Electrochemistry — A-Level Chemistry Study Guide

For: A-Level Chemistry candidates sitting A-Level Chemistry.

Covers: Redox reaction oxidation states, standard electrode potentials and EMF, electrolysis electrode products, Faraday's laws of electrolysis, and industrial electrochemical cells including fuel cells and electrolytic refining.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Chemistry style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Electrochemistry?

Electrochemistry is the study of the interconversion between chemical energy and electrical energy, driven by redox (oxidation-reduction) reactions. It is split into two core branches: spontaneous reactions that generate electrical energy (galvanic/voltaic cells, used in batteries) and non-spontaneous reactions that require electrical energy to proceed (electrolytic cells, used for metal refining and chemical synthesis). This topic accounts for 8-12 marks across A-Level Chemistry Papers 2 and 4, with frequent synoptic links to transition metal chemistry and energetics.

2. Redox reactions — oxidation states

Redox reactions involve the transfer of electrons between species: oxidation is the loss of electrons (increase in oxidation state), and reduction is the gain of electrons (decrease in oxidation state). Oxidation state (OS) is a hypothetical charge assigned to an atom in a compound, calculated using the following non-negotiable rules:

1. The OS of an uncombined element is 0 (e.g. $\text{Fe}$, ${\text{O}}_2$, ${\text{Cl}}_2$ all have OS = 0)
2. The OS of a monatomic ion equals its charge (e.g. ${\text{Na}}^{+}$ = +1, ${\text{S}}^{2-}$ = -2)
3. Oxygen has an OS of -2 except in peroxides (e.g. ${\text{H}}_2{\text{O}}_2$, OS = -1) and ${\text{F}}_2\text{O}$ (OS = +2)
4. Hydrogen has an OS of +1 except in metal hydrides (e.g. $\text{NaH}$, OS = -1)
5. The sum of OS in a neutral compound is 0; the sum of OS in a polyatomic ion equals its charge.

Worked Example

Calculate the oxidation state of manganese in ${\text{MnO}}_4^{-}$: Let $x$ = OS of Mn. Oxygen = -2, sum of OS = -1 (charge of ion): $$x+4(-2)=-1$$ $$x-8=-1$$ $$x=+7$$ Examiners frequently test oxidation states of transition metals in complex ions, so always cross-check your sum against the overall charge of the species to avoid arithmetic errors.

3. Standard electrode potentials and the EMF

The standard electrode potential ($E^{\circ }$) of a half-cell is the potential difference between that half-cell and the standard hydrogen electrode (SHE, defined as $E^{\circ }=0.00\text{V}$), measured under standard conditions: 298 K, $1.0{\text{mol dm}}^{-3}$ concentration of aqueous ions, 1 atm pressure for gaseous reactants, and pure solid electrodes.

The standard cell electromotive force (EMF, $E_{\text{cell}}^{\circ }$) is the maximum potential difference between two half-cells in a galvanic cell, calculated using: $$E_{\text{cell}}^{\circ }=E_{\text{cathode (reduction)}}^{\circ }-E_{\text{anode (reduction)}}^{\circ }$$ A positive $E_{\text{cell}}^{\circ }$ indicates a spontaneous reaction. The cathode is always the half-cell with the higher $E^{\circ }$ value (it undergoes reduction), while the anode is the half-cell with the lower $E^{\circ }$ value (it undergoes oxidation).

Worked Example

A galvanic cell is constructed using a $\text{Zn}/{\text{Zn}}^{2+}$ half-cell ($E^{\circ }=-0.76\text{V}$) and a $\text{Ag}/{\text{Ag}}^{+}$ half-cell ($E^{\circ }=+0.80\text{V}$). Calculate the standard cell EMF: Cathode = $\text{Ag}/{\text{Ag}}^{+}$ (higher $E^{\circ }$, reduction: ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$) Anode = $\text{Zn}/{\text{Zn}}^{2+}$ (lower $E^{\circ }$, oxidation: $\text{Zn}\to {\text{Zn}}^{2+}+2e^{-}$) $$E_{\text{cell}}^{\circ }=+0.80-(-0.76)=+1.56\text{V}$$ Exam tip: Never swap the sign of $E^{\circ }$ values and use the subtraction formula, as this will double count the sign change and give an incorrect result.

4. Electrolysis — products at electrodes

Electrolysis uses an external electrical current to drive non-spontaneous redox reactions in an electrolytic cell. The anode is the positive electrode (oxidation occurs here) and the cathode is the negative electrode (reduction occurs here, opposite to galvanic cells). Use the following rules to predict products:

1. Molten ionic compounds: No water present, so the cation is reduced at the cathode and the anion is oxidised at the anode. For example, molten $\text{NaCl}$ produces $\text{Na}$ metal at the cathode and ${\text{Cl}}_2$ gas at the anode.
2. Aqueous solutions: Water is present, so compare $E^{\circ }$ values of ions and water, plus concentration effects:

• Cathode: If the metal is below hydrogen in the reactivity series (its $E^{\circ }$ value is greater than 0.00 V), the metal ion is reduced to solid metal. If it is above hydrogen, ${\text{H}}^{+}$ from water is reduced to ${\text{H}}_2$ gas.
• Anode: If a halide ion is present at high concentration, it is oxidised to halogen gas. If dilute, ${\text{OH}}^{-}$ from water is oxidised to ${\text{O}}_2$ gas instead.

