1iElectrolysis

Separate Chemistry only.

Electrolysis is using electricity to break a compound down. Before any of it makes sense, one question has to be settled: what makes a substance conduct in the first place?

Conduction needs mobile charge

Two conditions, both required:

1. There must be charged particles.
2. Those particles must be free to move.

That pair explains every case on the specification.

Why most covalent substances do not conduct

Melting a simple molecular substance separates whole molecules, and molecules are electrically neutral. No charged particles are released at any point, so there is nothing to carry a current in any state.

This is the real difference from an ionic compound, which does conduct once melted.

Why an ionic solid does not conduct

It has ions. They simply cannot move, because they are fixed in the lattice.

Melting or dissolving breaks the lattice down and frees them, which is why the same compound conducts in one state and not the other.

Anions and cations

Opposite charges attract, so each ion travels to the electrode of opposite sign.

Two memory aids: the t in cation can be read as a plus sign, and anion goes to the anode.

And note the electrode charges: in electrolysis the cathode is negative. Getting that backwards invalidates every prediction that follows.

Separate Chemistry only.

Molten compounds are the simple case

With nothing present but the compound itself, the products are just its two elements.

Molten lead(II) bromide gives lead at the cathode and bromine at the anode. Metal to the cathode, non-metal to the anode, every time.

Aqueous solutions add a complication

Water is present, and water supplies H⁺ and OH⁻ ions of its own. So at each electrode there is a competition, and forgetting the water is why students mispredict these.

At the cathode: metal or hydrogen?

Compare the metal with hydrogen in the reactivity series.

• Metal more reactive than hydrogen → hydrogen is produced.
• Metal less reactive than hydrogen → the metal is produced.

So sodium chloride solution gives hydrogen, because sodium is more reactive. Copper(II) sulfate solution gives copper, because copper is less reactive.

The rule is that the less reactive one is discharged.

At the anode: which negative ion?

• A halide present in a concentrated solution (chloride, bromide, iodide) → that halogen is produced. In a very dilute halide solution there are too few halide ions to compete and oxygen comes off instead, which is why questions specify concentrated when they want the halogen.
• Sulfate or nitrate present → they are never discharged. Oxygen is produced from the water instead.

So dilute sulfuric acid gives oxygen at the anode, and copper(II) sulfate solution does too.

The concentration detail

For sodium chloride solution, chlorine is produced at the anode when the solution is concentrated. In a dilute solution, oxygen is produced instead.

If a question specifies concentrated brine, that word is doing work — it is telling you which product to expect.

Why reactive metals need molten compounds

Aluminium cannot be extracted from a solution, because hydrogen would be discharged in preference. Only a molten compound, with no water present, gives the metal.

That is the whole reason aluminium extraction is so expensive.

Separate Chemistry only.

The definitions

OIL RIG — Oxidation Is Loss, Reduction Is Gain — of electrons.

In electrolysis this maps directly onto the electrodes:

Reduction always happens at the cathode. Both words carry an odd sense here — an ion being reduced gains something — so it is worth learning the pairing rather than reasoning it out each time.

Writing half-equations

At the cathode, electrons are gained, so they appear on the left:

Pb²⁺ + 2e⁻ → Pb

At the anode, electrons are lost, so they appear on the right:

2Br⁻ → Br₂ + 2e⁻

That gives you a way to read any half-equation: electrons on the left means reduction; electrons on the right means oxidation.

Two things that go wrong

Diatomic products. Chlorine, bromine, hydrogen and oxygen come off as molecules, so two ions are needed. Writing Cl⁻ → Cl + e⁻ leaves the equation unbalanced.

Charges. Examiners report answers using H⁻ instead of H⁺, and electrons placed on the wrong side. Check the charges balance across the arrow as well as the atoms.

Practical points

Use direct current. With alternating current the electrodes swap polarity constantly and no separation happens.

Use inert electrodes — graphite or platinum. A reactive electrode would take part in the reaction itself, changing the products and being consumed.

Reading the observations

During the electrolysis of copper(II) sulfate solution, the blue colour slowly fades. That is because the blue comes from the copper ions in solution, and they are being removed and deposited as copper metal at the cathode.

