Question

How many grams of solid nickel will be plated if an aqueous nickel(II) sulfate solution is electroplated over 10.5 min with a constant current of 3.20 A

Answer 1

613 mg

Step-by-step explanation:

`$Ni^(2+) + 2e \rightarrow Ni(s)$`

Number of fargday's
`$=(It)/(96500)$`

Here, I = 9.20 A

t = 10.5 min

= 10.5 x 60 seconds

So,
`$(It)/(96500)$`

`$=(3.20 * 10.5 * 60)/(96500)$`

= 0.0208 F

Here, 2e, 2F

2F = 1 mol of Ni

`$0.0208 \ F = (0.0208)/(2) = 0.0104 \ mol \ Ni$`

1 mol = 59 gm of Ni

0.0104 mol = 59 x0.0104 gm Ni

= 0.613 gm Ni

= (0.613 x 1000 ) mg of Ni

= 613 mg of Ni