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When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO4 is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode. Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery.

User Abhishekkumar
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Answer: 3.73 grams

Step-by-step explanation:


Q=I* t

where Q= quantity of electricity in coloumbs

I = current in amperes = 96.0 A

t= time in seconds = 37.0 sec


Q=96.0A* 37.0s=3552C

The reaction at cathode is:


Pb^(2+)+2e^-\rightarrow Pb


96500* 2=193000Coloumb of electricity deposits 1 mole of

3552 C of electricity deposits =
(1)/(193000)* 3552=0.018moles of Pb


\text {mass of Pb}={\text{moles}* {\text{Molar mass}}


\text {mass of Pb}={0.018}* {207.2g/mol}=3.73g

Thus mass of lead deposited is 3.73 g

User Sikander
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