Worked Example

Predict the products at each electrode for electrolysis of concentrated aqueous copper(II) bromide using inert graphite electrodes:

• Cathode: ${\text{Cu}}^{2+}$ has $E^{\circ }=+0.34\text{V}$ (higher than 0.00 V for ${\text{H}}^{+}$), so solid copper metal deposits.
• Anode: Bromide ions are concentrated, so ${\text{Br}}_2$ gas forms instead of ${\text{O}}_2$.

5. Faraday's laws and charge calculations

Faraday's laws quantify the relationship between charge passed through an electrolytic cell and the amount of substance produced at the electrodes:

1. First Law: The mass of substance deposited or liberated at an electrode is directly proportional to the total charge passed through the cell.
2. Second Law: The number of moles of different substances deposited by the same amount of charge is inversely proportional to the charge on their ion.

Use these core formulas for all calculations:

1. Total charge passed: $Q=I\times t$, where $Q$ is charge in coulombs (C), $I$ is current in amperes (A), and $t$ is time in seconds (s).
2. 1 Faraday ($F$) = charge of 1 mole of electrons = $96500{\text{C mol}}^{-1}$.
3. Moles of electrons transferred: $n(e^{-})=\frac{Q}{F}$
4. Moles of substance produced: $n(\text{substance})=\frac{n(e^{-})}{z}$, where $z$ is the number of electrons transferred per ion of the substance.

Worked Example

Calculate the mass of silver metal deposited at the cathode when a current of 1.5 A is passed through aqueous silver nitrate for 20 minutes.

1. Convert time to seconds: $t=20\times 60=1200\text{s}$
2. Calculate charge: $Q=1.5\times 1200=1800\text{C}$
3. Moles of electrons: $n(e^{-})=\frac{1800}{96500}=0.01865\text{mol}$
4. Reduction reaction: ${\text{Ag}}^{+}+e^{-}\to \text{Ag}$, so $z=1$
5. Moles of Ag = 0.01865 mol, mass = $0.01865\times 108=2.01\text{g}$

6. Industrial cells — fuel cells, electrolytic refining

Electrochemical cells have widespread industrial applications, with two core examples tested in the A-Level Chemistry syllabus:

Hydrogen-Oxygen Fuel Cells

Fuel cells convert the chemical energy of a fuel (hydrogen) and oxidiser (oxygen) directly to electrical energy, with no combustion step, giving 40-60% efficiency (double that of internal combustion engines). The alkaline fuel cell reactions are:

• Anode (oxidation): $2{\text{H}}_2+4{\text{OH}}^{-}\to 4{\text{H}}_2\text{O}+4e^{-}$
• Cathode (reduction): ${\text{O}}_2+2{\text{H}}_2\text{O}+4e^{-}\to 4{\text{OH}}^{-}$
• Overall reaction: $2{\text{H}}_2+{\text{O}}_2\to 2{\text{H}}_2\text{O}$, $E_{\text{cell}}^{\circ }=+1.23\text{V}$ Advantages include zero toxic emissions (only water waste) and constant power output as long as fuel is supplied; disadvantages include high cost of platinum catalysts and difficulty storing compressed hydrogen.

Electrolytic Refining of Copper

Impure copper extracted from ore has high electrical resistance, so it is refined to 99.99% purity via electrolysis:

• Anode: Impure copper block, which oxidises: $\text{Cu (impure)}\to {\text{Cu}}^{2+}+2e^{-}$. Less reactive impurities (gold, silver) fall to the bottom as anode sludge, which is collected for separate processing.
• Cathode: Pure copper sheet, where copper ions deposit: ${\text{Cu}}^{2+}+2e^{-}\to \text{Cu (pure)}$
• Electrolyte: Aqueous copper(II) sulfate, which maintains a constant concentration of ${\text{Cu}}^{2+}$ ions during the process.

