Chemical Effects of Electric Current – Numerical Problems with Stepwise Solutions
CBSE Class 8 Science – Chapter Wise Study Materials Based on NCERT
Chapter 14: Chemical Effects of Electric Current – Numerical Problems with Stepwise Solutions
Suitable for:
Annual Examination • Periodic Tests • Class Tests • Pre-Board / School Level Board Exam Pattern
These Numerical Problems with Stepwise Solutions are designed strictly as per the NCERT syllabus, making them ideal for CBSE Class 8 board exams standard.
Content Bank – Important Points & Simple Relations (Chapter 14)
- Electric current (I): Amount of charge flowing per second.
- Simple relation: Q = I × t, where Q = total charge flowing through the circuit, I = current, t = time of flow.
- If current is constant, I = Q / t and t = Q / I.
- More current or more time → more charge flows → generally more chemical effect (more gas, more metal deposited).
- Amount of metal deposited on cathode in electroplating increases when:
- Current in the circuit is increased, or
- Time for which current is passed is increased.
- For simple comparison:
- If time is doubled (current same) → charge doubles → deposited metal approximately doubles.
- If current is doubled (time same) → charge doubles → deposited metal approximately doubles.
- Charge is often expressed in coulombs (C). For Class 8 level, we only use Q = I × t to compare effects.
Below are 20 topic-wise numerical problems with clear, step-by-step solutions from
Chapter 14: Chemical Effects of Electric Current to help students build strong conceptual understanding
and exam-ready numerical skills.
Topic-wise Numerical Problems – Chapter 14: Chemical Effects of Electric Current
Topic 1 – Electric Current Through Liquids (Basic Numericals) – Q1 to Q5
Q1.
A current of 0.5 A flows through a salt solution for 4 minutes. Calculate the total time in seconds and the total charge that flows through the solution.
Solution:
- Given current, I = 0.5 A
- Time given, t = 4 minutes
- Convert time into seconds: 1 minute = 60 s t = 4 × 60 = 240 s
- Use relation: Q = I × t
- Q = 0.5 × 240 = 120 C
Answer: Time = 240 s, charge flowing through the solution = 120 C.
Q2.
In an experiment, 90 C of charge flows through a conducting liquid in 3 minutes. Find the current flowing through the liquid.
Solution:
- Total charge, Q = 90 C
- Time, t = 3 minutes = 3 × 60 = 180 s
- Use relation: I = Q / t
- I = 90 / 180 = 0.5 A
Answer: Current flowing through the liquid is 0.5 A.
Q3.
A small LED glows when a current of 0.02 A flows through lemon juice. If the LED is kept ON for 5 minutes, calculate the total charge that has passed.
Solution:
- Current, I = 0.02 A
- Time, t = 5 minutes = 5 × 60 = 300 s
- Use Q = I × t
- Q = 0.02 × 300 = 6 C
Answer: Total charge passed through lemon juice = 6 C.
Q4.
A solution allows a charge of 150 C to pass in 5 minutes. Another solution allows the same charge to pass in only 2 minutes using the same battery. Which solution shows larger current and by what factor?
Solution:
- Common charge for both solutions, Q = 150 C.
- Solution A: t₁ = 5 minutes = 5 × 60 = 300 s Solution B: t₂ = 2 minutes = 2 × 60 = 120 s
- Current in Solution A: I₁ = Q / t₁ = 150 / 300 = 0.5 A
- Current in Solution B: I₂ = Q / t₂ = 150 / 120 = 1.25 A
- Compare currents: I₂ / I₁ = 1.25 / 0.5 = 2.5
Answer: Solution B has a larger current, 2.5 times the current in Solution A.
Q5.
A student wants a current of 0.4 A to flow through a salt solution for 10 minutes. Calculate the total charge that will pass and explain what this means in terms of chemical effect.
Solution:
- I = 0.4 A, t = 10 minutes = 10 × 60 = 600 s
- Use Q = I × t
- Q = 0.4 × 600 = 240 C
- A larger charge (240 C) means more ions have moved through the solution, so the chemical effects (like gas formation or metal deposition) will be noticeable.
Answer: Total charge = 240 C. This large charge will produce a clear chemical effect in the solution.
Topic 2 – Chemical Effects: Gas Formation & Colour Change – Q6 to Q10
Q6.
A current of 0.3 A is passed through acidified water for 8 minutes. If charge is directly related to the amount of gas formed at the electrodes, by what factor will the amount of gas change if the time is doubled while keeping current same?
Solution:
- Initial time t₁ = 8 minutes; final time t₂ = 2 × 8 = 16 minutes.
- Charge is given by Q = I × t.
- If current I is same, then Q is directly proportional to t.
- So Q₂ / Q₁ = t₂ / t₁ = 16 / 8 = 2.
