Faraday's laws of electrolysis: Class 12 practice

What you should know first

• Mole concept

The idea behind faraday's laws of electrolysis

Charge Q = It. Moles of electrons = Q/F (F = 96500 C/mol). Mass deposited = (Q/F) × (molar mass/n), where n is electrons per ion.

Step-by-step method

1. Q = I × t (seconds).
2. Moles of e⁻ = Q/96500.
3. Moles of metal = moles e⁻ ÷ n; mass = moles × M.

Worked example 1 (Basic)

A current of 7 A passes through a Ag⁺ solution for 36 min. Find the mass of metal deposited.

1. Q = 7 × 2160 = 15120 C
2. Moles e⁻ = 0.157
3. Mass = 0.157 × 108/1 = 16.922 g
4. Answer: 16.922 g

Worked example 2 (Standard)

A current of 4 A passes through a Zn²⁺ solution for 52 min. Find the mass of metal deposited.

1. Q = 4 × 3120 = 12480 C
2. Moles e⁻ = 0.129
3. Mass = 0.129 × 65/2 = 4.203 g
4. Answer: 4.203 g

Worked example 3 (Challenge)

How long must 5 A flow to deposit 2 g of metal from Ag⁺?

1. Moles = 2/108
2. Q = moles × 1 × 96500 = 1787.04 C
3. t = Q/I = 357.41 s
4. Answer: 357.41 s

Common mistakes and misconceptions

• Using minutes
• Forgetting to divide by n

Practice questions

1. (Basic) A current of 10 A passes through a Al³⁺ solution for 42 min. Find the mass of metal deposited.
2. (Basic) A current of 2 A passes through a Cu²⁺ solution for 31 min. Find the mass of metal deposited.
3. (Standard) A current of 7 A passes through a Zn²⁺ solution for 21 min. Find the mass of metal deposited.
4. (Standard) A current of 5 A passes through a Al³⁺ solution for 45 min. Find the mass of metal deposited.
5. (Challenge) How long must 9 A flow to deposit 4 g of metal from Ag⁺?
6. (Challenge) How long must 3 A flow to deposit 7 g of metal from Ni²⁺?

Answer key with full working

• Q1: Q = 10 × 2520 = 25200 C → Moles e⁻ = 0.261 → Mass = 0.261 × 27/3 = 2.35 g → Answer: 2.35 g
• Q2: Q = 2 × 1860 = 3720 C → Moles e⁻ = 0.039 → Mass = 0.039 × 63.5/2 = 1.224 g → Answer: 1.224 g
• Q3: Q = 7 × 1260 = 8820 C → Moles e⁻ = 0.091 → Mass = 0.091 × 65/2 = 2.97 g → Answer: 2.97 g
• Q4: Q = 5 × 2700 = 13500 C → Moles e⁻ = 0.14 → Mass = 0.14 × 27/3 = 1.259 g → Answer: 1.259 g
• Q5: Moles = 4/108 → Q = moles × 1 × 96500 = 3574.07 C → t = Q/I = 397.12 s → Answer: 397.12 s
• Q6: Moles = 7/58.7 → Q = moles × 2 × 96500 = 23015.33 C → t = Q/I = 7671.78 s → Answer: 7671.78 s

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
• Conductivity and molar conductivity
• Standard cell EMF and ΔG°
• Electrolysis cells in series
• 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 faraday's laws of electrolysis 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.