DSE Chemistry: Redox and Electrolysis — Tutor Lesson Plan (oxidation numbers, half-equations)

2026-10-28

Tutor lesson-prep guides · Secondary chemistry

Students hear oxidation and say gains oxygen, then freeze at permanganate. This lesson teaches redox as electron transfer plus oxidation number change, and makes them translate between the two languages.

Objectives — the student can assign oxidation numbers, balance a half-equation, and convert charge into deposited mass in electrolysis.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Ions	Difference between Fe2+ and Fe3+	One electron fewer
Moles	Particles in 0.0005 mol of KMnO4	3.01 x 10 to the 20
Charge	Units in Q = It	C = A x s

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Three definitions	Gain of oxygen, loss of electrons, rise in number	Show they agree
8-22 min	Oxidation numbers	Rules demonstrated for Mn, Cr and S	Get +7, +6, +6
22-38 min	Half-equations	Balance atoms, oxygen, hydrogen, charge	Write the permanganate half
38-52 min	Electrolysis	Q = It to moles of electrons to mass	Get 0.3175 g of copper
52-56 min	Titration link	Permanganate against iron(II), ratio 1 to 5	Get 0.0025 mol
56-60 min	Exit check	Three questions marked	Finish alone

03 Teaching script (read it out)

Scene 1: Teach all three definitions together

Oxidation is gain of oxygen, loss of electrons, or a rise in oxidation number; reduction is the reverse.
The oxidising agent is itself reduced, and that reversal is where most marks are lost.

Student states whether MnO4- to Mn2+ is oxidation and what the ion acts as.

Scene 2: Oxidation numbers: fix the known, solve the unknown

In KMnO4, K is +1 and O is -2, so with Mn as x: 1 + x – 8 = 0 and x = +7.
In SO4 2-: x – 8 = -2 and x = +6. A monatomic ion simply equals its charge.

Note: write let x equal, then the equation, then the answer, so the working marks are earned.

Scene 3: Electrolysis: charge is the bridge

Q = It. Half an amp for 1930 s gives 965 C, and 965 / 96 500 gives 0.01 mol of electrons.
Cu2+ plus two electrons gives copper, so only 0.005 mol deposits: 0.005 x 63.5 = 0.3175 g.

Student calculates the mass of silver deposited by the same charge (1.08 g).

04 Question types and worked steps

Type	Example	Steps	Answer
Oxidation number	Mn in KMnO4	Let x, sum to the ion charge	+7
Half-equation	MnO4- to Mn2+ in acid	Balance O with water, H with H+, charge with electrons	8H+ and 5e- on the left
Titration	permanganate against iron(II)	moles of KMnO4 times 5	0.0025 mol
Electrolysis	0.50 A for 1930 s on copper	Q to electrons, halve, times 63.5	0.3175 g

Sanity check (teach this): Charge must balance across a half-equation. If the left sums to -1 and the right to +2, electrons are missing.

05 Independent practice (3 questions)

Question	Answer	What to watch
Oxidation number of Mn in KMnO4	+7	Started from K +1 and O -2
Iron(II) in moles for 0.0200 M KMnO4, 25.0 cm3	0.0025 mol	Multiplied by 5
Copper from 0.50 A for 1930 s	0.3175 g	Divided by 2

06 Common mistakes and what to say

Mistake	Student writes	Say this
Oxidising agent called oxidised	MnO4- said to be oxidised	The agent is reduced; it oxidises the other species.
Ion charge ignored	S in SO4 2- written +4	The sum equals -2, not zero.
Electron count forgotten	copper written as 0.01 mol	Cu2+ needs two electrons, so divide by 2.
Titration ratio 1 to 1	iron(II) written as 0.0005 mol	One permanganate takes five electrons, so 1 to 5.

07 Homework (paper, section and question numbers)

DSE Chemistry practice paper (oxidation numbers, titration, electrolysis) — Section A (oxidation number of S in sulfate)
Secondary chemistry chapter index — All chapters

08 Exit check (last 5 minutes)

Question	Answer	Decision
Oxidation number of Mn in KMnO4	+7	All three right: move to cells
Ratio of permanganate to iron(II)	1 to 5	One wrong: re-teach half-equations
Copper from 0.50 A for 1930 s	0.3175 g	Two wrong: repeat scene 3

09 Tutor FAQ

Memorise the oxidation number rules?

Five rules cover it: element zero, oxygen -2, hydrogen +1, ion equals charge, sum matches.

Is a half-equation always acidic?

Only when the question says acidic; alkaline questions use hydroxide and water.

Only deposited mass?

Gas volumes appear too, and start the same way from moles of electrons.

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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.