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. 254 g
B. 127 g
C. 63.5 g
D. 42.3 g
E. 31.8 g
Correct
To determine the number of grams of copper deposited from the aqueous solution of copper sulfate, we need to use the concept of electrochemistry and Faraday's law of electrolysis.
The first step is to calculate the number of moles of electrons (n) that are needed to deposit one mole of copper (Cu):
1 mole Cu ↔ 2 moles e-
From the equation, we can see that it takes 2 moles of electrons (2e-) to deposit one mole of copper (Cu). Therefore, the value of 'n' is 2.
Next, we need to calculate the amount of charge (Q) that is associated with one mole of electrons. This can be done using Faraday's constant (F):
1 mole of electrons = Q coulombs
Q = n x F
The Faraday's constant (F) is given as 96,485 C/mol.
Q = 2 x (96,485 C/mol) = 192,970 C
Now, we can calculate the amount of copper deposited (m) using the equation:
m = Q x (molar mass of Cu / Faraday's constant)
The molar mass of copper (Cu) is 63.5 g/mol.
m = 192,970 C x (63.5 g/mol / 96,485 C/mol)
Simplifying the expression:
m ≈ 192,970 C x 0.001 g/C ≈ 192.97 g
Therefore, the correct answer is not among the given options.