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

Answer:

34672.96 s

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

$Q=\dfrac{m}{M}eF\\\RightarrowQ=\dfrac{6.32}{63.5}\times 2\times 96500\\\Rightarrow Q=19208.82\ \text{C}$

Charge is given by

$Q=It\\\Rightarrow t=\dfrac{Q}{I}\\\Rightarrow t=\dfrac{19208.82}{0.554}\\\Rightarrow t=34672.96\ \text{s}$

Time taken to deposit the copper is 34672.96 s.