Question

What mass (in g) of silver can be electroplated when 0.385 amps are used for 1.259 hours using a solution of silver nitrate? Answer:

Answer 1

Answer: 1.96 grams

Step-by-step explanation:

`Q=I* t`

where Q= quantity of electricity in coloumbs

I = current in amperes = 0.385 A

t= time in seconds = 1.259 hours = 4532.4 sec (1hour =3600 s)

`Q=0.385A* 4532.4s=1755C`

`AgNO_3\rightarrow Ag^++NO_3^-`

`Ag^++e^-\rightarrow Ag`

96500C of electricity deposits 1 mole of Ag

1755 C of electricity deposits =
`(1)/(96500)* 1755=0.018moles` of Ag

`Mass={\text {no of moles}}* {\text {Molar mass}}=0.018* 108=1.96g`

Thus 1.96 g of silver can be electroplated when 0.385 amps are used for 1.259 hours using a solution of silver nitrate.