A lead storage cell is discharge which causes H2SO4 electrolyte to change from a concentration of 40% by weight(density=1.25g/mL) to 30% by weight. The original volume of electrolyte is 1L,how many faraday has left the anode of battery ?
Amount of sulphuric acid initially = 40% by weight = 400 g in 1 L electrolyte
Amount of sulphuric acid after discharging = 30% by weight = 300 g in 1 L electrolyte
Amount reacted = 100g
Pb = Pb2+ + 2e- ( oxidation)
PbO2 + 4 H+ + 2e- = Pb2+ + 2 H2O ( reduction)
Pb + PbO2 + 2H2SO4 = 2 PbSO4 + 2 H2O
Basically two electrons are involved in the reaction that means the charge released is 2 Faraday (Q=nF).
This means that 2 moles of H2SO4 or 392 g of H2SO4 release 2 Faraday.
Therefore charge released by 100 g of H2SO4 =
Amount of sulphuric acid after discharging = 30% by weight = 300 g in 1 L electrolyte
Amount reacted = 100g
Pb = Pb2+ + 2e- ( oxidation)
PbO2 + 4 H+ + 2e- = Pb2+ + 2 H2O ( reduction)
Pb + PbO2 + 2H2SO4 = 2 PbSO4 + 2 H2O
Basically two electrons are involved in the reaction that means the charge released is 2 Faraday (Q=nF).
This means that 2 moles of H2SO4 or 392 g of H2SO4 release 2 Faraday.
Therefore charge released by 100 g of H2SO4 =