In a common car battery, six identical cells each carry out the following reaction:
Pb + PbO₂ + 2HSO₄^- + 2H⁺ → 2PbSO₄ + 2H₂O

Suppose that to start a car on a cold morning, 138 amperes is drawn for 12.0 seconds from such a cell. How many grams of Pb are consumed? (The atomic mass of Pb is 207.19 g/mol.)

Question

Quizplus · Accepted Answer

First, we need to find the amount of charge that passes through the cell using the equation Q = It, where Q is charge (coulombs), I is current (amperes), and t is time (seconds). 
Q = (138 A)(12.0 s) = 1656 C 
Next, we use the balanced equation to find the moles of Pb consumed per cell: 
1 mol Pb : 2 mol e- (from the half-reaction) 
2 mol e- : 2HSO4- 
1 mol Pb : 2HSO4- 
1 mol Pb : 2(98.08 g/mol) = 196.16 g/mol 
2 mol HSO4- : 1 mol Pb 
1656 C × 1 mol e-/96500 C (Faraday constant) × 1 mol HSO4- / 2 mol e- × 1 mol Pb / 2 mol HSO4- × 196.16 g/mol Pb = 1.78 g Pb

In a Common Car Battery, Six Identical Cells Each Carry