Electrochemistry, rechargeable batteries & fuel cells answers
1. (a) Lead metal(b) Anode (−): Pb + SO42− ⟶ PbSO4 + 2e−Cathode (+): PbO2 + 4H+ + SO42− + 2e− ⟶ PbSO4 + 2H2OOverall reaction: Pb + PbO2 + 4H+ + 2SO42− ⟶ 2PbSO4 + 2H2O (c) 2PbSO4 + 2H2O ⟶ Pb + PbO2 + 4H+ + 2SO42− (d) Since six cells in series gives 12.00 V each cell produces 2.00 V so the EMF = 2.00 VFrom Section 24: Pb ⟶ Pb2+ + 2e− E⦵ = − 0.13 VE⦵ for the half-reaction at the cathode = + 1.87 V(e) Any two...