The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is: (Molar mass of PbSO4 = 303 g mol–1)
For 2F current passed, PbSO4 deposited = 303 g/mol
For 0.05 F: PbSO4 deposited
The lead-acid battery involves electrochemical reactions during charging and discharging. The anodic half-cell reaction during charging is the reverse of the discharge reaction. For the anodic half-cell in a lead-acid battery, the reaction during charging is:
This equation shows that 2 moles of electrons are required to electrolyze 1 mole of PbSO4.
Given: Electricity used = 0.05 F (Faraday). Since 1 Faraday corresponds to 1 mole of electrons, 0.05 F corresponds to 0.05 moles of electrons.
From the reaction stoichiometry: 2 moles of electrons electrolyze 1 mole of PbSO4.
So, moles of PbSO4 electrolyzed = (moles of electrons) / 2 = 0.05 / 2 = 0.025 moles.
Molar mass of PbSO4 = 303 g/mol.
Therefore, mass of PbSO4 electrolyzed = moles × molar mass = 0.025 × 303 = 7.575 g, which is approximately 7.6 g.
Thus, the correct answer is 7.6 g.
Faraday's Laws of Electrolysis: These laws relate the amount of substance deposited or dissolved at an electrode during electrolysis to the quantity of electricity passed. The first law states that the mass of substance deposited is proportional to the charge passed. The second law states that when the same charge is passed through different electrolytes, the masses of substances deposited are proportional to their equivalent weights.
Stoichiometry in Electrochemistry: Electrochemical reactions involve transfer of electrons, and the stoichiometry can be used to relate the moles of electrons (Faradays) to moles of reactants and products.
Mass of substance electrolyzed, m = (Q × M) / (n × F)
Where:
Q = charge passed in coulombs (but here given directly in Faradays, so Q = number of Faradays × F)
M = molar mass
n = number of electrons involved in the reaction
F = Faraday constant (96485 C/mol, but when charge is in Faradays, it simplifies since 1 F = 1 mole of electrons)
In this case, since charge is given as 0.05 F, it directly gives moles of electrons = 0.05. Then m = (moles of electrons × M) / n = (0.05 × 303) / 2 = 7.575 g ≈ 7.6 g.