DSE Chemistry: chemical cells and electrolysis (tutor lesson plan)

2026-11-03

Tutor lesson-prep guides · Secondary Chemistry

Students memorise positive and negative plates, then meet an electrolytic cell and lose everything. This lesson uses one clue: which way the electrons go.

Objectives — Students can: (1) use electron flow to decide which electrode is oxidised and which is reduced, and state how a cell differs from an electrolytic cell; (2) write two half equations and combine them with the electrons cancelled; (3) use Q = It and the Faraday constant to find the mass deposited, for example 0.30 g of copper from 0.50 A for 30 minutes.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Oxidation numbers	What does a rise in oxidation number mean?	Oxidation, loss of electrons
Ionic equations	How are two half equations combined?	Match the electrons, then add term by term
Moles and mass	How do you turn moles into mass?	Multiply by molar mass

02 Lesson flow (45 minutes)

Time	Stage	Tutor does	Student does
0-5 min	Two cells side by side	Draw a discharging cell and an electrolytic cell	State the electron flow
5-15 min	One clue: electron flow	From the negative electrode through the circuit	Both half equations correct
15-25 min	Half equations to overall	Balance electrons, then add	Electron counts equal
25-40 min	Charge calculation	0.50 A, 30 minutes, Q = 900 C	0.30 g of copper
40-45 min	Exit check	Mark live and correct live	Three out of three

03 Teaching script (read it out)

Scene 1: One clue: electrons leave the negative electrode and travel to the positive

In a discharging cell the more reactive metal is the negative electrode, it loses electrons and is oxidised; the electrons pass through the external circuit to the positive electrode where the oxidising agent receives them. An electrolytic cell reverses this, because the power supply pulls electrons away from the anode, so the anode is where oxidation happens.
Keep two words apart from the plates: the anode is always oxidation and the cathode always reduction, in both kinds of cell. What flips is which plate is positive. Students who memorise positive and negative will confuse themselves; students who ask which electrode is oxidised will not.
Draw the two cells side by side with one arrow for electron flow and the difference becomes visible.

Ask the student: in an electrolytic cell, which terminal of the supply does the anode connect to?

Scene 2: Balance the electrons before adding half equations

Three steps per half equation: write the species change; balance atoms with water and either hydrogen ions in acid or hydroxide ions in alkali; then balance charge with electrons.
Before adding the two half equations the electron counts must match, so multiply to the lowest common multiple. Skipping this is why half an electron appears in a student answer.
The only check that matters: total charge on the left equals total charge on the right, and no electrons survive in the overall equation.

Ask the student: why does the overall equation never contain electrons?

Scene 3: Charge first, then moles of electrons, then mass

Three fixed steps: Q = It, so 0.50 A times 30 times 60 s gives 900 C; moles of electrons is 900 divided by 96500, which is 0.0093 mol; copper is divalent, so moles of copper is 0.0047 mol and the mass is 0.0047 times 63.5, which is 0.30 g.
Reverse the question to make it stick: to deposit 1.00 g of copper needs 3039 C, which at 0.50 A takes about 101 minutes. Doing both directions shows the student what is being calculated.
Units decide most of the marks here: minutes become seconds, amperes are coulombs per second, mass is in grams, and every one of them is written beside the number.

Ask the student: if the current doubles, how long does the same mass take?

04 Question types and worked steps

Type	Example	Steps	Answer
Electron flow	Which electrode is oxidised	Find the negative plate or terminal, follow the electrons	Anode oxidation, cathode reduction
Half equation	Write the anode and cathode reactions	Balance atoms, balance charge, add electrons	Charge equal on both sides
Overall cell reaction	Combine two half equations	Match electrons first, then add	No electrons in the overall equation
Mass or time calculation	Find mass deposited or time needed	Q = It, divide by 96500, divide by electron number, multiply by molar mass	900 C gives 0.30 g

Sanity check (teach this): Check three numbers: 900 C, 0.0093 mol of electrons, 0.30 g of copper. The sanity test is that one mole of electrons, 96500 C, deposits half a mole of copper, about 31.75 g. Here only 0.0093 mol of electrons passed, so the mass must be far below 31.75 g, and 0.30 g is consistent. A result above 31.75 g means the division by two was skipped.

05 Independent practice (3 questions)

Question	Answer	What to watch
0.50 A for 30 minutes: how many coulombs pass?	900 C	Whether the 60 was applied
In copper sulfate solution, what mass of copper is deposited?	0.30 g	Whether the division by two was applied
How long at 0.50 A to deposit 1.00 g of copper?	about 101 minutes	Whether the answer was converted back through coulombs

06 Common mistakes and what to say

Mistake	Student writes	Say this
Anode and cathode confused with positive and negative	Writing that the anode in electrolysis is negative	Remember the oxidised electrode, not the sign. Oxidation is always at the anode.
Adding half equations before balancing electrons	Half an electron appears in the answer	The electron counts must match before you add; multiply if they do not.
Minutes left as minutes	Computing Q = 0.50 times 30	Amperes are per second, so the time must be in seconds.
Forgetting copper is divalent	Quoting 0.59 g as the mass	One copper ion needs two electrons, so skipping the division doubles the answer.

07 Homework (paper, section and question numbers)

DSE Chemistry electrochemistry practice paper — Questions 2 to 4: draw the electron flow arrow first, then both half equations, and show that the overall equation has no electrons left.
Same paper, calculation question — Question 6: go from 0.50 A and a time to a mass, then from a mass back to a time, with units at every step.

08 Exit check (last 5 minutes)

Question	Answer	Decision
What happens at the anode	Oxidation	Answering reduction means reteach the first script
Mass of copper from one mole of electrons	Half a mole, about 31.75 g	Answering 63.5 g means the valency was missed
Mass of copper from 900 C	about 0.30 g	Answering 0.59 g means the division by two was missed

09 Tutor FAQ

Must cells and electrolytic cells be memorised separately?

No. One rule covers both: oxidation at the anode, reduction at the cathode. Then follow the electrons.

Must the value 96500 be memorised?

Yes, and the student should know it is the charge on one mole of electrons, in coulombs per mole.

Does dilute copper sulfate behave like concentrated copper sulfate?

Not always. In dilute solution water may be oxidised instead, so follow the hint given in the question.

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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.