IB Chemistry HLTopic 6 — Electron TransferPaper 1 & 2Core idea~12 min read
Electrolysis of Aqueous Solutions
Electrolysing a molten salt is easy to predict: there are only two ions, so metal goes to the cathode and non-metal goes to the anode. Add water and it gets interesting, because water can be reduced and oxidised. Now there is a competition at each electrode, and you have to work out who wins.
📚 What you need to know
In solution, water is always a competitor at both electrodes.
At the cathode, the species with the more positive Eθ is reduced.
At the anode, the species with the more negative Eθ is oxidised (it is the easiest to strip electrons from).
Three things decide the products: Eθ values, concentration, and the electrode material.
Reactive metals such as Na, K, Mg and Al are never produced from aqueous solution — hydrogen comes off instead.
Active electrodes take part in the reaction; passive (inert) electrodes such as graphite and platinum do not.
The two water half-equations
Learn to recognise these two. They turn up in every aqueous electrolysis question.
Water being reduced (cathode)
2H2O(l) + 2e− → H2(g) + 2OH−(aq) Eθ = −0.83 V
Water being oxidised (anode)
2H2O(l) → O2(g) + 4H+(aq) + 4e− Eθ = +1.23 V
Notice what each one leaves behind. Reducing water makes the solution around the cathode alkaline; oxidising water makes the solution around the anode acidic. If a question mentions a pH change or a colour change in universal indicator, this is why.
Who wins at each electrode
Same idea at both ends, opposite direction. The cathode wants whichever species holds on to electrons most eagerly; the anode wants whichever gives them up most easily.
Why sodium never appears
Compare the two candidates at the cathode in sodium chloride solution:
Na+(aq) + e− → Na(s) Eθ = −2.71 V
2H2O(l) + 2e− → H2(g) + 2OH−(aq) Eθ = −0.83 V
Water is nearly two volts more positive, so water is reduced every single time. This is exactly why sodium has to be extracted from molten sodium chloride, not from brine.
The three factors, one at a time
1. Standard electrode potentials
The starting point. Look both candidates up, compare, pick the winner using the rule above.
2. Concentration
When two Eθ values are close, concentration decides it. Chloride at −1.36 V and water at −1.23 V are only 0.13 V apart, so:
Dilute sodium chloride → mostly oxygen at the anode
Concentrated brine → mostly chlorine at the anode
The bit the tables do not tell you: oxygen also suffers from a large overpotential — it needs noticeably more voltage than its Eθ suggests before it will actually form. That extra hurdle is a second reason chlorine wins in concentrated brine, and it is worth a sentence if a question asks why the prediction from Eθ alone is unreliable.
3. The electrode itself
Passive (inert) electrodes — graphite, platinum. They only carry current.
Active electrodes — copper, silver, nickel. The electrode metal itself can be oxidised at the anode, and it is usually easier to oxidise than water.
Products you should be able to predict
Electrolyte
At the cathode
At the anode
Why
Water (plus a little acid)
Hydrogen
Oxygen
Only water is available at both electrodes; the acid just carries the current
Dilute sodium chloride
Hydrogen
Oxygen
Na+ is far too negative; dilute chloride loses to water
Concentrated sodium chloride
Hydrogen
Chlorine
High Cl− concentration plus the oxygen overpotential tips it
Copper sulfate, graphite electrodes
Copper
Oxygen
Cu2+ at +0.34 V beats water; sulfate cannot be oxidised further
Copper sulfate, copper electrodes
Copper
The anode dissolves
Copper metal is easier to oxidise than water, so the electrode goes instead
Sulfate is a dead end and it is worth knowing why: the sulfur in SO42− is already at +6, its highest oxidation state. It cannot lose any more electrons, so it just sits there as a spectator. Nitrate behaves the same way.
Purifying copper: active electrodes at work
The copper(II) concentration barely changes: every ion that plates out at the cathode is replaced by one dissolving at the anode. The anode slime is valuable enough to pay for the electricity.
Worked examples
WORKED EXAMPLE
Predict the products of electrolysing dilute sodium chloride solution with graphite electrodes, and write the half-equations.
Step 1: List the candidates at the cathodeNa+ (−2.71 V) against water (−0.83 V). Water is far more positive.2H2O(l) + 2e− → H2(g) + 2OH−(aq)Step 2: List the candidates at the anodeCl− (−1.36 V) against water (−1.23 V). They are close, but the solution is dilute, so water wins.2H2O(l) → O2(g) + 4H+(aq) + 4e−Hydrogen at the cathode, oxygen at the anodethe sodium and chloride ions are just spectators here
WORKED EXAMPLE
Concentrated brine is electrolysed. State the products and give the overall equation.
Step 1: Cathode is unchanged2H2O(l) + 2e− → H2(g) + 2OH−(aq)Step 2: Anode flips because chloride is now concentrated2Cl−(aq) → Cl2(g) + 2e−Step 3: Combine, remembering the Na+ left behind with the OH−2NaCl(aq) + 2H2O(l) → 2NaOH(aq) + H2(g) + Cl2(g)this is the chlor-alkali process — three useful products from salt water
WORKED EXAMPLE
Copper(II) sulfate solution is electrolysed with graphite electrodes. Explain what you would see and what happens to the pH.
Step 1: CathodeCu2+ is +0.34 V, water is −0.83 V. Copper wins easily.Cu2+(aq) + 2e− → Cu(s) — a brown coating formsStep 2: AnodeSulfate cannot be oxidised (sulfur is already +6), and graphite is inert, so water goes.2H2O(l) → O2(g) + 4H+(aq) + 4e− — gas bubblesStep 3: Follow the colour and the ionsBlue fades as Cu2+ is removed; H+ builds up so the pH fallsthe solution slowly turns into dilute sulfuric acid
💡 Exam tip
Always write both candidates down before choosing. Examiners give credit for the comparison, not just the answer.
Say which Eθ value is more positive or more negative in your explanation. “Water is reduced” alone is not an explanation.
Watch the wording: dilute or concentrated in the question stem is there for a reason.
Check the electrode material. “Copper electrodes” completely changes the anode answer.
Remember the pH clue: OH− at the cathode, H+ at the anode.
If asked why the prediction from Eθ can be wrong, mention overpotential and non-standard concentrations.
⚠ Common mix-up
Predicting sodium metal from brine. Never happens in water. Hydrogen comes off instead.
Using the “more positive wins” rule at the anode too. At the anode you want the species that is easiest to oxidise, so the more negative value wins.
Forgetting water exists. The whole point of aqueous electrolysis is that water is a candidate at both ends.
Expecting sulfate or nitrate to be oxidised. Their central atoms are already at maximum oxidation state.
Mixing up the electrode charges. In electrolysis the cathode is negative and the anode is positive — the opposite of a voltaic cell.
Assuming the ratio of gases is always 2:1. That is true for water, but not for brine, where you get equal moles of H2 and Cl2.
Up next: Electroplating — the same active-electrode idea as copper purification, but this time you are deliberately coating an object with a thin, even layer of metal.
Want this explained one-to-one?
Book a free session with an experienced IB Chemistry tutor and get your trickiest topics made simple.
