If 3.00×10^-3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.75 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.
A
One Response
I = ne/(change in time)
convert kg to g
Gold atom weight is 196.97
n = 3 g/196.97g/mol = .01523
Avagdros # = 6.02 e 23
.01523(6.02 x 10^23) = 9.168 x 10^21
plug and chug
I = [(9.168e21)(1.60e-19)]/9900
Make sure you check the calculations but this should be correct!