IB Chemistry SLTopic 6 — Electron TransferPaper 1 & 2Core idea~14 min read
Primary Cells
Drop zinc into copper sulfate and the electrons jump straight across, releasing their energy as heat. Separate the two halves into different beakers and the electrons have to go the long way round — through a wire, where you can put them to work.
📚 What you need to know
A voltaic (galvanic) cell converts the energy of a spontaneous redox reaction into electricity.
Each half of the reaction happens in its own half-cell, joined by a wire and a salt bridge.
Oxidation at the anode, reduction at the cathode. In a voltaic cell the anode is negative.
Electrons flow through the wire from anode to cathode; ions flow through the salt bridge to keep both solutions neutral.
The potential difference produced is the cell potential or EMF.
Cell diagram convention: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s), oxidation on the left.
Primary cells cannot be recharged; a fuel cell runs continuously while fuel is supplied.
A half-cell on its own
Dip a strip of metal into a solution of its own ions and two opposite processes start immediately. Atoms leave the rod as ions, abandoning their electrons on the metal; ions from solution collect electrons and deposit as atoms. An equilibrium is reached, and where it lies decides whether the rod ends up slightly negative or slightly positive.
Nothing useful happens here. No overall reaction takes place — there is only a potential difference between the rod and the solution, waiting for somewhere to go.
You cannot measure a single electrode potential, only a difference between two. It is like asking how strong an arm-wrestler is: the question only has an answer once you sit them opposite someone else.
Connecting two half-cells
Zinc holds its electrons more loosely than copper does. Join a zinc half-cell to a copper half-cell with a wire and the electrons take their chance: they flow from the zinc, round the external circuit, and onto the copper electrode, where Cu2+ ions are waiting to collect them.
Watch it run for long enough and the zinc electrode visibly thins while the copper one grows. The blue of the right-hand solution fades as Cu2+ is used up.
The parts and what each does
Part
What it does
Detail worth quoting
Anode
where oxidation happens
negative in a voltaic cell, because electrons pile up on it
Cathode
where reduction happens
positive in a voltaic cell, because electrons are drawn away from it
External wire
carries the electrons
they travel from anode to cathode
Salt bridge
completes the circuit
an inert electrolyte such as KNO3; anions move to the anode, cations to the cathode
Voltmeter
measures the EMF
high resistance, so it draws almost no current
The salt bridge is the part most often dismissed as decoration, and it is the part that makes the whole thing work. Without it, the left-hand solution would rapidly build up positive Zn2+ ions and the right-hand one would be left with excess negative sulfate. That charge separation would stop the electron flow within moments.
Potassium nitrate or potassium chloride are used because nitrates and chlorides are almost always soluble. A salt bridge that formed a precipitate in either half-cell would disturb the equilibrium it is supposed to be leaving alone.
Two mnemonics, and you need both. AN OX: anode is oxidation. RED CAT: reduction at the cathode. These hold in every cell. What flips is the polarity: the anode is negative in a voltaic cell but positive in an electrolytic one, which is the topic of a later page.
Writing a cell diagram
Rather than draw the apparatus every time, chemists use a shorthand. Once you know the four conventions it reads as easily as an equation.
Written the other way round, Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s), the same cell has E = –1.10 V. The negative sign tells you the right-hand electrode is the negative one.
Primary cells, and one that never runs down
A primary cell is one whose reaction cannot be reversed by charging: once a reactant is used up, the cell is finished. Ordinary alkaline batteries are primary cells, and so is the zinc–copper cell above.
A fuel cell escapes the problem entirely by having its reactants delivered continuously rather than stored inside. In the hydrogen–oxygen cell, hydrogen is oxidised at one electrode and oxygen is reduced at the other.
no carbon dioxide, and no nitrogen oxides because there is no high-temperature combustion
Efficient
chemical energy goes to electrical energy directly, rather than through heat
Runs continuously
energy is not stored in the cell; it lasts as long as the fuel supply
Storage is difficult
hydrogen is highly flammable and has a low energy density by volume, so tanks are heavy
The hydrogen has to come from somewhere
most is currently made from fossil fuels, which offsets the environmental gain
WORKED EXAMPLE
A voltaic cell is made from a magnesium electrode in magnesium sulfate solution and a copper electrode in copper(II) sulfate solution. (a) Write the two electrode half-equations. (b) State the polarity of each electrode and the direction of electron flow. (c) Write the cell diagram.
(a) which metal is oxidised?Magnesium is well above copper in the reactivity series, so magnesium loses electrons.Mg → Mg²⁺ + 2e⁻Cu²⁺ + 2e⁻ → Cu(b) polarity and flowMg is the negative anode; Cu is the positive cathodeElectrons flow through the external wire from the magnesium to the copper.(c) cell diagramMg(s) | Mg²⁺(aq) || Cu²⁺(aq) | Cu(s)Oxidation on the left, so magnesium goes first. The double line is the salt bridge.
WORKED EXAMPLE
An aluminium electrode is connected to a zinc electrode. The voltmeter reads 0.94 V and the aluminium is the negative electrode. Write the conventional cell diagram, including the cell potential.
Step 1 — what does “negative” tell you?Electrons pile up on the negative electrode, so that is where oxidation is happening. Aluminium is the anode.Step 2 — place it on the leftThe oxidation half-cell is always written first, and aluminium forms Al³⁺.Al(s) | Al³⁺(aq) || Zn²⁺(aq) | Zn(s) E = +0.94 VThe sign is positive because the right-hand electrode, zinc, is the positive one. Write it the other way round and E becomes −0.94 V.
WORKED EXAMPLE
Explain what would happen to the reading on the voltmeter if the salt bridge were removed from a working voltaic cell.
What the salt bridge is doingIt completes the circuit and lets ions move so that neither solution builds up a net charge.Without itThe anode half-cell would accumulate positive metal ions and the cathode half-cell would be left with excess negative ions.the reading falls to zero almost immediatelyThe charge build-up opposes further electron flow, so the reaction stops. Nothing is wrong with the chemistry — the circuit is simply broken.
💡 Exam tip
Use the reactivity series to decide which electrode is the anode before you write anything else.
Say “electrons flow through the external circuit from anode to cathode”. Electrons never travel through the solution.
In a voltaic cell, the anode is negative. Learn it as a pair with the electrolytic case to avoid mixing them up.
When explaining the salt bridge, mention completing the circuit and balancing the charge — both are needed.
In a cell diagram, put oxidation on the left and use || for the salt bridge.
For a fuel cell, remember the cell operates continuously and stores no energy of its own.
⚠️ Common mix-up
Assuming the anode is always positive. In a voltaic cell it is negative.
Having electrons travel through the salt bridge. Ions move there, not electrons.
Writing the cell diagram backwards and then quoting a positive E anyway.
Saying the salt bridge “lets electricity through” without explaining the charge balance.
Calling a fuel cell rechargeable. It is refuelled, which is not the same thing.
Claiming hydrogen fuel cells are entirely clean without mentioning how the hydrogen was produced.
Up next: Secondary Cells — a primary cell dies when a reactant runs out. Run the same chemistry backwards with an external voltage and you have a battery you can use again.
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