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 ?

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Answer 1 · 2015-07-31T15:05:00

#1.2"A"#

#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"#