Question

How many minutes would be required to reduce 0.900 g of Au3+ to metallic gold with a current of 0.300 amperes?

Answer 1

According to Faraday's law of electrolysis:

the weight of substance liberated is directly proportional to the quantity of electricity passed

W=Z*Q (Where z is electrochemical constant)

W=Z*i*t (Q=i*t)

from the question,

W=0.9gm , i=0.3A, Z=(equivalent weight of substance)/96500 where equivalent weight= (atomic weight)/net charge gain or loss

so, Z=197/(3*96500)

putting all values,

(0.9/1000)=0.3*t*{197/(3*96500)}

so, t=4.408 sec hence it will 0.07346667 minutes to deposit required Au.