Last updated: 14 October 2026
In a spontaneous electrochemical cell, a redox reaction produces electrical energy. Electrolysis turns that relationship around. An external power supply provides electrical energy and drives a redox reaction in a direction that would not occur spontaneously.
This topic therefore brings together oxidation and reduction, electrode potentials, ionic movement and mole calculations. Rather than memorising a long list of rules, keep asking two questions: what can gain electrons at the cathode, and what can lose electrons at the anode?
Try these five questions before reading the chapter. Focus on the electrode reactions, the direction of electron transfer and the relationship between electrical charge and chemical amount.
- 5 diagnostic questions
- Electrode reactions and calculations
- Instant explanation for every answer option
What is electrolysis?
Electrolysis is the use of electrical energy to drive a non-spontaneous redox reaction. It takes place in an electrolytic cell connected to an external direct-current power supply.
The electrolyte contains mobile ions, while the electrodes provide conducting surfaces at which electron transfer occurs.
Oxidation always occurs at the anode. Reduction always occurs at the cathode.
The cathode and anode
The cathode is connected to the negative terminal of the power supply. Electrons are supplied to the cathode, so species there can gain electrons and undergo reduction.
The anode is connected to the positive terminal. Electrons are removed from species at the anode, so oxidation occurs there.
Cathode: negative; attracts cations; reduction occurs.
Anode: positive; attracts anions; oxidation occurs.
How charge moves through the cell
Electrons move through the external wires and electrodes. Within the electrolyte, charge is carried by moving ions.
Cations migrate towards the cathode, while anions migrate towards the anode.
A useful memory aid is AN OX, RED CAT: oxidation at the ANode and REDuction at the CAThode.
Electrolysis of molten electrolytes
A molten ionic compound contains mobile ions but no water. Only the ions belonging to the ionic compound therefore need to be considered.
Example: molten sodium chloride
Sodium metal is produced at the cathode and chlorine gas at the anode.
Electrolysis of aqueous solutions
Aqueous electrolysis is more complicated because water is present. Species from water can therefore compete with dissolved ions for discharge at the electrodes.
In an aqueous electrolyte, always consider whether water can take part in the electrode reactions.
Selective discharge at the cathode
Reduction occurs at the cathode. Standard electrode potentials can be used as a thermodynamic guide when comparing possible reductions.
Under comparable conditions, the oxidised species belonging to the couple with the more positive E° value has the greater thermodynamic tendency to be reduced.
More positive E° → greater thermodynamic tendency for reduction.
Selective discharge at the anode
Oxidation occurs at the anode. For metal/redox couples under comparable conditions, a metal belonging to a couple with a lower E° value is generally more readily oxidised.
Lower E° → the reduced form generally has a greater tendency to be oxidised.
Analyse the cathode and anode separately rather than trying to apply one discharge rule to the whole electrolytic cell.
Why concentration can affect the products
E° values apply under standard conditions. In an actual electrolytic cell, concentrations may be different, and electrode kinetics and overpotential can also influence which reaction is observed.
Concentrated aqueous sodium chloride is an important example. Hydrogen is formed at the cathode, while the high chloride concentration favours chlorine formation at the anode.
Inert and reactive electrodes
An inert electrode, such as platinum or graphite in many situations, mainly provides a conducting surface.
A reactive electrode can itself take part in the electrode reaction. Copper electrorefining is an important example because the impure copper anode deliberately undergoes oxidation.
Faraday calculations in electrolysis
The amount of chemical change during electrolysis is related to the amount of electrical charge transferred.
Q is charge in coulombs, I is current in amperes and t is time in seconds. If time is given in minutes or hours, convert it into seconds before using the equation.
The Faraday constant
The Faraday constant, F, is the magnitude of charge carried by one mole of electrons:
It is related to the Avogadro constant, L, and the magnitude of the charge on one electron, e:
Hence:
From moles of electrons to moles of product
The balanced electrode half-equation gives the mole ratio between electrons and the substance formed.
