Electrolysis cells in series: Class 12 practice
Cells in series carry the same charge, so the masses deposited are in the ratio of equivalent masses (molar mass ÷ electrons per ion): m₁/m₂ = E₁/E₂.. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Faraday's laws
The idea behind electrolysis cells in series
Cells in series carry the same charge, so the masses deposited are in the ratio of equivalent masses (molar mass ÷ electrons per ion): m₁/m₂ = E₁/E₂.
Step-by-step method
- Find each equivalent mass E = M/n.
- Same charge → m ∝ E.
- Scale from the known mass.
Worked example 1 (Basic)
Two cells with Zn²⁺ and Al³⁺ solutions are in series. If 14 g of Zn deposits, how much Al deposits?
- E(Zn²⁺) = 32.5, E(Al³⁺) = 9
- m = 14 × 9/32.5
- Answer: 3.88 g
Worked example 2 (Standard)
Two cells with Ag⁺ and Al³⁺ solutions are in series. If 20 g of Ag deposits, how much Al deposits?
- E(Ag⁺) = 108, E(Al³⁺) = 9
- m = 20 × 9/108
- Answer: 1.67 g
Worked example 3 (Challenge)
17 g of Al is deposited. How much charge passed?
- Moles of electrons = 17/9
- Q = 182278 C
- Answer: 182278 C
Common mistakes and misconceptions
- Using molar masses directly
- Wrong charge on the ion
Practice questions
- (Basic) Two cells with Ag⁺ and Ni²⁺ solutions are in series. If 8 g of Ag deposits, how much Ni deposits?
- (Basic) Two cells with Ag⁺ and Al³⁺ solutions are in series. If 2 g of Ag deposits, how much Al deposits?
- (Standard) Two cells with Zn²⁺ and Ni²⁺ solutions are in series. If 4 g of Zn deposits, how much Ni deposits?
- (Standard) Two cells with Ni²⁺ and Zn²⁺ solutions are in series. If 17 g of Ni deposits, how much Zn deposits?
- (Challenge) 8 g of Cu is deposited. How much charge passed?
- (Challenge) 6 g of Al is deposited. How much charge passed?
Answer key with full working
- Q1: E(Ag⁺) = 108, E(Ni²⁺) = 29.35 → m = 8 × 29.35/108 → Answer: 2.17 g
- Q2: E(Ag⁺) = 108, E(Al³⁺) = 9 → m = 2 × 9/108 → Answer: 0.17 g
- Q3: E(Zn²⁺) = 32.5, E(Ni²⁺) = 29.35 → m = 4 × 29.35/32.5 → Answer: 3.61 g
- Q4: E(Ni²⁺) = 29.35, E(Zn²⁺) = 32.5 → m = 17 × 32.5/29.35 → Answer: 18.82 g
- Q5: Moles of electrons = 8/31.75 → Q = 24315 C → Answer: 24315 C
- Q6: Moles of electrons = 6/9 → Q = 64333 C → Answer: 64333 C
Syllabus note
This skill is listed under "Electrochemistry" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
- Electrochemistry chapter notes (Class 12)
- Cell EMF and the Nernst equation
- Faraday's laws of electrolysis
- Conductivity and molar conductivity
- Standard cell EMF and ΔG°
- Molality and mole fraction
- Boiling point elevation and freezing point depression
- First-order reactions and half-life
- Arrhenius equation
- All Class 12 practice
Frequently asked questions
Which chapter is electrolysis cells in series in?
"Electrochemistry", Class 12 Chemistry (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.