Question

A common car battery consists of six identical cells each of which carries out the reaction: Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O Suppose that in starting a car on a cold morning a current of 350 amperes is drawn for 12.2 seconds from a cell of the type described above. How many grams of Pb would be consumed? (The atomic weight of Pb is 207.19.)

Answer 1

Answer:

The correct answer is 4.58 grams.

Explanation:

Based on the Faraday's law of electrolysis, at the time of electrolysis, the amount of deposited substance is directly equivalent to the concentration of the flow of charge all through the solution. If current, I, is passed for time, t, seconds and w is the concentration of the substance deposited, then w is directly proportional to I*t or w = zIt (Here z refers to the electrochemical equivalent or the amount deposited when 1 C is passed).  

For the reaction, n * 96500 C = molar mass

1C = molar mass/n*96500 = Equivalent wt / 96500

w = Equivalent wt / 96500 * I * t

In the given reaction,  

Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O, n = 2, the current or I drawn is 350 A, for time, t 12.2 seconds.  

Now putting the values in the equation we get,  

w = 207.19 / 2 * 96500 * 350 * 12.2 ( The molecular weight of Pb is 207.19 and the equivalent weight of Pb is 207.19 / 2)

w = 4.58 gm.