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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.

Respuesta :

Answer: 3.73 grams

Explanation:

[tex]Q=I\times t[/tex]

where Q= quantity of electricity in coloumbs

I = current in amperes = 96.0 A

t= time in seconds = 37.0 sec

[tex]Q=96.0A\times 37.0s=3552C[/tex]

The reaction at cathode is:

[tex]Pb^{2+}+2e^-\rightarrow Pb[/tex]

[tex]96500\times 2=193000Coloumb[/tex] of electricity deposits 1 mole of  

3552 C of electricity deposits =[tex]\frac{1}{193000}\times 3552=0.018moles[/tex] of Pb

[tex]\text {mass of Pb}={\text{moles}\times {\text{Molar mass}}[/tex]

[tex]\text {mass of Pb}={0.018}\times {207.2g/mol}=3.73g[/tex]

Thus mass of lead deposited is 3.73 g

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