Senior Chemistry | Electrochemistry (Cells and Electrolysis) Prep Guide (for tutors) | Potential difference, electron flow and Faraday in three steps

2026-11-06

Tutor lesson-prep guides · Secondary Chemistry

Students can recite that the anode is oxidised, but stall when asked which way the electrons travel or what the salt bridge does, and forget to divide by the number of electrons when calculating electrolysis. This lesson follows one electron path through cells and electrolysis.

Objectives — Students can: (1) identify the electrodes, half-equations and electron flow in a chemical cell and explain the salt bridge; (2) calculate cell voltage from two standard electrode potentials and use dG = -nFE to judge spontaneity; (3) use charge, the Faraday constant and moles of electrons to find the mass and gas volume produced at the electrodes.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Redox	Is Zn to Zn2+ plus 2e- oxidation or reduction	Oxidation, electrons lost
Faraday constant	How much charge one mole of electrons carries	96 485 C
Electrode potential	What a more negative E value means	Loses electrons more easily, stronger reducing agent

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Opening diagnostic	Ask which electrode is negative in a zinc-copper cell and which way electrons flow	Hands up, then draw arrows
8-22 min	The electron path	Zinc oxidised, copper ions reduced, voltage 1.10 V, salt bridge balances charge	Draw the path once
22-36 min	Voltage and free energy	dG = -2 x 96 485 x 1.10 = -212.3 kJ/mol	Substitute once themselves
36-50 min	Faraday calculation	2.0 A x 1800 s = 40 C, copper 1.185 g, chlorine 448 cm3	Calculate once themselves
50-56 min	Fuel cells and everyday batteries	Compare primary, rechargeable and fuel cells	State one difference
56-60 min	Exit check	Collect and mark three items	Work independently

03 Teaching script (read it out)

Scene 1: Separate oxidation and reduction, then draw the electron path

In a zinc-copper cell the zinc strip is the negative electrode where oxidation happens: Zn to Zn2+ plus 2e-. The copper strip is the positive electrode where reduction happens: Cu2+ plus 2e- to Cu.
Electrons travel through the external circuit from zinc to copper, so conventional current runs the other way. Inside the solution, ions carry the charge: cations move towards the positive electrode side through the salt bridge and anions move towards the negative side.
Give students one line to hold on to: in a cell that is discharging, the anode is the negative electrode, because the anode is where oxidation happens.

Ask the student: what happens to the bulb if the salt bridge is removed, and why? If they cannot answer, go back to charge balance.

Scene 2: The voltage is just the potential difference

E cell = E cathode minus E anode = 0.34 minus (-0.76) = 1.10 V. A positive voltage means the reaction is spontaneous under standard conditions.
Quantify it with dG = -nFE: n = 2, F = 96 485 C/mol, so dG = -2 x 96 485 x 1.10 = -212267 J, that is -212.3 kJ/mol. The negative sign agrees with spontaneity.
Swap the two electrodes and the voltage becomes -1.10 V with a positive dG, so the reaction is no longer spontaneous. This reversal is a favourite exam question.

Ask the student: replace zinc with silver (E = +0.80 V) and state how the voltage and spontaneity change.

Scene 3: Faraday law: turn current into product

A current of 2.0 A for 1800 seconds gives Q = It = 40 C. Moles of electrons = Q divided by F = 40 divided by 96 485 = 0.0373 mol.
Copper at the cathode follows Cu2+ plus 2e- to Cu, so 2 mol of electrons give 1 mol of copper: 0.0187 mol, mass = 0.0187 x 63.5 = 1.185 g.
Chlorine at the anode follows 2Cl- to Cl2 plus 2e-, so also 0.0187 mol. At room temperature the molar volume is 24.0 dm3, giving 0.448 dm3, that is 448 cm3.

Ask the student: double the current and halve the time, then explain why the yield is unchanged (the charge is the same).

04 Question types and worked steps

Type	Example	Steps	Answer
Identify electrodes and electron flow	Zinc-copper cell	Write both half-equations, then mark the oxidation side	Zinc negative, electrons zinc to copper
Cell voltage	E(Cu2+/Cu) = 0.34, E(Zn2+/Zn) = -0.76	Cathode minus anode	1.10 V
Free energy change	Same cell, n = 2	dG = -nFE, convert J to kJ	-212.3 kJ/mol
Electrolysis yield	2.0 A, 1800 s, copper sulfate solution	Q = It, divide by F, divide by n	Copper 1.185 g, chlorine 448 cm3

Sanity check (teach this): Sanity check: the voltage 1.10 V is positive, consistent with copper having the higher E value; dG of -212.3 kJ/mol is negative, so both routes agree. Moles of electrons 0.0373 is below 0.05 mol, so the copper mass must be below 0.05 x 63.5, about 3.2 g, and 1.185 g fits. The chlorine volume 448 cm3 matches moles times 24 000 cm3 per mol exactly.

05 Independent practice (3 questions)

Question	Answer	What to watch
Electrodes and electron flow, three items	Zinc-copper, iron-copper and magnesium-zinc	Does the student confuse anode with positive electrode
Voltage and free energy, two items	Different electrode pairs, one asking about spontaneity	Are the minus sign and n dropped
Faraday calculations, three items	Different current and time, one asking for gas volume	Is the division by n forgotten

06 Common mistakes and what to say

Mistake	Student writes	Say this
Anode and positive electrode swapped	Writing that the positive electrode is the anode where oxidation happens	On discharge the anode is the oxidation side, which is the negative electrode in a cell. Write the half-equations first and the labels follow.
Free energy sign dropped	Writing dG = +212.3 kJ/mol	dG = -nFE, so a positive voltage must give a negative dG; the minus sign means energy is released.
Dividing by n forgotten	Using 0.0373 mol of electrons as 0.0187 mol of copper	Copper is divalent, so 2 mol of electrons give 1 mol of copper and you must divide by n; silver is monovalent and needs no division.
Units not converted	Treating mA as A or minutes as seconds	Convert to A and seconds before multiplying, then convert the answer to grams or cm3.

07 Homework (paper, section and question numbers)

Secondary chemistry chapter index — All chapters at a glance
Prep guide master index — Compare similar chapters
Redo this chapter — Two voltage items and three Faraday items

08 Exit check (last 5 minutes)

Question	Answer	Decision
In a zinc-copper cell, which metal do electrons leave and which do they reach	From zinc to copper	Answering the reverse sends you back to act one
If E cell is 1.10 V and n = 2, what is dG	-212.3 kJ/mol	A positive value sends you back to act two
How many moles of copper form when 40 C passes	0.0187 mol	Answering 0.0373 means dividing by 2 was missed

09 Tutor FAQ

Do cells and electrolytic cells use the same electrode names?

The names are the same but the roles reverse: while a cell discharges the negative electrode is the anode, whereas in electrolysis with an external supply the positive electrode is the anode. First ask whether there is an external power source.

Must students memorise the E table?

No, only a few common values are useful, such as zinc -0.76, copper +0.34 and silver +0.80. Exams supply the data; the skill is using cathode minus anode correctly.

What is the most common Faraday mistake?

Forgetting to divide by the number of electrons. Write the half-equation first, read the electron coefficient, then divide.

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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.