Question

The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of $PbS{O}_4$ electrolysed in g during the process is:(molar mass of $PbS{O}_4$=303 g ${\mathrm{mol}}^{-1}$)

• A. 22.8
• B. 15.2
• C. 7.6
• D. 11.4

Answer 1

The correct option is B 15.2

According to the question, the reaction undergone by the half-cell is as follows: -

PbSO4 undergoes dissociation as:

At anode, the reaction is as follows :

$P{b}_{\left(s\right)}+S{O}_4^{2-}⟶PbS{O}_{4\left(s\right)}+2e^{-}$

From the equation, 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:

2Faraday→303$gmo{l}^{-1}$

Then,

1Faraday→3032$gmo{l}^{-1}$ =151.5$gmo{l}^{-1}$

1 Faraday liberates 151.5$gmo{l}^{-1}$ 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,

0.05Faraday→151.5g×0.05=7.575g