7. Common Pitfalls (and how to avoid them)

• Pitfall 1: Labelling the anode as negative in electrolytic cells. Why it happens: Students memorize "anode = negative" for galvanic cells and generalize incorrectly. Correct move: Always define electrodes by reaction, not charge: anode = oxidation, cathode = reduction. Anode is positive in electrolytic cells, negative in galvanic cells.
• Pitfall 2: Swapping the sign of the oxidation half-cell $E^{\circ }$ and using the subtraction formula for $E_{\text{cell}}^{\circ }$. Why it happens: Confusion over different calculation methods. Correct move: Either keep all $E^{\circ }$ values as given and use $E_{\text{cell}}^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$, or reverse the sign of the oxidation half-cell and add to the reduction $E^{\circ }$, never do both.
• Pitfall 3: Using time in minutes instead of seconds for Faraday calculations. Why it happens: Rushing questions and forgetting SI unit conventions. Correct move: As soon as you see a time value given in minutes, cross it out and write the equivalent in seconds next to it before starting any calculation.
• Pitfall 4: Predicting oxygen as the anode product for concentrated halide solutions. Why it happens: Over-reliance on $E^{\circ }$ values without checking concentration. Correct move: Circle the word "concentrated" in any electrolysis question if it appears, and prioritize halogen production over oxygen for the anode.
• Pitfall 5: Assigning hydrogen an oxidation state of +1 in metal hydrides. Why it happens: Defaulting to the common OS rule for hydrogen. Correct move: If hydrogen is bonded to a Group 1 or Group 2 metal, its OS is -1, always.

8. Practice Questions (A-Level Chemistry Style)

Question 1

(a) Give the oxidation state of chlorine in each of the following species: (3 marks) (i) ${\text{Cl}}_2{\text{O}}_7$ (ii) ${\text{ClO}}_2^{-}$ (iii) $\text{HCl}$

(b) A galvanic cell is constructed using $\text{Mg}/{\text{Mg}}^{2+}$ ($E^{\circ }=-2.37\text{V}$) and $\text{Pb}/{\text{Pb}}^{2+}$ ($E^{\circ }=-0.13\text{V}$). Calculate the standard cell EMF and write the overall spontaneous reaction. (3 marks)

Solution

(a) (i) O = -2, $2x+7(-2)=0$ → $x=+7$ (ii) O=-2, $x+2(-2)=-1$ → $x=+3$ (iii) H=+1, $+1+x=0$ → $x=-1$ (b) Cathode = $\text{Pb}/{\text{Pb}}^{2+}$ (higher $E^{\circ }$), reduction: ${\text{Pb}}^{2+}+2e^{-}\to \text{Pb}$; Anode = $\text{Mg}/{\text{Mg}}^{2+}$, oxidation: $\text{Mg}\to {\text{Mg}}^{2+}+2e^{-}$ $E_{\text{cell}}^{\circ }=-0.13-(-2.37)=+2.24\text{V}$; Overall reaction: $\text{Mg}+{\text{Pb}}^{2+}\to {\text{Mg}}^{2+}+\text{Pb}$

Question 2

Predict the products formed at the anode and cathode when dilute aqueous potassium iodide is electrolysed using inert platinum electrodes. Explain your answer. (4 marks)

Solution

• Cathode: ${\text{K}}^{+}$ has $E^{\circ }=-2.93\text{V}$, much lower than $E^{\circ }$ for ${\text{H}}^{+}/{\text{H}}_2$ (0.00 V), so ${\text{H}}^{+}$ from water is reduced to form ${\text{H}}_2$ gas.
• Anode: Iodide ions are dilute, so ${\text{OH}}^{-}$ from water is oxidised to form ${\text{O}}_2$ gas (concentration is too low for ${\text{I}}_2$ to form preferentially). Products: Cathode = hydrogen gas, Anode = oxygen gas.

Question 3

Calculate the volume of chlorine gas (measured at RTP, 1 mol of gas = $24{\text{dm}}^3$) produced when a current of 2.5 A is passed through concentrated aqueous sodium chloride for 1 hour. (4 marks)

Solution

1. Time in seconds: $t=1\times 60\times 60=3600\text{s}$
2. Total charge: $Q=2.5\times 3600=9000\text{C}$
3. Moles of electrons: $n(e^{-})=9000/96500=0.0933\text{mol}$
4. Anode reaction: $2{\text{Cl}}^{-}\to {\text{Cl}}_2+2e^{-}$, so 2 mol electrons produce 1 mol ${\text{Cl}}_2$
5. Moles of ${\text{Cl}}_2=0.0933/2=0.04665\text{mol}$
6. Volume of ${\text{Cl}}_2=0.04665\times 24=1.12{\text{dm}}^3$

9. Quick Reference Cheatsheet

10. What's Next

Electrochemistry is a foundational topic that connects to multiple later sections of the A-Level Chemistry syllabus, including transition metal chemistry (where redox reactions of complex ions are a core assessment objective), reaction kinetics (you may be asked to measure the rate of electrolysis reactions to determine rate equations), and organic chemistry (oxidation and reduction of functional groups use the same electron transfer rules you learned here). It is also a common synoptic question topic in Paper 4, where you may be asked to combine electrochemistry calculations with energetics or equilibrium concepts for extended response questions.

If you have gaps in your understanding of any section of this guide, or want to practice more exam-style questions tailored to your weak areas, you can ask Ollie, our AI tutor, for personalized support at any time. You can also find more topic guides and full mock exams for A-Level Chemistry on the OwlsPrep homepage to get fully prepared for your exam.

Aligned with the Cambridge International AS & A Level Chemistry 9701 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.