The anodic half-cell of lead-acid battery is recharged unsing electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process in: (Molar mass of PbSO4 = 303 g mol-1)

Question

The anodic half-cell of lead-acid battery is recharged unsing electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process in: (Molar mass of PbSO4 = 303 g mol-1) (A) 22.8 (B) 15.2 (C) 7.6 (D) 11.4

Tardigrade Admin · Accepted Answer

A) underset0.05 / 2 text mode PbSO 4+2 OH -arrow PbO 2+ H 2 SO 4+ underset0.05 F 2 e- B) underset0.05 / 2 text mole PbSO 4+ underset0.05 F +2 H + arrow Pb (s)+ H 2 SO 4 nT( PbSO 4)=(0.05/2) mole m PSO 4=(0.05/2) × 303=7.6 gm

Q. The anodic half-cell of lead-acid battery is recharged unsing electricity of $0.05$ Faraday. The amount of $PbSO X_{4}$ electrolyzed in g during the process in : (Molar mass of $PbSO X_{4} = 303 g mol X^{- 1}$ )

22.8

15.2

7.6

11.4

A) $0.05/2 mode P b S O_{4} + 2 O H - \to P b O_{2} + H_{2} S O_{4} + 0.05 F 2 e^{-}$
B) $0.05/2 mole P b S O_{4} + 0.05 F + 2 H^{+} \to P b (s) + H_{2} S O_{4}$
$n_{T} (P b S O_{4}) = \frac{0.05}{2}$ mole
$m_{PS O_{4}} = \frac{0.05}{2} \times 303 = 7.6 g m$

Q. The anodic half-cell of lead-acid battery is recharged unsing electricity of 0.05 Faraday. The amount of PbSOX4​ electrolyzed in g during the process in : (Molar mass of PbSOX4​=303 g molX−1)

22.8

15.2

7.6

11.4

A) 0.05/2 mode PbSO4​​+2OH−→PbO2​+H2​SO4​+0.05F2e−​ B) 0.05/2 mole PbSO4​​+0.05F​+2H+→Pb(s)+H2​SO4​ nT​(PbSO4​)=20.05​ mole mPSO4​​=20.05​×303=7.6gm

Q. The anodic half-cell of lead-acid battery is recharged unsing electricity of $0.05$ Faraday. The amount of $PbSO X_{4}$ electrolyzed in g during the process in : (Molar mass of $PbSO X_{4} = 303 g mol X^{- 1}$ )

A) $0.05/2 mode P b S O_{4} + 2 O H - \to P b O_{2} + H_{2} S O_{4} + 0.05 F 2 e^{-}$
B) $0.05/2 mole P b S O_{4} + 0.05 F + 2 H^{+} \to P b (s) + H_{2} S O_{4}$
$n_{T} (P b S O_{4}) = \frac{0.05}{2}$ mole
$m_{PS O_{4}} = \frac{0.05}{2} \times 303 = 7.6 g m$