CBSE 2026 · Region 5 · Set 1 · Q30 · 4 marks

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. [] Answer the following questions :

(a)

Predict the products of electrolysis in each of the following :

(i)

An aqueous solution of $\displaystyle \mathrm{CuCl}_{2}$ with platinum electrodes.

(ii)

A concentrated solution of $\displaystyle \mathrm{H}_{2} \mathrm{SO}_{4}$ with platinum electrodes.

(b)

(i)OR(b)

How much charge in faraday is required for the reduction of $\displaystyle 1$ mol of $\displaystyle \mathrm{Ag}^{+}$to Ag ?

(c)

The following reactions occur at the anode during the electrolysis of aqueous sodium chloride solution : \[\begin{aligned} & \mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow \frac{1}{2} \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{e}^{-} \mathrm{E}_{(\text {cell })}^{\mathrm{o}}=1.36 \mathrm{~V} \\ & 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} \mathrm{E}_{(\text {cell })}^{\mathrm{o}}=1.23 \mathrm{~V} \end{aligned} \]

(I)

(II)

Which reaction is feasible at the anode and why ?

CBSE Class 12 Chemistry past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.