Calculation of Electricity Required to Deposit Aluminum from Aluminum Sulphate Solution

To determine the amount of electricity required to deposit 9 grams of aluminum from aluminum sulphate solution, we need to consider the principles of electrochemistry and Faraday's laws of electrolysis.

1. Faraday's First Law of Electrolysis:
According to Faraday's first law of electrolysis, the amount of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The equation representing this law is:

m = Z * I * t

Where:
m = mass of substance deposited
Z = electrochemical equivalent (amount of substance deposited by one coulomb of electricity)
I = current in amperes
t = time in seconds

2. Determining the Electrochemical Equivalent:
To calculate the electrochemical equivalent of aluminum, we need to consider its molar mass and the number of electrons involved in the reduction reaction. The reduction reaction for aluminum is:

Al3+ + 3e- -> Al

From the balanced equation, we can see that 3 moles of electrons are required to deposit 1 mole of aluminum. The molar mass of aluminum is 26.98 g/mol. Therefore, the electrochemical equivalent of aluminum is:

Z = molar mass of aluminum / (3 * Faraday's constant)

3. Calculation of Electricity Required:
To calculate the amount of electricity required to deposit 9 grams of aluminum, we can rearrange the equation from Faraday's first law of electrolysis:

t = m / (Z * I)

Substituting the values, we get:

t = 9 g / (Z * I)

4. Faraday's Second Law of Electrolysis:
Faraday's second law of electrolysis states that the same quantity of electricity is required to deposit different substances if the quantities of the substance are in the ratio of their equivalent weights. The equivalent weight of a substance is the mass of the substance that is equivalent to one mole of electrons.

5. Calculation of the Equivalent Weight of Aluminum:
The equivalent weight of aluminum can be calculated using the molar mass and the number of electrons involved in the reduction reaction. From the balanced equation, we know that 3 moles of electrons are required to deposit 1 mole of aluminum. Therefore, the equivalent weight of aluminum is:

Equivalent weight of aluminum = molar mass of aluminum / 3

6. Final Calculation:
Using the equivalent weight of aluminum, we can determine the amount of electricity required to deposit 9 grams of aluminum. We substitute the equivalent weight into the equation derived from Faraday's first law of electrolysis:

t = 9 g / (Z * I)
t = 9 g / (Equivalent weight of aluminum * I)

This equation will give us the time required to deposit 9 grams of aluminum. To convert this time into electricity, we need to calculate the charge (Q) in coulombs using the equation:

Q = I * t

Summary:
To determine the amount of electricity required to deposit 9 grams of aluminum from aluminum sulphate solution, we need to calculate the time required using Faraday's first law of electrolysis. This time can then be used to calculate