₂/H₂O +1.23 V; Cl₂/Cl⁻ +1.36 V. What forms at each electrode?

Cathode: hydrogen; anode: chlorine
Cathode: sodium; anode: chlorine
Cathode: hydrogen; anode: oxygen
Cathode: chlorine; anode: hydrogen

A current of 0.500 A is passed through aqueous copper(II) sulfate for 30.0 minutes. What mass of copper is deposited on the cathode? (Ar: Cu = 63.5; F = 96 500 C mol⁻¹)

0.296 g
0.592 g
0.00494 g
1.18 g

Why are hydrogen–oxygen fuel cells and improved batteries being developed for electric vehicles?

They can be smaller, have a lower mass or give a higher voltage than older cells, and a hydrogen–oxygen fuel cell forms only water at the point of use.
A fuel cell never needs a supply of fuel once it has been built.
Improved batteries are designed to have a lower voltage, which makes vehicles faster.
Heavier batteries are preferred, because a greater mass stores more energy per kilogram.

Check my progress

When you feel ready, answer on your own to show what you can do.

About 10 minutes

Answer 7 questions. You'll see your score, the answers and explanations at the end. Your result can count towards your course progress.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

Check again

After practising what your progress check showed, check those skills again.

About 10 minutes

Answer 7 questions. You'll see your score, the answers and explanations at the end. Your result can count towards your course progress.

Recent attempts

History is stored only in this browser.

No completed attempts are saved yet.

Review

Come back later to see whether your learning has lasted.

About 10 minutes

Answer 7 questions. You'll see your score, the answers and explanations at the end. A scheduled review counts towards your course progress only when it is due.

Recent attempts

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No completed attempts are saved yet.

Connections and targeted revision

Electrochemistry brings together ideas from Energetics and Thermodynamics and feeds directly into redox systems in Transition Elements.

If one error persists, return to the matching lesson:

setup, reference or standard conditions → Standard Potentials;
sign, direction, concentration or feasibility → Cell Potentials;
coefficients or electron balance → Redox Equations;
n, units or energy sign → Δ G⦵ = -nFE⦵;
products, charge or industrial process → Electrolysis;
device comparison → Batteries and Fuel Cells.

Electrochemistry

What You’ll Learn

Galvanic cell vs electrolysis cell

Galvanic (voltaic): spontaneous

Electrolysis: non-spontaneous

Lessons (Recommended Order)

Quick Reference

Hub Quiz

Practise

Practice complete

Your answers

Practise after feedback

Practice complete

Your answers

Check what I know

Starting check complete

Your answers

Check my progress

Progress check complete

Your answers

Check again

Progress check complete

Your answers

Review

Review complete

Your answers

Connections and targeted revision