How many grams of copper is deposited from an aqueous solution of copper sulfate, (CuSO4(aq)), by passing one mole of electrons into a copper plating cell?
A. 254g
B. 127g
D. 42.3g
E. 31.8g
Answers did not post.
To determine the number of grams of copper deposited, we need to use the concept of Faraday's law of electrolysis. According to this law, the mass of a substance deposited or liberated during electrolysis is proportional to the quantity of electricity (in moles of electrons) passed through the electrolytic cell.
The equation relating the moles of electrons passed (n), the Faraday constant (F), and the moles of substance (mol) deposited or liberated is given by:
mol = n/F
In this case, we are given that one mole of electrons is passed into the copper plating cell. The Faraday constant (F) is the charge of one mole of electrons and is equal to 96,485 Coulombs.
So, by plugging in these values into the equation, we can determine the moles of copper deposited:
mol = 1/96,485
Now, to find the mass of copper deposited, we need to use the molar mass of copper. The molar mass of copper is approximately 63.55 g/mol.
mass of copper = mol * molar mass of copper
mass of copper = (1/96,485) * 63.55
Simplifying this expression, we get:
mass of copper ≈ 0.000654 g
Rounding off to the appropriate number of significant figures, the answer is approximately 0.001 g.
Therefore, none of the options provided (A, B, D, or E) are correct.