The lead acid storage battery is commonly used in cars. In dilute solutions, the overall reaction of — Electrochemistry / Thermodynamics Chemistry Question
Problem Context
The lead acid storage battery is commonly used in cars. In dilute solutions, the overall reaction of a cell in this battery is:
Pb(s) + PbO2(s) + 4 H+(aq) + 2 SO4 2–(aq) 2 PbSO4(s) + 2 H2O(l)
The variation of the standard cell potential Eº with temperature is shown below:
Sketch the setup of a cell of the lead acid storage battery, clearly showing the location of all components. Label the cathode and the anode and indicate the direction of flow of electrons through the external circuit as the cell is discharged.
Model Answer
PbO2 cathode Pb anode aq. H2SO4 e– = PbSO4(s)
Calculate ∆Gº for the reaction at 298 K.
Model Answer
Eº = 1.64 V + (0.00138 V K-1)T
At 298 K, Eº = 2.05 V
∆Gº = –nFEº = –(2)(96500 J V-1 mol-1)(2.05 V) = –396 kJ mol-1
Calculate ∆Hº for the reaction.
Model Answer
Since ∆Gº = ∆Hº – T∆Sº = –nFEº:
Eº = –(∆Hº/nF) + T(∆Sº/nF) = 1.64 V + T(0.00138 V K-1)
Thus ∆Hº = –(2)(96500 J V-1 mol-1)(1.64 V) = –317 kJ mol-1
Calculate ∆Sº for the reaction.
Model Answer
From the analysis in (c), ∆Sº = (2)(96500 J V-1 mol-1)(0.00138 V K-1) = +266 J mol-1 K-1
Calculate the cell potential E for the reaction at 5 ºC at pH = 5.0 and 0.10 M sulfate ion concentration.
Model Answer
E = Eº – RT / nF ln(Q)
E = (1.64 V + (0.00138 V K-1)•278 K) – (8.314 J mol-1 K-1)(278 K) / (2)(96500 J V-1 K-1) ln( 1 / [H+]4[SO4 2-]2 )
E = 2.024 V – (0.0120 V)ln( 1 / [1.0´10-5]4[0.10]2 )
E = 1.42 V