How many ions of copper wil be deposited by half Faraday of electricity through an aqueous solution of CuSO4?

To determine how many ions of copper will be deposited by half a Faraday of electricity through an aqueous solution of CuSO4, we need to understand a few key concepts related to electrolysis and Faraday's laws of electrolysis.

Understanding Electrolysis and Faraday's Laws

Electrolysis is a process that uses electrical energy to drive a non-spontaneous chemical reaction. In the case of copper sulfate (CuSO4) solution, when an electric current passes through it, copper ions (Cu²⁺) in the solution are reduced to solid copper (Cu) at the cathode.

Faraday's First Law of Electrolysis

Faraday's first law states that the amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity passed through the electrolyte. The relationship can be expressed mathematically as:

m = Z × Q

Where:

• m = mass of the substance deposited (in grams)
• Z = electrochemical equivalent of the substance (in grams per coulomb)
• Q = total electric charge (in coulombs)

Calculating the Charge

One Faraday (F) is equivalent to approximately 96485 coulombs. Therefore, half a Faraday would be:

Q = 0.5 × 96485 C = 48242.5 C

Electrochemical Equivalent of Copper

The electrochemical equivalent (Z) for copper can be calculated using its molar mass and the number of electrons transferred during the reduction of copper ions. Copper has a molar mass of about 63.5 g/mol and each Cu²⁺ ion requires 2 electrons to be reduced to Cu:

Z = (Molar mass of Cu) / (n × F)

Where:

• n = number of electrons transferred (2 for Cu²⁺)
• F = Faraday's constant (96485 C/mol)

Substituting the values:

Z = 63.5 g/mol / (2 × 96485 C/mol) = 0.000329 g/C

Calculating the Mass of Copper Deposited

Now, we can find the mass of copper deposited using the formula:

m = Z × Q

Substituting the values we have:

m = 0.000329 g/C × 48242.5 C ≈ 15.85 g

Determining the Number of Copper Ions Deposited

To find out how many copper ions are deposited, we need to convert the mass of copper deposited into moles and then into the number of ions. The number of moles (n) can be calculated as:

n = m / Molar mass

Substituting the values:

n = 15.85 g / 63.5 g/mol ≈ 0.249 moles

Now, using Avogadro's number (approximately 6.022 × 10²³ ions/mol), we can find the total number of copper ions:

Number of ions = n × Avogadro's number

Number of ions ≈ 0.249 moles × 6.022 × 10²³ ions/mol ≈ 1.5 × 10²³ ions

Final Thoughts

In summary, by passing half a Faraday of electricity through an aqueous solution of CuSO4, approximately 1.5 × 10²³ copper ions will be deposited. This illustrates the powerful relationship between electricity and chemical reactions in electrolysis, showcasing how we can quantify the effects of electrical energy on ionic solutions.