3. What mass of copper could be deposited from a copper(II) Sulphate solution using a current of 5.0 A over 100 seconds? ( F =96
1 answer:
Mass of copper : 0.165 g
<h3>Further explanation</h3>Given
5.0 A over 100 seconds
Required
Mass of copper
Solution
Faraday's law:
<em>The mass of the substance formed at each electrode is proportional to the electric current flowing in the electrolysis</em>
<em /><em />
e = Ar / valence = eqivalent weight
i = current
t = time
W = weight
CuSO₄ ----> Cu²⁺ + SO₄²⁻
Cu ----> Cu²⁺ + 2e
e = Ar/2
= 63,5/2 = 31,75
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Hey there! Hello!
Not sure if you still need the answer to this question, but I'd love to help out if you do.
So, the way to balance this equation is pretty simple. First, you need to keep in mind that molecules of hydrogen and oxygen do not come in single molecules, but in bonded pairs, represented by H2 and O2.
But, that's incorrect. The combination of 2 hydrogen molecules with 1 oxygen molecule yields water, but that leaves one oxygen molecule leftover. When broken down, this is how many of each molecule is on each side of the previously stated equation:
Left:
H: 2
O: 2
Right:
H: 2
O: 1
So we have to multiply H2O on the right side by 2 in order to get this:
Left:
H: 2
O: 2
Right:
H: 4
O: 2
The last step is to multiply H2 on the left by two to make it match up with the right side, balancing the equation:
Left:
H: 4
O: 2
Right:
H: 4
O: 2
That makes our equation balanced! I hope this helped you out, feel free to ask any additional questions if you need further clarification. :-)
Not sure if you still need the answer to this question, but I'd love to help out if you do.
So, the way to balance this equation is pretty simple. First, you need to keep in mind that molecules of hydrogen and oxygen do not come in single molecules, but in bonded pairs, represented by H2 and O2.
But, that's incorrect. The combination of 2 hydrogen molecules with 1 oxygen molecule yields water, but that leaves one oxygen molecule leftover. When broken down, this is how many of each molecule is on each side of the previously stated equation:
Left:
H: 2
O: 2
Right:
H: 2
O: 1
So we have to multiply H2O on the right side by 2 in order to get this:
Left:
H: 2
O: 2
Right:
H: 4
O: 2
The last step is to multiply H2 on the left by two to make it match up with the right side, balancing the equation:
Left:
H: 4
O: 2
Right:
H: 4
O: 2
That makes our equation balanced! I hope this helped you out, feel free to ask any additional questions if you need further clarification. :-)