In a molten compound, only its own ions are present. In an aqueous solution, water adds H⁺ and OH⁻ as well, so each electrode has a choice. This lesson gives a short set of rules for simple cases, and you should check your syllabus for the exact scope.
It compares directly with molten products and uses half-equations to write the results. Knowing which metals are more reactive helps, as in metals and reactivity.
What is the decision routine?
Use the same two questions every time.
At the cathode (negative): is the metal less reactive than hydrogen? If yes, the metal forms. If no, hydrogen forms.
At the anode (positive): is a halide ion (chloride, bromide, iodide) present in a suitable concentration? If yes, the halogen forms. If no, oxygen forms.
With active electrodes such as copper in copper(II) sulfate, the anode itself can dissolve. That changes the anode rule, and electroplating uses it.
Worked example
Compare three cells, all with inert electrodes.
| Electrolyte | Ions present | Cathode | Anode |
|---|---|---|---|
| Molten sodium chloride | Na⁺, Cl⁻ | sodium | chlorine |
| Concentrated sodium chloride solution | Na⁺, Cl⁻, H⁺, OH⁻ | hydrogen | chlorine |
| Copper(II) sulfate solution | Cu²⁺, SO₄²⁻, H⁺, OH⁻ | copper | oxygen |
Reasoning for the second row: sodium is more reactive than hydrogen, so H⁺ is discharged: 2H⁺ + 2e⁻ → H₂. Chloride is a halide and concentrated, so Cl⁻ is discharged: 2Cl⁻ → Cl₂ + 2e⁻. The Na⁺ and OH⁻ ions stay in the solution, which becomes sodium hydroxide solution.
Reasoning for the third row: copper is less reactive than hydrogen, so Cu²⁺ + 2e⁻ → Cu. No halide is present, so hydroxide ions give up electrons: 4OH⁻ → 2H₂O + O₂ + 4e⁻. The solution loses copper ions and becomes more acidic.
Hazard idea in words: chlorine is toxic and corrosive, which is why industrial chlorine cells are designed to keep the gas contained. This page covers the reasoning only.
A short note on dilute acid
For dilute sulfuric acid, the ions present are H⁺, OH⁻ and SO₄²⁻. Hydrogen forms at the cathode (4H⁺ + 4e⁻ → 2H₂) and oxygen at the anode (4OH⁻ → 2H₂O + O₂ + 4e⁻).
Four electrons give 2 H₂ and 1 O₂, so the gases are in the volume ratio 2 : 1. The net change is that water is split: 2H₂O → 2H₂ + O₂.
The mistake to watch for
A typical slip is to carry the molten answer into the aqueous case.
Mistaken answer: “Concentrated sodium chloride solution gives sodium at the cathode.”
Sodium is more reactive than hydrogen, so hydrogen ions are discharged, not sodium ions.
The correction is to list all four ions, then apply the two questions. Write “molten” or “aqueous” at the start of your working.
Check yourself
1. Predict the products of copper(II) sulfate solution with inert electrodes.
Show answer
Cathode: copper (less reactive than hydrogen). Anode: oxygen (no halide present).
2. Predict the products of concentrated sodium chloride solution with inert electrodes, and name the substance left in solution.
Show answer
Cathode: hydrogen. Anode: chlorine. Left in solution: sodium hydroxide, from Na⁺ and OH⁻.
3. What gases form in dilute sulfuric acid electrolysis, and in what volume ratio?
Show answer
Hydrogen at the cathode and oxygen at the anode, in a 2 : 1 volume ratio, because 4 electrons release 2 H₂ and 1 O₂.
Where this leads next
Next, interpret an electroplating diagram safely, which applies active electrodes. The mole and equation-ratio tutor helps when a question asks for amounts of products.
If the aqueous rules are still a blur, our teachers can test your reasoning on fresh cells in online one-to-one Chemistry tuition.