DSE Chemistry: Electrolysis and Industrial Applications (Tutor Lesson Plan)

DSE Chemistry: Electrolysis and Industrial Applications (Tutor Lesson Plan) | Faraday maths and plating thickness

2026-11-04

Tutor lesson-prep guides · Secondary Chemistry

Students can recite oxidation at the anode and reduction at the cathode, then freeze at a question such as how many grams of copper 2 A deposits in half an hour. Split Faraday into three steps and the question collapses.

Objectives — Students can write both half-equations for a given cell, find a deposited mass with the three-step Faraday method, correct it with current efficiency, and explain the industrial design of plating and refining.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Moles and mass	Mass of one mole of copper	63.55 g
Redox	Is Cu2+ to Cu oxidation or reduction	Reduction, gains two electrons
Charge and current	Charge in 60 seconds at 1 A	60 C

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Electrode reactions	Anode oxidises, cathode reduces	Write three cells
8-20 min	Three-step Faraday	Q=It, n(e)=Q/F, divide by valency	1.19 g of copper
20-32 min	Current efficiency	Actual over theoretical	1.09 g at 92 percent
32-44 min	Plating and refining	Thin even coating, impure anode	About 6.1 micrometres
44-54 min	Energy and cost	E = QV	12.6 kJ
54-60 min	Exit check	Three questions marked at once	Independent

03 Teaching script (read it out)

Scene 1: Electrode reactions: cell first, then electrode

Inert electrodes in copper sulphate: cathode Cu2+ plus 2e- gives Cu, anode water gives oxygen, protons and electrons, so the solution turns acidic.
Copper electrodes: the anode dissolves and the cathode plates, so the concentration stays the same. That is exactly how electrorefining works.
Brine: hydrogen and hydroxide at the cathode, chlorine at the anode, so the cathode region turns alkaline.

Tutor move: write cell on the left of the board and electrode on the right, and insist students name the cell before naming the electrode, so the two sets of reactions never get mixed.

Scene 2: Three-step Faraday: turning charge into moles

Step one, Q = It = 2.0 times 1800 = 3600 C.
Step two, moles of electrons = 3600 divided by 96485 = 0.0373 mol.
Step three, copper is divalent, so moles of copper = 0.0187, mass = 0.0187 times 63.55 = 1.186 g.
At 92 percent efficiency the real mass is 1.091 g.

Ask the student: for the same charge, how many times heavier is the copper deposit than an aluminium one (about 3.5 times)?

Scene 3: Plating, refining and cost

Plating: the object is the cathode, the plating metal is the anode, and the electrolyte holds that metal ion. Use a small current for a long time so the coat is even.
Thickness: 1.091 g of copper is 1.217e-07 cubic metres, which over 0.02 square metres is 6.09 micrometres.
Refining: impure anode, pure cathode strip, impurities fall as anode sludge that still contains gold and silver.
Energy: E = QV = 12.6 kJ, and cutting voltage saves more money than cutting current.

Reminder: the most common plating question asks which electrode is the anode, and the answer is the plating metal, never the object being plated.

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Faraday mass	Q=It, moles of electrons, divide by valency	1.09 g
Calculation	Coating thickness	Mass over density over area	About 6.1 micrometres
Explanation	Why plating uses a small current	Slow growth gives a fine even coat	Large current gives a rough coat
Decision	Electrodes in refining	Impure anode, pure cathode	Electrolyte holds copper ions

Sanity check (teach this): A mass larger than 1.2 g means the valency division was skipped. Charge divided by 96500 gives moles of electrons, not moles of metal. Efficiency is below one, so the real mass can never exceed the theoretical value.

05 Independent practice (3 questions)

Question	Answer	What to watch
Charge passed by 2.0 A for 30 minutes	3600 C	Q = It with seconds
Theoretical copper from that charge	1.186 g	Divide by two, then multiply by 63.55
Coating thickness at 92 percent efficiency	About 6.1 micrometres	Mass, volume, then thickness

06 Common mistakes and what to say

Mistake	Student writes	Say this
Skipping the valency division	Answering 2.37 g	“Divide the moles of electrons by the ion charge; copper is divalent.”
Using minutes in Q=It	Substituting 30 for time	“Current is per second, so convert the time to seconds first.”
Swapping the electrodes	Saying the object is the anode	“The object is the cathode, where ions gain electrons.”
Adding efficiency	Answering 2.28 g	“Efficiency is a discount, so multiply the theoretical mass by it.”

07 Homework (paper, section and question numbers)

DSE chemistry practice paper (electrolysis and redox) — Section A calculations and Section B plating explanations
Secondary chemistry chapter index — All chapters

08 Exit check (last 5 minutes)

Question	Answer	Decision
Charge for 2.0 A for 30 minutes	3600 C	All three right: move to energy
Theoretical copper mass at valency two	1.19 g	One wrong: reteach Faraday next lesson
Which electrode is the plated object	The cathode	Two wrong: restart act one

09 Tutor FAQ

Must students memorise 96500?

Yes, DSE uses 96500 C per mole, and either value works as long as it is used consistently.

Why add sulphuric acid when electrolysing water?

It carries the current; sulphate ions do not react at the electrodes.

Why does refining keep the concentration constant?

The anode dissolves exactly the moles the cathode deposits.

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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.