Question
During electrolysis, how do we predict what happens at the anode and cathode for different electrolytes and electrode types?
Solution — Step by Step
When an electrolyte dissolves, it dissociates into cations (+) and anions (-). Water itself provides and ions.
For example, aqueous NaCl gives: , , ,
At the cathode (-), cations compete for reduction. At the anode (+), anions compete for oxidation.
The species with the higher (less negative) reduction potential gets reduced first. Use the electrochemical series:
- Between ( V) and ( V): H+ is reduced because its reduction potential is higher.
Na+ stays in solution. This is why electrolysis of aqueous NaCl gives hydrogen gas, not sodium metal. To get sodium metal, we must electrolyse molten NaCl (no water, no competing H+).
The species with the lower (less positive) oxidation potential gets oxidised first. But there is a complication — the electrode material matters.
With inert electrodes (Pt, graphite): anions compete.
- Between and : despite OH having a lower oxidation potential, Cl gets oxidised in concentrated solution due to overpotential effects.
- In dilute solution, is oxidised:
With active electrodes (Cu, Ag): the electrode itself dissolves (gets oxidised) instead of the anions.
graph TD
A{Electrolysis setup} --> B[Identify all ions in solution]
B --> C{At Cathode: which cation?}
B --> D{At Anode: which anion?}
C --> E{Compare reduction potentials}
E -->|Higher E0| F[Gets reduced]
D --> G{Inert or active electrode?}
G -->|Inert: Pt, graphite| H{Compare oxidation potentials}
G -->|Active: Cu, Ag| I[Electrode dissolves]
H -->|Concentrated halide| J[Halide oxidised]
H -->|Dilute / no halide| K[OH- oxidised, O2 released]Why This Works
Electrolysis is essentially a competition: among all the possible reduction reactions at the cathode and oxidation reactions at the anode, the thermodynamically most favourable ones (or kinetically most accessible ones) proceed first.
The electrochemical series is the cheat sheet — higher reduction potential means easier to reduce (cathode), and the reverse tells us what is easier to oxidise (anode).
| Electrolyte | Cathode product | Anode product (inert electrode) |
|---|---|---|
| Molten NaCl | Na metal | Cl2 gas |
| Aqueous NaCl (conc.) | H2 gas | Cl2 gas |
| Aqueous NaCl (dilute) | H2 gas | O2 gas |
| Aqueous CuSO4 | Cu metal | O2 gas |
| Aqueous H2SO4 | H2 gas | O2 gas |
Alternative Method
For quick prediction, use these two rules:
- At cathode: if the cation’s reduction potential is more negative than V (i.e., metals above zinc in the series), water gets reduced instead. Otherwise, the metal deposits.
- At anode with inert electrode: if concentrated halide is present, halide gets oxidised. Otherwise, water/OH gets oxidised to give O2.
Common Mistake
Students assume that electrolysis of aqueous CuSO4 with copper electrodes gives the same products as with platinum electrodes. With copper electrodes, the copper anode dissolves () while copper deposits at the cathode. The net effect is copper transfer from anode to cathode — this is the basis of electrorefining. With Pt electrodes, O2 is produced at the anode instead. The electrode material completely changes the anode product.