An observation like that is usually a question in disguise: name the ion responsible, and say where it went.

Separate Chemistry only.

The standard example is molten lead(II) bromide with inert graphite electrodes.

Apparatus

Crucible on a pipeclay triangle, two graphite electrodes, d.c. supply, Bunsen, fume cupboard.

Method

1. Heat the solid lead(II) bromide until it melts.
2. Dip the electrodes in, without letting them touch.
3. Connect to a d.c. supply and watch both electrodes.

What you see

Cathode: Pb²⁺ + 2e⁻ → Pb

Anode: 2Br⁻ → Br₂ + 2e⁻

Positive lead ions are attracted to the negative electrode, and negative bromide ions to the positive one. Metal at the cathode, non-metal at the anode.

Why it must be molten

This is the point the experiment exists to make.

In the solid, the ions are locked in a lattice. They are charged, but they cannot move, so no current flows and nothing is decomposed.

Melting frees the ions. They can now move to the electrodes, and only then does electrolysis happen.

So a lamp in the circuit stays off while the compound is solid and lights as it melts — evidence that mobile ions, not merely charged ones, are what carry the current.

Why these choices

Graphite electrodes conduct and are inert, so they take no part in the reaction. A reactive metal electrode would corrode and change the products.

Direct current. Each electrode must keep the same charge throughout. With a.c. the polarity swaps many times a second and the products would form alternately at both electrodes.

Safety

Bromine vapour is toxic and corrosive, so the experiment is done in a fume cupboard. The melt is also extremely hot.

Predicting products for other melts

For any molten binary ionic compound: metal at the cathode, non-metal at the anode. There is no water present, so nothing else can be discharged — which is exactly what makes molten electrolysis simpler to predict than the aqueous case.

Separate Chemistry only.

The difference from a molten compound is that water is present, supplying H⁺ and OH⁻ ions. Four kinds of ion now compete, and the products are no longer simply the metal and the non-metal.

Apparatus

Electrolysis cell or beaker, two carbon electrodes, d.c. supply, inverted test tubes to collect the gases.

Method

1. Fill the cell with the solution and fit the electrodes.
2. Fill two test tubes with the solution and invert one over each electrode.
3. Switch on and let the gases collect.
4. Test each gas, and note any change at the electrodes or in the colour of the solution.

The rules

At the cathode (−) — the less reactive of the two positive ions is discharged:

• Metal less reactive than hydrogen (copper, silver) → the metal is deposited
• Metal more reactive than hydrogen (sodium, potassium, calcium) → hydrogen is given off

At the anode (+):

• A concentrated halide present → the halogen (chlorine, bromine, iodine)
• No halide, or a very dilute one → oxygen from the OH⁻ ions

The concentration is part of the rule, not a footnote to it. Make the solution very dilute and there are too few halide ions to compete, so oxygen comes off instead — which is why questions specify concentrated sodium chloride solution when they want chlorine.

Safety

Chlorine is toxic. Electrolysing concentrated sodium chloride must be done in a fume cupboard, or on a very small scale in a well-ventilated room with the gas collected rather than allowed to escape. Do not sniff the electrode.

Testing the products

Worked examples

Dilute sulfuric acid gives hydrogen and oxygen in a 2 : 1 volume ratio — the water is being decomposed, and the ratio is the formula of water made visible.

Why sodium chloride does not give sodium

Sodium is far more reactive than hydrogen, so the H⁺ ions from the water are discharged instead, and hydrogen bubbles off. Any sodium metal formed would react violently with the water around it in any case.

This is the most common error in the topic: reading Na⁺ and Cl⁻ and forgetting the water is a reagent too.

Half equations

Cathode: 2H⁺ + 2e⁻ → H₂ or Cu²⁺ + 2e⁻ → Cu

Anode: 2Cl⁻ → Cl₂ + 2e⁻ or 4OH⁻ → O₂ + 2H₂O + 4e⁻

Reduction at the cathode (electrons gained), oxidation at the anode (electrons lost).

Where marks are lost

• Predicting a reactive metal at the cathode from an aqueous solution.
• Writing oxygen at the anode when a halide is present.
• Leaving electrons out of a half equation, or putting them on the wrong side.