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MaRussiya [10]
3 years ago
12

If an electric current of 8.50 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, t

hen 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.)
Physics
2 answers:
muminat3 years ago
7 0

Answer:

75.5g

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=\frac{mass}{molar mass} \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g

Hence the amount of copper produced is 75.5g

Ann [662]3 years ago
5 0

Answer:

75.5 g.

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.

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