Faraday's law states that the amount of substance deposited at an electrode during electrolysis is directly proportional to the amount of charge passed through the electrode. The amount of charge is determined by the number of electrons transferred.
The formula for Faraday's law is:
$$m = \frac{MIt}{nF}$$
where:
- m is the mass of the substance deposited (in grams)
- M is the molar mass of the substance (in grams per mole)
- I is the current (in amperes)
- t is the time (in seconds)
- n is the number of electrons transferred per atom or molecule of the substance
- F is Faraday's constant (96,485 coulombs per mole)
In the case of copper, the molar mass is 63.55 grams per mole and each copper atom requires two electrons to be deposited.
Substituting the given values into the formula, we get:
$$6.35 g = \frac{63.55 g/mol \times I \times t}{2mol \times 96,485 C/mol}$$
Solving for I, we get:
$$I = \frac{6.35 g \times 2 mol \times 96,485 C/mol}{63.55 g/mol \times t}$$
This equation gives us the current required to deposit 6.35 grams of copper in a given amount of time. The number of electrons required can be calculated by multiplying the current by the time and dividing by Faraday's constant:
$$n = \frac{I \times t}{F}$$
Substituting the calculated value of I, we get:
$$n = \frac{(6.35 g \times 2 mol \times 96,485 C/mol)/(63.55 g/mol \times t) \times t}{96,485 C/mol}$$
Simplifying, we get:
$$n = \frac{6.35 g \times 2 mol}{63.55 g/mol}$$
$$n = 0.2 mol$$
Therefore, 0.2 moles of electrons would be required to deposit 6.35 grams of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate.
