DSE Chemistry — Electrolysis and Chemical Cells Tutor Prep Guide | Charge, discharge order, four calculations

2026-11-03

Tutor lesson-prep guides · Secondary Chemistry

Students can recite that the cathode gains metal and the anode gives oxygen, but freeze when a current and a time are given. This lesson builds one chain: current times time gives charge, charge divided by 96500 gives moles of electrons, and the half-equation ratio turns that into mass or gas volume. At 2.0 A for thirty minutes, 1.18 g of copper is deposited.

Objectives — Students can: (1) convert charge to moles of electrons with Q = It and n = Q/96500 (2.0 A for 1800 s gives 3600 C and 0.0373 mol); (2) use half-equation ratios to get deposited mass and gas volume (1.18 g of copper, 224 cm3 of oxygen); (3) use discharge order to name the products when copper sulphate solution is electrolysed, and state how a chemical cell differs from an electrolytic cell in energy direction.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Mole relation	How much charge does one mole of electrons carry?	96500 C
Half-equation	How many electrons reduce Cu2+ to copper?	Two
Molar volume	What volume does one mole of gas occupy at room conditions?	About 24 dm3
Redox	Does reduction happen at the cathode or the anode?	At the cathode, where electrons are gained

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0–8 min	Open with the chain	Write Q = It and n = Q/F	Compute 2 A for 30 minutes themselves
8–20 min	Charge to mass	Copper deposit follows the half-equation ratio	Get 1.18 g and compare with 1.09 g measured
20–32 min	Discharge order	Compare standard electrode potentials of four ions	Name both products with reasons
32–44 min	Cell versus electrolysis	The Daniell cell is spontaneous; electrolysis is driven	Draw electron flow and the energy direction
44–52 min	Question practice	Two calculations, one explanation	Eight minutes working alone
52–60 min	Exit check	Mark three answers on the spot	Work alone

03 Teaching script (read it out)

Scene 1: Charge is not voltage: multiply by time

Students see 2 A and think the mass is decided. Current only gives a rate. 2.0 A for 1800 seconds is 3600 C; divided by 96500 that is 0.0373 mol of electrons. Require three lines before any mass calculation: I, t, Q. Miss one and the answer cannot be marked.

Students should explain why halving the time halves the mass at the same current.

Scene 2: Ratios come from the half-equation, not from dividing by charge

Cu2+ + 2e- gives Cu, so 0.0373 mol of electrons deposits only 0.0187 mol of copper, which is 1.18 g. At the other electrode, 4e- release 1 mol of O2, a volume of 224 cm3. The same charge gives different mole counts on the two sides, and only the coefficients explain why. The measured gain of 1.09 g is an efficiency of 92.0 per cent; the rest went to side reactions and losses.

Students should explain why copper and oxygen differ by a factor of four in moles.

Scene 3: A cell and an electrolytic cell are one reaction, two directions

The Daniell cell gives 1.10 V because zinc oxidation and copper ion reduction are spontaneous; electrons flow from zinc to copper. Electrolysis is pushed the other way by an external supply, which forces electrons onto the cathode. Answers must separate spontaneous from driven: a cell converts chemical energy to electrical, an electrolytic cell does the reverse.

Students should say when the same pair of half-reactions acts as a cell and when as electrolysis.

04 Question types and worked steps

Type	Example	Steps	Answer
Charge calculation	2.0 A for 30 minutes: find the moles of electrons passed	1. Q = It 2. Convert to seconds 3. n = Q/F	3600 C, 0.0373 mol of electrons
Deposited mass	Same conditions in copper sulphate solution: cathode gain	1. Write the half-equation 2. Electrons to copper is 2:1 3. Multiply by molar mass	1.18 g theoretical
Gas volume	Same conditions: oxygen at the anode at room conditions in cm3	1. 4e- gives 1 mol O2 2. Multiply by 24 dm3 3. Convert to cm3	224 cm3
Explanation	Why is the measured copper gain below the theoretical value?	1. Compare the two values 2. Name a side reaction 3. Name a physical loss	1.18 g theoretical against 1.09 g, 92.0 per cent efficiency

Sanity check (teach this): Checks: 3600 C over 96500 is always of order 0.037 mole of electrons; an answer of 37 mol means minutes were never converted to seconds. Copper at 63.5 g per mole cannot give a gain near 12 g from this charge, so such an answer must be rejected. The oxygen volume must be two orders of magnitude larger in number than the copper mass (224 cm3 against 1.18 g), because 24 dm3 is a large volume.

05 Independent practice (3 questions)

Question	Answer	What to watch
Charge from 2.0 A for 30 minutes	3600 C	Did they multiply minutes by 60
Moles of electrons	0.0373 mol	Did they divide by 96500
Theoretical copper gain	1.18 g	Is the ratio 2:1 for electrons to copper
Oxygen volume	224 cm3	Is the ratio 4:1 for electrons to oxygen

06 Common mistakes and what to say

Mistake	Student writes	Say this
Minutes treated as seconds	Q = 2 x 30 = 60 C	Time in Q = It is always in seconds; 30 minutes is 1800 s, and that factor of 60 is the whole error.
Electrons to metal one to one	0.0373 mol of electrons gives 0.0373 mol of copper	Read the coefficient: Cu2+ takes two electrons, so copper is half the moles of electrons.
Anode described as reduction	The anode gains electrons so metal forms there	The anode is oxidation and loses electrons; metal forms at the cathode. This is a definition, not a fact to memorise loosely.
Cell and electrolytic cell confused	During electrolysis electrons flow spontaneously through the solution	Electrolysis is driven by the supply; spontaneity belongs to the cell, not the electrolytic cell.

07 Homework (paper, section and question numbers)

Chemistry subject index — Revise the redox chapter on electron transfer, then run the three-step charge chain
Electrolysis (S5 chapter) — Redo questions 2 and 3 on that page and check the discharge order
Chemical cells (S5 chapter) — Draw the electron flow in the Daniell cell and in an electrolytic cell

08 Exit check (last 5 minutes)

Question	Answer	Decision
Charge for 2.0 A and 1800 s	3600 C	Error: using 30 as seconds
Moles of electrons	0.0373 mol	Error: dividing by 9650
Theoretical copper mass	1.18 g	Error: 1:1 electron to copper
Efficiency at 1.09 g measured	92.0 per cent	Error: dividing theoretical by measured

09 Tutor FAQ

Must every coefficient be memorised?

No, but four must be instant: two electrons per copper, one per silver, four per oxygen molecule and two per hydrogen molecule. Everything else follows from the half-equation on the page.

How many reasons does an efficiency question need?

Two: one side reaction, such as hydrogen from water, and one physical loss, such as plating that flakes off. Three versions of the same reason earn nothing extra.

Can the formula replace the half-equation?

For the calculation, yes; for the explanation, no. When the question asks why, the marks sit in the half-equation and the ratio.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.