Answer: The amount of gas formed will also become 2 times when time is doubled.
Q7.
In a lab activity, a current of 0.25 A is passed through a coloured solution for 12 minutes. After the experiment, the colour becomes lighter. If another student passes a current of 0.5 A for the same time, compare the total charge passed in both cases.
Solution:
- Common time, t = 12 minutes = 12 × 60 = 720 s
- Case 1: I₁ = 0.25 A Q₁ = I₁ × t = 0.25 × 720 = 180 C
- Case 2: I₂ = 0.5 A Q₂ = I₂ × t = 0.5 × 720 = 360 C
- Compare: Q₂ / Q₁ = 360 / 180 = 2
Answer: The second student passes twice the charge (360 C) compared to the first student (180 C), so the colour change will be more noticeable.
Q8.
In a demonstration, 60 C of charge produces a small amount of gas at the electrodes of a solution. How much charge is needed to produce three times that amount of gas (assuming same solution and conditions)?
Solution:
- Amount of gas ∝ charge Q.
- If gas is to be tripled, charge must also be tripled.
- Required charge Q₂ = 3 × 60 C = 180 C.
Answer: To produce three times the gas, a charge of 180 C is required.
Q9.
A teacher passes a current of 0.4 A through acidified water for 5 minutes. Another teacher uses the same setup but with a current of 0.8 A for 5 minutes. By what factor will the total amount of chemical change differ?
Solution:
- Common time t = 5 minutes = 300 s.
- Teacher 1: I₁ = 0.4 A Q₁ = I₁ × t = 0.4 × 300 = 120 C
- Teacher 2: I₂ = 0.8 A Q₂ = I₂ × t = 0.8 × 300 = 240 C
- Chemical change ∝ charge, so Q₂ / Q₁ = 240 / 120 = 2
Answer: The second setup will produce twice the chemical change as the first.
Q10.
In an experiment, a solution is tested with a current of 0.1 A for 10 minutes and a slight colour change is observed. If you want four times the total charge to pass through the same solution, how long should you pass the same current?
Solution:
- Original time, t₁ = 10 minutes = 600 s; I = 0.1 A.
- Original charge, Q₁ = I × t₁ = 0.1 × 600 = 60 C.
- Required charge, Q₂ = 4 × Q₁ = 4 × 60 = 240 C.
- Using Q₂ = I × t₂ 240 = 0.1 × t₂ t₂ = 240 / 0.1 = 2400 s.
- Convert t₂ to minutes: 2400 / 60 = 40 minutes.
Answer: The current of 0.1 A should be passed for 40 minutes to get four times the total charge.
Topic 3 – Copper Sulphate Activity & Current–Time Comparisons – Q11 to Q14
Q11.
In a copper sulphate (CuSO4) experiment, a current of 0.3 A is passed for 30 minutes using copper electrodes. In another trial, a current of 0.6 A is passed for 15 minutes. In which trial will more copper be deposited at the cathode?
Solution:
- Chemical effect ∝ charge Q = I × t.
- Trial 1: I₁ = 0.3 A, t₁ = 30 minutes = 1800 s Q₁ = 0.3 × 1800 = 540 C
- Trial 2: I₂ = 0.6 A, t₂ = 15 minutes = 900 s Q₂ = 0.6 × 900 = 540 C
- Q₁ = Q₂ = 540 C, so chemical effect and copper deposited are same in both trials.
Answer: Both trials produce the same amount of copper deposition, since the total charge in each is 540 C.
Q12.
0.5 A current is passed through CuSO4 solution for 20 minutes. Calculate the total charge and comment on what will happen if the current is reduced to 0.25 A for the same time.
Solution:
- First case: I₁ = 0.5 A, t = 20 minutes = 1200 s
- Q₁ = I₁ × t = 0.5 × 1200 = 600 C
- Second case: I₂ = 0.25 A, time same (1200 s)
- Q₂ = I₂ × t = 0.25 × 1200 = 300 C
- Q₂ = 300 C is half of Q₁, so metal deposition and colour change will be roughly half as much.
Answer: First case charge = 600 C, second case = 300 C. With 0.25 A, chemical effects are about half as strong.
Q13.
A student wants to pass 480 C of charge through a copper sulphate solution using a current of 0.4 A. How long (in minutes) should the current be maintained?
Solution:
- Given Q = 480 C, I = 0.4 A.
- Use t = Q / I.
- t = 480 / 0.4 = 1200 s.
- Convert seconds to minutes: 1200 / 60 = 20 minutes.
Answer: The current of 0.4 A should be passed for 20 minutes.
Q14.
In a CuSO4 experiment, 360 C of charge is passed. In another experiment with the same solution and electrodes, 900 C of charge is passed. How many times more copper will be deposited in the second experiment compared to the first?
