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

Explanation:

$Q=I\times t$

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)

$Q=0.385A\times 4532.4s=1755C$

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

$Ag^++e^-\rightarrow Ag$

96500C of electricity deposits 1 mole of Ag

1755 C of electricity deposits =$\frac{1}{96500}\times 1755=0.018moles$ of Ag

$Mass={\text {no of moles}}\times {\text {Molar mass}}=0.018\times 108=1.96g$

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.