Two moles of electrons are therefore required to deposit one mole of copper.
| Half-equation | Electron ratio | Meaning |
|---|---|---|
| Ag⁺ + e⁻ → Ag | 1 mol e⁻ : 1 mol Ag | 1 mol of electrons deposits 1 mol of Ag |
| Cu²⁺ + 2e⁻ → Cu | 2 mol e⁻ : 1 mol Cu | 2 mol of electrons deposit 1 mol of Cu |
| Al³⁺ + 3e⁻ → Al | 3 mol e⁻ : 1 mol Al | 3 mol of electrons deposit 1 mol of Al |
A reliable calculation sequence
Step 1: calculate Q using Q = It.
Step 2: calculate n(e⁻) using n(e⁻) = Q/F.
Step 3: use the balanced electrode half-equation to calculate moles of the substance.
Step 4: convert the amount into the required mass or gas volume.
Worked example: mass of copper deposited
A current of 2.00 A is passed through a solution containing Cu²⁺ for 30.0 min. Calculate the mass of copper deposited. Take F = 9.65 × 10⁴ C mol⁻¹ and Aᵣ(Cu) = 63.5.
Q = It = 2.00 × (30.0 × 60) = 3600 C
n(e⁻) = 3600 / (9.65 × 10⁴) = 0.0373 mol
Cu²⁺ + 2e⁻ → Cu
n(Cu) = 0.0373 / 2 = 0.0187 mol
mass of Cu = 0.0187 × 63.5 = 1.18 g
Electrolytic purification of copper
Copper used for electrical wiring must be very pure because impurities reduce its electrical conductivity. Electrolytic purification, or electrorefining, transfers copper from an impure copper anode to a pure copper cathode while separating it from many impurities.
The impure copper is the anode, a thin sheet of pure copper is the cathode, and the electrolyte contains Cu²⁺ ions, commonly in aqueous copper(II) sulfate.
Copper atoms leave the impure anode as Cu²⁺ ions. At the cathode, Cu²⁺ ions from the electrolyte gain electrons and are deposited as copper.
Copper is removed from the impure anode and deposited as purer copper at the cathode.
Why do zinc, copper and silver behave differently?
Suppose the impure copper contains zinc and silver. Their standard electrode potentials can be compared with that of copper:
Reversible arrows are used here because these equations represent electrode redox couples associated with E° values. When describing an actual electrode reaction taking place during electrolysis, a single arrow is used.
Step 1: what happens at the impure anode?
Oxidation occurs at the anode.
Zinc has a much lower E° value than copper, so zinc metal is more readily oxidised. Zinc impurities can therefore dissolve:
Copper also undergoes oxidation:
Silver behaves differently. The Ag⁺/Ag couple has a more positive E° value than the Cu²⁺/Cu couple, so metallic silver is less readily oxidised than copper.
As the surrounding copper dissolves, silver remains mainly as solid metal. The silver particles become detached from the anode and collect at the bottom as anode sludge or anode mud. Gold behaves similarly.
Zinc: readily oxidised → enters the electrolyte as Zn²⁺.
Copper: oxidised → enters the electrolyte as Cu²⁺.
Silver: not readily oxidised → remains mainly metallic and forms anode sludge.
Step 2: what happens at the cathode?
The electrolyte may now contain both Cu²⁺ and Zn²⁺ ions. At the cathode, reduction occurs.
The Cu²⁺/Cu couple has the much more positive E° value. Cu²⁺ therefore has a much greater thermodynamic tendency to be reduced than Zn²⁺.
Copper is therefore preferentially deposited at the cathode, while Zn²⁺ tends to remain in the electrolyte.
Zinc can dissolve from the impure anode without significantly contaminating the cathode because Cu²⁺ is much more readily reduced than Zn²⁺.
Why doesn’t zinc eventually replace copper at the cathode?
Cu²⁺ ions are continuously being removed and replaced during electrorefining.
At the cathode:
At the same time, at the anode:
Therefore, Cu²⁺ ions removed from the electrolyte at the cathode are continually replenished by copper dissolving from the anode.
