Learnosha CBSE 12
Chemistry Electrochemistry 4 Marks
Appeared in: 2026 (3 sets)
Electrochemistry is the study of the relationship between chemical energy and electrical energy. Many spontaneously occurring chemical reactions liberate electrical energy. In electrolysis, electrical energy is converted directly into chemical energy. The product of an electrolytic reaction depends on the nature of the material being electrolysed and the type of electrode used. Oxidising and reducing species present in the electrolytic cell and their standard electrode potential too, affect the products of electrolysis. Electrolysis plays an important role in most people's daily lives, whether it is for the manufacturing of aluminium, electroplating of metals, or the synthesis of chemical compounds. Michael Faraday was the first scientist who proposed two laws to explain the quantitative aspects of electrolysis, popularly known as Faraday's laws of electrolysis. Faraday's laws of electrolysis provide a basis for mathematical analysis of the mass deposited at electrodes and the amount of charge passed through them. Faraday's laws are fundamental in various applications, including electroplating, metal extraction, battery technology and chemical synthesis. These laws also help in environmental monitoring and in various chemistry experiments.
✓ Verified CBSE Marking Scheme Official Evaluation Steps & EL10 Insight

CBSE Class 12 Chemistry: Quantitative Aspects of Electrolysis

Problem Statement: A solution of $CuSO_4$ is electrolysed for 10 minutes with a current of 1.5 amperes. Calculate the mass of copper deposited at the cathode. (Atomic mass of $Cu = 63.5 \, \text{g/mol}$, $1F = 96500 \, \text{C/mol}$)

Step 1 (Formula / Governing Law & Given Data): [1.0 Mark]

According to Faraday’s First Law of Electrolysis, the mass ($w$) of the substance deposited is given by:

$$w = \frac{M \cdot I \cdot t}{n \cdot F}$$

Given Data:

  • Current ($I$) = $1.5 \, \text{A}$
  • Time ($t$) = $10 \, \text{min} = 10 \times 60 = 600 \, \text{s}$
  • Molar mass of $Cu$ ($M$) = $63.5 \, \text{g/mol}$
  • Valency factor ($n$) for $Cu^{2+} + 2e^- \rightarrow Cu(s)$ is $2$
  • Faraday's constant ($F$) = $96500 \, \text{C/mol}$

Step 2 (Step-by-Step Mathematical Working): [2.0 Marks]

First, calculate the total charge ($Q$) passed through the cell:

$$Q = I \times t = 1.5 \, \text{A} \times 600 \, \text{s} = 900 \, \text{C}$$

Now, substitute the values into the Faraday's equation:

$$w = \frac{63.5 \, \text{g/mol} \times 900 \, \text{C}}{2 \times 96500 \, \text{C/mol}}$$ $$w = \frac{57150}{193000} \, \text{g}$$ $$w \approx 0.2961 \, \text{g}$$

Step 3 (Final Concluded Answer): [1.0 Mark]

The mass of copper deposited at the cathode is $0.296 \, \text{g}$.

💡 EL10 Concept Insight: Think of electricity as a delivery service for electrons; since each copper ion needs exactly two electrons to turn into a solid metal atom, the more "delivery time" (current × time) you provide, the more copper atoms get dropped off at the electrode!

🚩 Report an error in this question Open & Practice in Interactive Hub ➔