Question

What current must pass through a solution of silver nitrate to deposit 0.8 g of silver if the current is passed for 10 minutes ?

Answer 1

`1.2"A"`

Explanation:

`Ag^(+) +erarr Ag`

No. moles `Ag` deposited `=(m)/(Ar)=0.8/108=0.0074`

From the equation we can see that the no. moles of electrons must be the same.

So the no. of electrons`=0.0074xx6.02xx10^(23)`

The. charge on an electron is `-1.6xx10^(-19)"C"`

So the total charge passed `Q` is given by:

`Q=0.0074xx6.02xx10^(23)xx(-1.6xx10^(-19))"C"`

`Q=-710"C"`

Electric current is the rate of flow of charge so:

`I=Q/t=(710)/(10xx60)=1.2"A"`