DSE Chemistry | Electrochemical and Fuel Cells Lesson-Prep Guide (for tutors) | half-equations, cell voltage, four common errors

2026-11-05

Tutor lesson-prep guides · Secondary Chemistry

Students can recite that oxidation happens at the negative electrode, yet freeze when asked which electrode releases electrons. This lesson treats the cell as an electron motorway: the negative electrode is the start, the wire is the road, the positive electrode is the finish, and the salt bridge only moves ions.

Objectives — Students can: (1) split a cell reaction into oxidation and reduction half-equations and state that electrons leave the negative electrode and travel through the external circuit to the positive one; (2) compute the zinc-copper cell voltage as E(positive) – E(negative) = 1.10 V and explain why the larger potential must be the positive electrode; (3) use Q = It and the Faraday constant to find the amount electrolysed (2.0 A for 30 minutes plates 1.18 g of copper), and explain that a fuel cell keeps running because reactants are supplied continuously.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Redox	Does oxidation gain or lose electrons?	It loses electrons (OIL RIG)
Standard electrode potential	E for Zn2+/Zn is -0.76 V. Does that mean zinc oxidises easily?	Yes, a more negative value means it loses electrons more readily
Moles and current	How many coulombs does one mole of electrons carry?	About 96500 C, the Faraday constant

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Prior check	Three quick items, especially the meaning of a negative potential	Answer orally, wrong ones go on the board corner
8-20 min	Split the half-equations	Separate Zn + Cu2+ into two lines with electrons	Students split a second cell reaction themselves
20-32 min	Voltage and electron flow	Show where 1.10 V comes from and which way electrons travel	Students draw electron and current arrows
32-44 min	Salt bridge and internal circuit	Explain ion migration and why a broken bridge kills the cell	Students state the bridge role in one sentence
44-56 min	Fuel cells and Faraday maths	Work the 1.18 g of copper example and the H2/O2 cell	Pairs check each other units
56-60 min	Exit check	Collect three items, mark at once, tag the act	Work independently

03 Teaching script (read it out)

Scene 1: Act 1: a cell is two half-reactions running side by side

The commonest error is memorising the overall equation, which leaves students unable to say what each electrode does. Write Zn + Cu2+ -> Zn2+ + Cu, then force two lines: Zn -> Zn2+ + 2e- (oxidation, negative electrode) and Cu2+ + 2e- -> Cu (reduction, positive electrode). Every line needs its electrons, and the counts must match or the cell cannot run.
Use the motorway picture: the two half-reactions are separate rooms, and electrons may only travel through the external wire from the negative electrode to the positive one. They cannot jump across the solution, so as zinc dissolves the electrons are pushed through the wire and the lamp lights. Ask what happens if the two metals touch: that is a short circuit, electrons take the shortcut and no useful work is done.
Put the misconception on the board: electrons do not leave the positive electrode, and conventional current points the opposite way to electron flow.

Students repeat: two half-equations, and electrons go from the negative electrode through the wire to the positive one.

Scene 2: Act 2: voltage is one potential minus the other

Act two is a single move: E(cell) = E(positive) – E(negative). The zinc-copper cell is +0.34 – (-0.76) = 1.10 V. Students should first write both potentials, then choose the electrode with the larger value as positive (reduction) and the smaller as negative (oxidation), so the subtraction always gives a positive number. A negative result means the polarity was reversed.
Make clear that a standard potential is measured against the hydrogen electrode, so it is not an energy that can be added; only subtraction has meaning. That is also why cell voltage does not depend on electrode size or solution volume under standard conditions: two 1.5 V cells in parallel still give 1.5 V, and only a series connection gives 3 V.
During practice, students fill four boxes at once: which electrode, oxidation or reduction, electron direction, current direction.

Students repeat: larger minus smaller, positive minus negative, and a negative answer means you swapped them.

