What mass of copper would be formed when a current of 10.0 A is passed through a solution of CuSO₄ for 1 hour? [Cu = 63.5; 1F = 96500C]

Yaseer924

10 years ago

The selected answer is wrong:

To find the mass of copper deposited you will use Q=It i.e.10:3600=36000C.Then you will report that 1F will deposit 63.5g of copper therefore 36000C will deposit 23.7g to 1 d.p

s.samuel

3 years ago

Yes the answer is B

Kevin_Aro

3 years ago

The amount of copper deposited during electrolysis can be calculated using Faraday's law, which states that the amount of substance produced at an electrode is directly proportional to the amount of electrical charge passed through the cell and the equivalent weight of the substance.

The equivalent weight of copper is its atomic weight divided by its valence, which is 63.5/2 = 31.75 g/equiv.

First, let's calculate the total charge passed through the cell:
Q = It = (10.0 A)(3600 s) = 36000 C

Next, let's calculate the number of equivalents of copper produced:
n = Q/F = 36000 C/96500 C/equiv = 0.373 equiv

Finally, let's calculate the mass of copper produced:
mass = n x equivalent weight = 0.373 equiv x 31.75 g/equiv = 11.84 g

Therefore, the mass of copper produced when a current of 10.0 A is passed through a solution of CuSO4 for 1 hour is approximately 11.8 g (option B).

MarkCotterell

9 months ago

I feel the answer is not correct because from my solvings I deduced that
m=Q\|z|f
36000\96500
=0.3730
recall;
number of moles=reacting mass\molar mass
i.e m\M
0.3730*63.5=23.7 3sf

Mercyebube

4 years ago

The answer is correct. 1 hour to seconds is 36000