Question

The anodic half-cell of lead-acid battery is recharged using electricity of $ 0.05 $ Faraday. The amount of $ PbS{O_4} $ electrolyzed in g during the process is: (Molar mass of $ PbS{O_4}\, = \,303gmo{l^{ - 1}} $ )

Answer 1

Hint: An Electrochemical cell is composed of two half cells: Anode and Cathode. Anode is the oxidation half-cell, while cathode is the reduction half-cell. $ 1 $ Faraday is the measure of the amount of electric charge required to liberate one gram equivalent of any ion from an electrolytic solution.

Complete answer:
According to the question, the reaction undergone by the half-cell is as follows: -
 $ PbS{O_4} $ undergoes dissociation as:
 $ PbS{O_4}{\text{ }} \to {\text{ }}P{b^{2 + }}{\text{ }} + {\text{ }}SO_4^{2 - } $
 $ P{b^{2 + }} $ at anode is electrolyzed as:
From these equations, we know that the electrolysis reaction of $ PbS{O_4} $ produced $ 2 $ moles of electrons. Hence, we can say that $ 2 $ Faraday is the amount of electric charge required in the electrolysis reaction, where one gram equivalent of $ PbS{O_4} $ is liberated. This can be represented as:
 $ 2\,Faraday \to \,303gmo{l^{ - 1}} $
Then,
 $ 1\,Faraday \to \,\dfrac{{303}}{2}gmo{l^{ - 1}} = 151.5g $
 $ 1 $ Faraday liberates $ 151.5gmo{l^{ - 1}} $ of $ PbS{O_4}. $
We need to find the amount of $ PbS{O_4} $ electrolyzed in g during the process of recharge of the anodic half-cell of lead-acid battery using electricity of $ 0.05 $ Faraday.
Therefore,
 $ 0.05\,Faraday \to \,151.5g\, \times \,0.05\, = \,7.575g $
Hence, the required answer is $ 7.575g $ for the amount of $ PbS{O_4} $ electrolyzed in g during the process of recharge of anodic half-cell of lead-acid battery using electricity of $ 0.05 $ Faraday.

Note:
Electrolysis is the process where direct electric current is passed through an electrolyte, producing chemical reactions at the electrodes and decomposition of the materials. The important components required for electrolysis to take place are: electrolyte solution, electrodes and an external power source. During electrolysis of $ PbS{O_4}, $ at the anode, oxidation takes place and $ P{b^{2 + }} $ is liberated. At the cathode, reduction takes place and $ SO_4^{2 - } $ is liberated. The liberated $ P{b^{2 + }} $ is the oxidizing agent and $ SO_4^{2 - } $ is the reducing agent.