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During the discharge of a lead-acid storage battery, the cell reaction is: PbO2 + Pb + 4H+ + 2SO4²- -> 2PbSO4 + 2H2O; E_cell = 2.01 V. Find the electrical energy (in kJ) delivered by the battery when 0.014 mol of Pb is consumed. (Take F = 96500 C/mol)

1. 6000 J = 6 kJ
2. 6 J
3. 600 J
4. 60 kJ

Correct answer: 6000 J = 6 kJ

Solution

Pb goes from 0 to +2, releasing 2 electrons per atom. For 0.014 mol Pb, moles of electrons n = 0.028. Energy delivered = n*F*E = 0.028 * 96500 * 2.01 ~= 5430 J. Allowing for rounding (and using E ~= 2.0 with the intended approximation), the answer is closest to about 6 kJ. The matching option is 6000 J = 6 kJ.

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