Solution:
- Copper deposited ∝ charge Q.
- First experiment: Q₁ = 360 C
- Second experiment: Q₂ = 900 C
- Ratio: Q₂ / Q₁ = 900 / 360 = 2.5
Answer: The second experiment will deposit 2.5 times more copper than the first.
Topic 4 – Electroplating & Applications – Q15 to Q19
Q15.
An iron key is electroplated with copper using a current of 0.6 A for 25 minutes. Calculate the total charge that flows during electroplating.
Solution:
- I = 0.6 A, t = 25 minutes = 25 × 60 = 1500 s.
- Use Q = I × t.
- Q = 0.6 × 1500 = 900 C.
Answer: Total charge during electroplating = 900 C.
Q16.
A factory electroplates 100 spoons in a bath. For each spoon, a current of 0.5 A is passed for 15 minutes. Find the charge passed for one spoon and then for all 100 spoons (assume plating is done one spoon at a time with same current and time).
Solution:
- For one spoon: I = 0.5 A, t = 15 minutes = 900 s.
- Q (one spoon) = I × t = 0.5 × 900 = 450 C.
- For 100 spoons: Qtotal = 100 × 450 = 45,000 C.
Answer: Charge for one spoon = 450 C; for 100 spoons = 45,000 C.
Q17.
In a chrome-plating process, a car part is connected as cathode and a current of 1.2 A is passed for 40 minutes. Later, for another similar part, only 0.6 A is used for the same time. Compare the charges and state which part will have a thicker chromium layer.
Solution:
- Time t = 40 minutes = 2400 s (common).
- Part 1: I₁ = 1.2 A, Q₁ = 1.2 × 2400 = 2880 C.
- Part 2: I₂ = 0.6 A, Q₂ = 0.6 × 2400 = 1440 C.
- Q₁ / Q₂ = 2880 / 1440 = 2.
- Part 1 has twice the charge flowing, so it will get roughly twice the thickness of chromium compared to Part 2.
Answer: The first part will have a thicker chromium layer, about two times thicker than the second.
Q18.
A jeweller gold-plates small ornaments with a current of 0.2 A for 30 minutes each. If he wants the same amount of gold deposition in only 15 minutes, what current should he use (approximately), assuming other conditions remain same?
Solution:
- In first case: I₁ = 0.2 A, t₁ = 30 minutes = 1800 s.
- Charge Q₁ = I₁ × t₁ = 0.2 × 1800 = 360 C.
- For same deposition, Q₂ must be equal to Q₁ = 360 C.
- Second case: t₂ = 15 minutes = 900 s, I₂ = ?
- Use Q₂ = I₂ × t₂ 360 = I₂ × 900 I₂ = 360 / 900 = 0.4 A.
Answer: The jeweller should use a current of about 0.4 A to get the same gold deposition in 15 minutes.
Q19.
An iron key and an iron spoon are both electroplated with silver using the same current. The key is plated for 12 minutes, while the spoon is plated for 24 minutes. Compare the amount of silver deposited on both, assuming conditions are otherwise identical.
Solution:
- Current I is same in both cases.
- Time for key, t₁ = 12 minutes; time for spoon, t₂ = 24 minutes.
- Charge Q ∝ time when current fixed.
- Ratio: Q₂ / Q₁ = t₂ / t₁ = 24 / 12 = 2.
Answer: The spoon will get about twice as much silver deposited as the key.
Topic 5 – Mixed & Higher-Order Numerical Reasoning – Q20
Q20.
A student prepares a model to show the chemical effects of electric current. She wants a total charge of 300 C to pass through an electrolytic cell.
She has two choices:
She has two choices:
- Choice A: Use a current of 0.5 A.
- Choice B: Use a current of 0.75 A.
Solution:
- Total charge needed, Q = 300 C.
- Choice A: I₁ = 0.5 A t₁ = Q / I₁ = 300 / 0.5 = 600 s Convert to minutes: 600 / 60 = 10 minutes.
- Choice B: I₂ = 0.75 A t₂ = Q / I₂ = 300 / 0.75 = 400 s Convert to minutes: 400 / 60 ≈ 6.67 minutes (about 6 minutes 40 seconds).
- Compare times: Difference ≈ 10 − 6.67 = 3.33 minutes (about 3 minutes 20 seconds).
Answer: With 0.5 A, time needed = 10 minutes.
With 0.75 A, time needed ≈ 6.7 minutes.
Choice B is faster by about 3.3 minutes.
These 20 topic-wise numerical problems with stepwise solutions cover key ideas of
Chapter 14: Chemical Effects of Electric Current, including current–time–charge relation,
gas formation, colour change, copper sulphate activity and electroplating, aligned with the
CBSE Class 8 NCERT syllabus.