Scene 3: Act 3: fuel cells and electrolysis use the same electron ledger

Act three extends the idea to two applications. In a hydrogen fuel cell, hydrogen is oxidised at the negative electrode (H2 -> 2H+ + 2e-) and oxygen is reduced at the positive one (O2 + 4H+ + 4e- -> 2H2O), giving 2H2 + O2 -> 2H2O with a theoretical 1.23 V. The key difference is that reactants flow in and products flow out, so the cell keeps supplying power, whereas an ordinary cell dies when its reactants are used up. Energy follows: change in G = -nFE = -2 x 96500 x 1.23, about -237 kJ per mole, matching the textbook value, which shows voltage and energy are the same coin.
Then the electrolysis calculation: 2.0 A for 30 minutes gives Q = It = 3600 C; electrons are 3600 / 96500 = 0.0373 mol; Cu2+ + 2e- -> Cu means two moles of electrons plate one mole of copper, so 0.0187 mol, which is 0.0187 x 63.5 = 1.18 g. One operation per line, units written every time, because skipped steps are where the marks vanish.

Students repeat: charge first, then divide by Faraday, then use the electron ratio.

04 Question types and worked steps

Type	Example	Steps	Answer
Split half-equations	Write the two half-equations for a Zn/Cu cell	1 write reactants and products 2 balance atoms 3 add electrons until charge balances	Zn -> Zn2+ + 2e-; Cu2+ + 2e- -> Cu
Compute cell voltage	Voltage of a Zn2+/Zn and Cu2+/Cu cell	1 write both potentials 2 subtract larger minus smaller 3 unit V and label the positive electrode	1.10 V
Electroplate copper	How much copper is plated by 2.0 A for 30 minutes	1 Q = It 2 n(e-) = Q/F 3 divide by 2 and multiply by 63.5	1.18 g
Compare with a fuel cell	How a hydrogen fuel cell differs from a zinc-copper cell	1 are reactants supplied continuously 2 are products removed 3 is it reversible	Continuous supply, theoretical 1.23 V

Sanity check (teach this): After any voltage, ask whether it is plausible: real cells sit between 1 and 2 V, so 1.10 V is fine, while 3 V or more usually means the two potentials were added. Plating 1.18 g from 3600 C matches 2.0 A for half an hour; tens of grams means the factor of two or the Faraday value was dropped. Efficiency is always below 100 per cent, so the electrical energy cannot exceed the combustion energy of the hydrogen used.

05 Independent practice (3 questions)

Question	Answer	What to watch
Write the two half-equations for a Zn/Cu cell and label the electrodes.	Zn -> Zn2+ + 2e- (negative); Cu2+ + 2e- -> Cu (positive)	Are electrons added to both lines and the polarity labelled
A current of 2.0 A passes through copper sulphate solution for 30 minutes. Find the cathode mass gain.	1.18 g	Did they use Q = It and divide by two electrons
What is the positive-electrode reaction in a hydrogen fuel cell, and why does it not run out?	O2 + 4H+ + 4e- -> 2H2O, with reactants supplied continuously	Are electrons shown and continuous supply explained

06 Common mistakes and what to say

Mistake	Student writes	Say this
Electrons leave the positive electrode	Answering that electrons travel from copper to zinc	The negative electrode is the oxidation one, so electrons start there and current runs the other way.
Adding the two potentials	-0.76 + 0.34 = -0.42 V	Voltage is a subtraction. Standard potentials are relative, so adding them means nothing.
Forgetting the electron count	Treating 0.0373 mol of electrons as 0.0373 mol of copper	Cu2+ needs two electrons, so the mole count of electrons is halved.
Thinking a broken salt bridge just weakens the cell	Answering that the current merely drops a little	Without the bridge, charge cannot be balanced, so the cell stops and the current is zero.

07 Homework (paper, section and question numbers)

Redo this chapter — Redo practice items 1 to 3, writing both half-equations and units beside each
Chemistry chapter index — Open the redox chapter and do question 2, labelling electron flow
Electrolysis table — Build a three-step Q to n(e-) to n(metal) table and paste it in your notes

08 Exit check (last 5 minutes)

Question	Answer	Decision
Which electrode releases electrons	The negative one, through the external circuit	Saying positive means replay Act 1
How is cell voltage found	E(positive) – E(negative), larger minus smaller	Adding them means replay Act 2
Why does a fuel cell keep running	Reactants are supplied and products removed continuously	Answering that it is bigger means replay Act 3

09 Tutor FAQ

Must standard potentials be memorised?

The paper supplies the data. The skill is choosing the larger value as positive, not recalling numbers.

What can replace a salt bridge?

Filter paper soaked in electrolyte does the same job: it lets ions migrate and keeps the solutions neutral.

Why is a real fuel cell below 1.23 V?

Internal resistance and activation overpotential turn part of the energy into heat, so the terminal voltage is lower.

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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.