Question
How do you predict what gets oxidised at the anode and what gets reduced at the cathode during electrolysis? Apply this to the electrolysis of (a) molten NaCl, (b) aqueous NaCl, and (c) aqueous CuSO₄ with copper electrodes.
(JEE Main 2023 asked about electrolysis of aqueous NaCl; CBSE 12 regularly tests all three cases)
Solution — Step by Step
At the cathode (negative electrode): the species with the highest reduction potential gets reduced preferentially. At the anode (positive electrode): the species with the lowest reduction potential (most easily oxidised) gets oxidised.
When water is the solvent, water itself competes for both reduction and oxidation.
No water present, so no competition. Only Na⁺ and Cl⁻ ions exist.
- Cathode: (sodium metal deposited)
- Anode: (chlorine gas evolved)
Now water competes. At the cathode, both Na⁺ and H₂O can be reduced. Water has a higher reduction potential than Na⁺, so:
- Cathode: (hydrogen gas, NOT sodium metal)
- Anode: (chlorine gas — due to overvoltage effect, Cl⁻ is oxidised instead of water)
This is the chlor-alkali process — it produces Cl₂, H₂, and NaOH simultaneously.
With active electrodes (copper), the anode itself dissolves:
- Cathode: (copper deposited)
- Anode: (copper electrode dissolves)
Net effect: copper transfers from anode to cathode. This is the basis of electrorefining of copper.
flowchart TD
A[Electrolysis Setup] --> B{Molten or Aqueous?}
B -->|Molten| C[Only ions present<br/>Direct discharge]
B -->|Aqueous| D{Water competes}
D --> E{At Cathode: Who has<br/>higher E° reduction?}
D --> F{At Anode: Electrode<br/>active or inert?}
E -->|Metal ion E° > water| G[Metal deposited]
E -->|Water E° > metal ion| H[H₂ gas evolved]
F -->|Active electrode| I[Electrode dissolves]
F -->|Inert: check ions vs water| J[Lower E° oxidised]Why This Works
Electrolysis is essentially a competition. Multiple species are present at each electrode, and the one requiring the least energy to undergo the electrode reaction wins.
The tricky part is the overvoltage effect — some reactions require extra voltage beyond what thermodynamics predicts due to kinetic barriers at the electrode surface. This is why Cl⁻ is oxidised at the anode during brine electrolysis even though water should theoretically be easier to oxidise.
Alternative Method
Quick prediction rule for aqueous solutions with inert electrodes: If the cation is from a reactive metal (Na, K, Ca, Mg, Al), water gets reduced instead (H₂ forms). If the cation is from a less reactive metal (Cu, Ag, Au), the metal ion gets reduced. At the anode, if concentrated halide ions are present, they get oxidised; otherwise water is oxidised (O₂ forms).
Common Mistake
Students assume sodium metal is deposited during electrolysis of aqueous NaCl. Sodium is far too reactive — its reduction potential ( V) is much more negative than water’s ( V). Water is always reduced preferentially, producing H₂ gas. Sodium metal can only be obtained from molten NaCl (Downs process) where water is absent.