A 100 W and 110 V incandescent lamp is connected in series with an electrolytic cell containing `CdSO_(4)` solution. What mass of cadmium will be deposited at the cathode after 4 hours of electrolysis ?
[Atomic mass of Cd=112.2]

Explanation

Step I. Calculation of the quantity of charge passed.
We know that `" " "Watt"="ampere"xx"volt"`
`:. " " "Ampere"=("watt")/("volt")=(100)/(110)`
Now charge`="current"xx"time"=((100)/(110)"amp")xx(4xx60xx60s)`=13091C
Step II. Calculation of mass of cadmium deposited
The cathodic reaction is :
`underset(112.2 g)(Cd^(2+))(aq)+underset(2xx96500C)(2e^(-))toCd(s)`
`2xx96500 C` of charge deposit Cd=122.2 g.
13091 C of charge deposit Cd`=((112.2 g))/((2xx96500C))xx(13091C)=7.61 g`.