Electrochemistry: IGCSE Chemistry Exam Guide

Electrochemistry covers two opposite processes: using electricity to drive reactions (electrolysis) and using reactions to generate electricity (simple cells). Both appear regularly on 0620, and both require you to apply the reactivity series and write half-equations. The detailed electrolysis decision tree is covered in electrolysis exam technique; this guide covers the broader electrochemistry topic and its exam patterns. The full content is under electrochemistry.

Electrolysis: the essentials for exam day

Electrolysis questions ask you to predict electrode products, write half-equations, and explain observations. The decision tree in the electrolysis technique guide handles product prediction. Here are the exam-day essentials.

What you must know cold

• Cathode = negative electrode = where cations go = reduction (gain of electrons).
• Anode = positive electrode = where anions go = oxidation (loss of electrons).
• OIL RIG: Oxidation Is Loss, Reduction Is Gain.
• The electrolyte must be molten or in aqueous solution for electrolysis to work (ions must be free to move).

The three electrolysis scenarios the exam repeats

1. Molten binary compound (e.g., molten PbBr2): metal at cathode, non-metal at anode. Simple because no water competes.
2. Aqueous solution (e.g., aqueous CuSO4): water competes at both electrodes. Use the reactivity series to predict whether the metal or hydrogen is deposited at the cathode, and whether a halogen or oxygen is produced at the anode.
3. Purification of copper (copper electrodes in CuSO4 solution): the impure copper anode dissolves, pure copper is deposited at the cathode. The mass of the anode decreases and the mass of the cathode increases by the same amount.

Writing half-equations

At the cathode (reduction): Cu2+ + 2e- -> Cu At the anode (oxidation): 2Cl- -> Cl2 + 2e-

Check: electrons on the left at the cathode, electrons on the right at the anode. The charge must balance on both sides. The technique for writing and balancing these is in ionic equations and balancing.

Simple cells (electrochemical cells)

A simple cell consists of two different metals dipped in an electrolyte (often a salt solution or dilute acid), connected by a wire. The voltage is produced because the two metals have different reactivities.

How it works

1. The more reactive metal loses electrons more readily and becomes the negative terminal. Its atoms are oxidised: Zn -> Zn2+ + 2e-.
2. The electrons flow through the external wire to the less reactive metal.
3. At the less reactive metal (positive terminal), ions in solution gain electrons and are reduced: Cu2+ + 2e- -> Cu.

What the exam tests

• Predicting terminals: The more reactive metal is the negative terminal. The less reactive metal is the positive terminal.
• Predicting voltage: The greater the gap in the reactivity series, the higher the voltage. A magnesium-copper cell produces a higher voltage than a zinc-copper cell.
• Direction of electron flow: Electrons flow from the more reactive metal to the less reactive metal through the external wire. Conventional current flows in the opposite direction.
• What happens to the metals: The more reactive metal dissolves (loses mass). In some setups, the less reactive metal gains a deposit.

Hydrogen fuel cells (Extended)

The hydrogen fuel cell produces electricity from the reaction: 2H2 + O2 -> 2H2O. The only product is water, making it clean. The advantages and disadvantages the exam expects:

Advantages: No polluting gases produced, water is the only product, very efficient, continuous as long as fuel is supplied. Disadvantages: Hydrogen is difficult and expensive to store (flammable, needs high pressure or low temperature), hydrogen production often uses fossil fuels, fuel cells are expensive to manufacture.

Electroplating

Electroplating is a direct application of electrolysis. The setup:

• Cathode: the object to be plated.
• Anode: the plating metal.
• Electrolyte: a solution of a soluble salt of the plating metal.

The anode dissolves (plating metal atoms lose electrons and enter the solution as ions), and the ions are deposited on the cathode (ions gain electrons and become metal atoms on the surface of the object).

Example: To plate a steel fork with nickel:

• Cathode: the steel fork.
• Anode: a piece of nickel.
• Electrolyte: nickel sulfate solution.

Quantitative electrolysis (Extended)

Extended candidates may be asked to calculate the mass deposited or the volume of gas produced during electrolysis.

Key relationship: Charge (C) = current (A) x time (s). Faraday constant: 96500 C per mole of electrons. Moles of electrons = charge / 96500.

Then use the half-equation to convert moles of electrons to moles of product, and moles to mass (or volume for gases).

Worked example: A current of 2.0 A is passed through molten lead(II) bromide for 965 seconds. Calculate the mass of lead deposited. (Ar: Pb = 207)

1. Charge = 2.0 x 965 = 1930 C
2. Moles of electrons = 1930 / 96500 = 0.020 mol
3. Half-equation: Pb2+ + 2e- -> Pb. So 2 mol e- deposits 1 mol Pb.
4. Moles of Pb = 0.020 / 2 = 0.010 mol
5. Mass of Pb = 0.010 x 207 = 2.07 g

Worked exam question

Q (Paper 4): Two cells are set up. Cell A uses magnesium and copper electrodes in dilute sulfuric acid. Cell B uses zinc and copper electrodes in dilute sulfuric acid. (a) In Cell A, which metal is the negative terminal? (1) (b) Which cell produces the higher voltage? Explain your answer. (2) (c) State the direction of electron flow in the external circuit of Cell A. (1)

Model answer: (a) Magnesium (1). (b) Cell A produces the higher voltage (1) because the difference in reactivity between magnesium and copper is greater than between zinc and copper (1). (c) Electrons flow from the magnesium electrode to the copper electrode through the external wire (1).

The explanation mark in (b) is specifically for linking voltage to the reactivity difference. Stating “magnesium is more reactive” without comparing the gaps in reactivity earns no marks for the explanation.

Electrochemistry ties together electrolysis, reactivity, and redox. If half-equations or product prediction are causing consistent losses, a trial lesson can target the specific reasoning step where marks are dropping.