Question

If an electric current of 8.50 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, then how much copper is deposited on the cathode (the negative electrode) of the cell? (Copper ions carry two units of positive elementary charge, and the atomic mass of copper is 63.5 g/mol.)

Answer 1

75.5 g.

Step-by-step explanation:

Dissociation equation:

Cu2+ + SO4^2- --> CuSO4

Q = I * t

Where,

I = current

= 8.5 A

t = time

= 3.75 hours

= 13500 s

Q = 8.5 * 13500

= 114750 C

1 faraday represent one mole of electron which equal 96500C

Number of mole of Cu2+

= 114750/96500

= 1.19 mol.

Mass = number of moles * molar mass

= 1.19 * 63.5

= 75.5 g.

Answer 2

75.5g

Step-by-step explanation:

From the ionic equation, we can write

`CU^(2+)+SO^(2-)_(4)\\`

next we find the number of charge

Note Q=it

for i=8.5A, t=3.75 to secs 3.75*60*60=13500secs

hence

`Q=8.5*13500\\Q=114750C`

Since one faraday represent one mole of electron which equal 96500C

Hence the number of mole produced by 114750C is

114750/96500=1.2mol

The mass of copper produced is

`mol=(mass)/(molar mass) \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g`

Hence the amount of copper produced is 75.5g