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

Answer:

613 mg

Explanation:

$Ni^{2+} + 2e \rightarrow Ni(s)$

Number of fargday's  $=\frac{It}{96500}$

Here, I = 9.20 A

        t = 10.5 min

          = 10.5 x 60 seconds

So, $\frac{It}{96500}$

  $=\frac{3.20 \times 10.5 \times 60}{96500}$

 = 0.0208 F

Here, 2e, 2F

2F = 1 mol of Ni

$0.0208 \ F = \frac{0.0208}{2} = 0.0104 \ mol \ Ni$

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