Question

How long will it take to deposit 6.32 g of copper from a CuSO4(aq) solution using a current of 0.554 amps

Answer 1

34672.96 s

Step-by-step explanation:

m = Mass of copper = 6.32 g

M = Molar mass of copper = 63.5 g/mol

F = Faraday constant = 96500 C/mol

The electrode equation would be

`Cu^(2+)(aq)+2e^(-)=Cu(s)`

Number of electrons = 2 = e

I = Current = 0.554 A

t = Time taken

Charge would be

`Q=(m)/(M)eF\\\RightarrowQ=(6.32)/(63.5)* 2* 96500\\\Rightarrow Q=19208.82\ \text{C}`

Charge is given by

`Q=It\\\Rightarrow t=(Q)/(I)\\\Rightarrow t=(19208.82)/(0.554)\\\Rightarrow t=34672.96\ \text{s}`

Time taken to deposit the copper is 34672.96 s.