Cu²⁺ remains available for reduction and is much more readily reduced than Zn²⁺. Copper therefore remains the principal cathode deposit, while Zn²⁺ tends to remain in solution.
Think of electrorefining as a continuous process: copper dissolves at the anode while copper is deposited at the cathode. Zn²⁺ does not simply wait for Cu²⁺ to disappear.
If Ag⁺ is easier to reduce than Cu²⁺, why is silver not deposited?
Compare the standard electrode potentials:
If appreciable Ag⁺ were present in the electrolyte, Ag⁺ would actually have a greater thermodynamic tendency to be reduced than Cu²⁺.
The important separation occurs earlier, at the anode. Metallic silver is not readily oxidised, so very little Ag⁺ enters the electrolyte. Silver instead remains mainly as solid metal and forms anode sludge.
Zinc impurity: enters the electrolyte as Zn²⁺, but Zn²⁺ is difficult to reduce relative to Cu²⁺.
Silver impurity: does not readily enter the electrolyte in the first place, so little Ag⁺ is available to compete at the cathode.
The two stages of copper purification
Copper electrorefining is easiest to understand as two stages of selectivity.
| Species | At the anode | What happens afterwards? |
|---|---|---|
| Zn | Readily oxidised to Zn²⁺ | Zn²⁺ tends to remain in the electrolyte |
| Cu | Oxidised to Cu²⁺ | Cu²⁺ is preferentially reduced at the cathode |
| Ag | Not readily oxidised | Remains mainly metallic and forms anode sludge |
Electrorefining is not the same as electroplating
The two processes can look similar because both involve metal deposition at a cathode. Their purposes, however, are different. Electrorefining uses electrodeposition to purify a metal, whereas electroplating uses electrodeposition to coat an object with a metal.
Do we simply choose a voltage between +0.34 V and +0.80 V?
No. E° values should not be interpreted as simple voltage settings on the external power supply.
Values such as +0.34 V for Cu²⁺/Cu and +0.80 V for Ag⁺/Ag are electrode potentials measured relative to the standard hydrogen electrode.
Actual electrode potentials depend on conditions such as concentration, and real electrolytic cells also involve effects such as overpotential and electrical resistance.
At A-Level, use E° values mainly to compare the relative tendencies for oxidation and reduction.
E° values help compare redox tendencies. They are not direct power-supply voltage settings.
Can mass changes be used to calculate the purity of the copper anode?
This depends entirely on what the measured masses represent. There is no general rule saying that cathode mass gain divided by anode mass loss gives the percentage purity of the original anode.
Three contrasting examples make the distinction clearer.
Example 1: when a purity calculation is valid
An impure copper sample has a mass of 5.00 g. The question states that all the copper in the sample is transferred to the cathode and that none of the impurities is deposited. The cathode gains 4.60 g.
Because the question explicitly tells us that the 4.60 g cathode gain is the total mass of copper originally present:
% Cu = (4.60 / 5.00) × 100% = 92.0%
This calculation works because the question has established a direct link:
A purity calculation is valid when the question clearly tells you that the measured mass corresponds to all the copper in the original sample.
Example 2: why cathode mass gain ÷ anode mass loss can be misleading
Now suppose that during electrorefining, both copper and zinc dissolve from the impure anode:
Imagine that 0.0100 mol of Cu and 0.0100 mol of Zn are oxidised.
Each mole of Cu and each mole of Zn releases two moles of electrons. Therefore:
If Cu²⁺ is preferentially reduced at the cathode:
then 0.0400 mol of electrons can deposit:
If we incorrectly tried to calculate purity using:
we would obtain:
But that does not mean the material lost from the anode was 98.5% copper. The actual copper fraction of the material lost was:
Why is there such a large difference? Because electrons released when zinc is oxidised can also contribute to the current that reduces Cu²⁺ already present in the electrolyte. Copper deposited at the cathode therefore does not have to correspond one-for-one to copper atoms that have just left the anode.
