Question

The cell diagram for the lead-acid cell that is used in automobile and truck batteries is Pb(s)|PbSO4(s)H, SO, (aq)|P60, (8), PbSO4) Pb(s) The comma between PbO,(s) and PbSO (3) denotes a heterogeneous mixture of the two solids. The right-hand lead electrode is nonreactive. Write the balanced equation for the net cell reaction. equation: Look up standard potentials for the oxidation and reduction half-reactions, and then calculate the value of Econ call = Calculate the value of AG an AG in = ku-mol-1 Calculate the value of Ecell at 25°C if [H, S02] = 10.0 M. Ecell = How many lead-acid cells are in a 12 V car battery? Round to the nearest integer. number of lead-acid cells:

Answer 1

The given cell is, Pb(s) PbSO. (s)H,90 (aq)PbO2(),PbSO. (s) Pb(s) (a). The balanced reaction for the given cell is written as follows: Anode half reaction: Oxidation takes place at anode. Pb(s) + S0 - (aq) + PbSO4 + 2e Reduction potential for anode half-cell is -0.356 V. Cathode half reaction: Reduction takes place at cathode. PbO2 (s)+S02 (aq) + 4H* + 2e → PbSO4(s)+ 2H,0 Reduction potential of cathode half-cell is +1.685 V. Therefore, the overall balanced reaction is given as follows: Pb(s)+PbO2 (s)+2H2SO4 (aq) → PbSO. (s)+2H,0(1) (b). Calculate the value of Es for the given cell as follows: Ecell = Ecathode-Eanoda Here, Esthede is standard reduction potential of cathode and Ec is standard reduction potential of anode. Eat = EcathodeEnde = (1.685V)-(-0.356V) = 2.04 V Therefore, the value of Ell for the given cell is 2.04V.

(c). Calculate the value of AGº for the given cell as follows: AG° =-nFEA Here, n is number of electrons used in the reaction, F is faradays constant, and Ellis standard electrode potential of cell. AG° =-nFECT! =-2x(96485.33C/mol)x(2.04V) =-393660.14J/mol =-393.66 kJ/mol Therefore, the value of AGº for the given cell is -393.66 kJ/mol. (d). calculate the value of Ecell, if [H 30.] is 10 M as follows: E = T 103 [H.80.J) Here, Enis potential of cell, R is gas constant, T is temperature, n is number of electrons used, and F is faraday constant. RT, Fall E nF H.SO.) (8.314 JK-mol-) (298K). 2x(96485.33C/mol) = 2.04 V- (1) [10?) = 2.07 V Therefore, the value of Ecell, if [H 30.) is 10 M is 2.07 V.

(e). calculate the number of lead acid cells in 12 V car battery as follows: 2.07 V is produced by: 1 cell 12 V is produced by: 2.077*12 x12 V cell =5.86cells Therefore, the number of lead acid cells in 12 V car battery is 6.