Prashnikaप्रश्निका
‹ Back to the paper

Answer the following.

In the year 1832, Michael Faraday performed many experiments on electrolysis. He established a…

Chemistry20263 marksCase based
In the year 1832, Michael Faraday performed many experiments on electrolysis. He established a relationship between the amount of products liberated at the electrodes and the quantity of electricity passed through the solution to carry out electrolysis. He formulated two laws of electrolysis that are known as Faraday’s First Law and Second Law of Electrolysis. Eva, a class XII student, studied these laws and tried to find a particular amount of charge in faraday to get $27\text{ g}$ of aluminium in an experiment involving $\text{Al}^{3+} / \text{Al}$ cell.
(i)[1.0]
State Faraday’s First Law of Electrolysis.
(ii)[1.0]
How much charge in faraday would Eva need to get a coating of $27\text{ g}$ of aluminium at electrode, if she follows the following reaction? $\text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al}$ (Atomic mass of $\text{Al} = 27\text{ g mol}^{-1}$)
(iii)[1.0]
How much mass of zinc in grams will be obtained if the amount of charge, calculated in subpart (ii), is passed through $\text{Zn}^{2+} / \text{Zn}$ cell for the same duration? $\text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn}$ (Atomic mass of $\text{Zn} = 65\text{ g mol}^{-1}$)

Answer

Answer (i)

AI
The mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte, $w \propto Q$.

Answer (ii)

AI

Formula used: Faraday's first law: w = M*Q/(n*F) with F = 96500 C mol^-1, M = 27 g mol^-1, n = 3.0, w = 27 g, so Q = 289500 C

$\mathrm{Al^{3+} + 3e^- \rightarrow Al}$: 3 mol of electrons (3 F) deposit 1 mol (27 g) of Al.

Final answer: 3 F

Answer (iii)

AI

Formula used: Faraday's first law: w = M*Q/(n*F) with F = 96500 C mol^-1, M = 65 g mol^-1, Q = 289500 C, n = 2.0, so w = 97.5 g

Charge = 3 F. For $\mathrm{Zn^{2+} + 2e^- \rightarrow Zn}$, 2 F deposit 65 g, so 3 F deposit $\frac{65}{2} \times 3 = 97.5$ g.

Final answer: 97.5 g

Electrochemistry

From ISC 2026 Improvement Chemistry Paper 1, question 1(C).

Check your working with the Chemistry numericals calculator: solutions, electrochemistry and kinetics formulas: fill in what you know and get the rest.