If other metals such as Zn are also oxidised at the anode, cathode mass gain ÷ anode mass loss does not give the copper purity.
Example 3: when there is simply not enough information
An impure copper anode initially has a mass of 10.0 g. After electrolysis for some time, the cathode has gained 2.00 g. Calculate the percentage purity of copper in the original anode.
Answer: the percentage purity cannot be determined from the information given.
The 2.00 g cathode gain tells us only how much copper has been deposited during that period. It does not tell us how much copper remains in the anode or how much copper was originally present.
Before using any mass in a percentage-purity calculation, first ask: what does this measured mass actually represent?
What should you do in an examination question?
Do not begin by looking for a formula involving anode mass loss and cathode mass gain. Begin by interpreting the experiment.
If the question states that all the Cu has been transferred: the cathode gain may represent the total Cu originally present.
If reactive impurities such as Zn also dissolve: cathode gain and anode loss cannot be equated directly with copper purity.
If electrolysis has only been carried out for some time: the cathode gain tells you only how much Cu has been deposited so far.
Exam habit: identify what each measured mass represents chemically before doing any percentage calculation.
Anodising aluminium
Aluminium naturally forms a thin aluminium oxide layer on its surface. This oxide layer adheres strongly to the metal and protects the aluminium from further corrosion.
Anodising is an electrolytic process used to make this protective oxide layer much thicker. The aluminium object is made the positive anode, while an inert material can act as the cathode. Dilute sulfuric acid is commonly used as the electrolyte.
How does the oxide layer become thicker?
At the aluminium anode, oxidation occurs. A useful school-level model is that water is oxidised at the anode to produce oxygen:
The oxygen produced at the aluminium surface can react with exposed aluminium to form aluminium oxide:
Aluminium oxide itself is strongly protective. If a completely impervious coating simply formed and remained unchanged, it would prevent further contact with the aluminium beneath it and limit further growth.
The acidic electrolyte is important because it partially dissolves the aluminium oxide as it forms. This creates microscopic pores in the outer oxide layer and allows the anodising process to continue, producing a much thicker coating than the thin natural oxide film.
During anodising, oxide formation and partial dissolution occur together. This produces a thick oxide coating with a porous outer structure.
What happens at the cathode?
Hydrogen ions in the acidic electrolyte gain electrons at the cathode and form hydrogen gas:
Why anodise aluminium?
The thicker aluminium oxide coating increases resistance to corrosion and wear. The freshly anodised outer layer contains microscopic pores, so dyes can be introduced into the pores to produce coloured aluminium surfaces.
The pores can then be sealed, producing a durable protective finish.
Copper purification: copper dissolves from the impure anode and is deposited as purer copper at the cathode.
Aluminium anodising: the aluminium object itself is the anode, where a thicker protective oxide layer is formed.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Thinking the anode is always negative | In an electrolytic cell, the anode is positive. | Remember oxidation at the anode and reduction at the cathode first. |
| Ignoring water in an aqueous electrolyte | Water can participate in electrode reactions. | Consider all realistic species that can be oxidised or reduced. |
| Using the most positive E° value for both electrodes | A more positive E° favours reduction; oxidation must be analysed separately. | Treat the cathode and anode separately. |
| Treating E° values as direct power-supply settings | E° values are electrode potentials measured relative to a reference electrode. | Use E° values to compare redox tendencies. |
| Thinking Zn²⁺ deposits before Cu²⁺ during copper purification | Cu²⁺/Cu has a much more positive E° value than Zn²⁺/Zn. | Cu²⁺ is preferentially reduced. |
| Thinking Ag is excluded because Ag⁺ is difficult to reduce | Ag⁺ is actually more readily reduced than Cu²⁺. | Explain that Ag metal is not readily oxidised at the anode, so very little Ag⁺ enters the electrolyte. |
| Thinking Zn²⁺ simply waits for Cu²⁺ to run out | Cu²⁺ is continually replenished as copper dissolves from the anode. | Think of copper dissolution and copper deposition as simultaneous processes. |
| Using cathode mass gain ÷ anode mass loss automatically as copper purity | Other metals may also dissolve from the anode, and the cathode deposit can come from Cu²⁺ already present in the electrolyte. | First determine what each measured mass actually represents. |
| Using a partial cathode mass gain as the total amount of Cu originally present | Copper may still remain in the anode. | Only equate cathode gain with total Cu when the question explicitly justifies this. |
| Using time in minutes directly in Q = It | Current in amperes is coulombs per second. | Convert time to seconds first. |
What you should be able to do
Identify the electrodes
State that oxidation occurs at the positive anode and reduction occurs at the negative cathode in an electrolytic cell.
