In electroplating, the object to be coated is the cathode, the coating metal is the anode, and the electrolyte contains ions of the coating metal. Metal ions gain electrons at the cathode and deposit. Metal atoms at the anode lose electrons and enter the solution.
This lesson applies electrode roles, half-equations and the active-electrode idea from aqueous and molten cases.
How do I read an electroplating diagram?
Answer four questions in order.
- Which electrode is joined to the negative terminal? That is the cathode, and the object being plated.
- Which is joined to the positive terminal? That is the anode, made of the plating metal.
- What ions are in the solution? They should be ions of the plating metal.
- What happens at each electrode? The cathode gains metal. The anode loses metal.
Because the amount dissolved at the anode matches the amount deposited at the cathode, the concentration of the solution stays about the same.
Worked example
A diagram shows a steel key joined to the negative terminal and a copper bar joined to the positive terminal. Both are in copper(II) sulfate solution. Explain what happens and write the half-equations.
Step 1, electrodes: the key is the cathode. The copper bar is the anode.
Step 2, cathode: Cu²⁺ ions from the solution move to the key and gain electrons: Cu²⁺ + 2e⁻ → Cu. A layer of copper forms on the key.
Step 3, anode: copper atoms in the bar lose electrons and enter the solution: Cu → Cu²⁺ + 2e⁻. The bar becomes thinner.
Step 4, consistency: both half-equations involve 2 electrons, so each Cu²⁺ that leaves the solution is replaced by one that dissolves. The blue colour stays about the same.
Invented data, for practice only: suppose the key gains 0.32 g of copper. The moles of copper are 0.32 ÷ 64 = 0.005 mol. The bar loses the same 0.005 mol, which is 0.005 × 64 = 0.32 g. The electrons transferred are 2 × 0.005 = 0.010 mol.
Check: 0.32 ÷ 64 = 0.005, and 0.005 × 2 = 0.010. The arithmetic holds.
Hazard idea in words: plating solutions and cleaning steps use chemicals that need proper controls, so this lesson is about reading the diagram, not doing the process.
The mistake to watch for
A common slip is to put the object at the anode or to use an unrelated electrolyte.
Mistaken answer: “The key is the anode, so copper deposits on it.”
Metal ions are positive and move to the negative cathode. An object at the anode would lose metal, not gain it.
The correction is to start from the terminals. Negative means cathode, and the object to be coated always sits at the cathode. Then check the electrolyte contains ions of the coating metal.
Check yourself
1. A spoon is to be silver-plated. Name the cathode, the anode and a suitable electrolyte.
Show answer
Cathode: the spoon. Anode: a bar of silver. Electrolyte: a solution containing silver ions, such as silver nitrate solution.
2. Write the half-equation at the cathode for silver plating.
Show answer
Ag⁺ + e⁻ → Ag. Charge: 0 on both sides.
3. (Invented data.) A spoon gains 0.216 g of silver. How many moles of silver is that? (Ar of Ag = 108.)
Show answer
0.216 ÷ 108 = 0.002 mol. Check: 0.002 × 108 = 0.216. Because Ag⁺ needs one electron, 0.002 mol of electrons were also transferred.
Where this leads next
Try the whole module in electrolysis reasoning: mixed practice. The mole and equation-ratio tutor supports the amount calculations.
If you want someone to check your diagram reasoning as you go, that is what our teachers do in online one-to-one Chemistry tuition.