Predict electrode products
Distinguish between molten and aqueous electrolytes and identify possible competing electrode reactions.
Use electrode potentials
Use E° values appropriately to compare tendencies for reduction and oxidation.
Use Faraday's constant
Relate charge to moles of electrons using Q = It and n(e⁻) = Q/F.
Explain copper purification
Explain why Zn can enter the electrolyte, Cu is deposited at the cathode and Ag forms anode sludge.
Interpret mass data correctly
Decide what anode mass loss and cathode mass gain represent before attempting a percentage-purity calculation.
These ten questions test whether you can predict electrode products, apply Faraday calculations and explain copper electrorefining and anodising.
- 10 mastery questions
- Selective discharge, electrorefining and Faraday calculations
- Instant explanation for every answer option
Frequently asked questions
Questions students often ask about electrolysis
Most confusion in electrolysis comes from mixing up the two electrodes, applying electrode potentials too mechanically, or assuming that every measured mass has an obvious chemical meaning.
Why are reversible arrows used when E° values are shown?
An E° value refers to an electrode redox couple at equilibrium, so Cu²⁺ + 2e⁻ ⇌ Cu is appropriate when quoting its standard electrode potential. When describing an actual electrode reaction during electrolysis, a single arrow is used.
Why doesn't zinc eventually deposit during copper electrorefining?
Cu²⁺ is much more readily reduced than Zn²⁺. In addition, Cu²⁺ removed at the cathode is continually replenished as copper dissolves from the anode. Copper therefore remains the principal cathode deposit while Zn²⁺ tends to remain in the electrolyte.
If Ag⁺ is easier to reduce than Cu²⁺, why doesn't silver contaminate the cathode?
Metallic silver is not readily oxidised at the anode, so very little Ag⁺ enters the electrolyte. Silver instead remains mainly as solid metal and forms anode sludge.
Is copper electrorefining the same as electroplating?
No. Both involve deposition of a metal at the cathode, but their purposes are different. Electrorefining uses electrodeposition to purify a metal, whereas electroplating uses electrodeposition to coat an object with a layer of metal.
Can cathode mass gain divided by anode mass loss give the copper purity?
Not in general. More reactive impurities such as zinc may also be oxidised at the anode, and the electrons released can contribute to the deposition of Cu²⁺ already present in the electrolyte. Therefore, cathode mass gain does not necessarily correspond only to copper removed from the anode.
When can cathode mass gain be used in a percentage-purity calculation?
It can be used when the question explicitly establishes that the cathode mass gain represents all the copper originally present in the sample and that impurities do not contribute to the cathode deposit. Always interpret what the measured mass represents before applying a percentage-purity formula.
Why is sulfuric acid used during anodising?
Dilute sulfuric acid acts as an electrolyte and also helps produce the porous structure of the anodised oxide layer by partially dissolving aluminium oxide as it forms. This allows the coating to continue growing and creates pores that can later be dyed and sealed.
Electrolysis becomes much easier when each electrode is analysed separately. Copper electrorefining brings the ideas together especially well: copper dissolves and is redeposited continuously, zinc can enter the electrolyte but tends to remain as Zn²⁺, and less readily oxidised metals such as silver remain behind as anode sludge. Faraday calculations then connect the flow of electrical charge to the amount of chemical change, while anodising shows how electrolysis can also be used deliberately to modify and protect a metal surface.