constant electric current flows between two electrochemical cells connected in series for 3.25 seconds. one contains a solution of AgNO2 and the other contains CuCl2.During this time 2g of silver are deposited on the first cell. How any grams of copper are deposited on the second cell?

Answer 1

To solve this problem, we will use Faraday's Law of Electrolysis, which states that the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the solution.

First, we need to calculate the quantity of electricity that passed through the solution in the first cell using the equation:

Q = I * t

Where:
Q is the quantity of electricity (coulombs)
I is the current (amps)
t is the time (seconds)

Given that the electric current is constant and equal in both cells, we can use the same value for both cells. Let's assume the electric current is I amps.

Therefore, Q = I * 3.25 seconds.

Next, we need to calculate the quantity of electricity required to deposit 2g of silver in the first cell. To do this, we use the equation:

Q = (m/M) * z * F

Where:
m is the mass of the substance (in grams)
M is the molar mass of the substance (in grams/mol)
z is the number of electrons transferred in the balanced redox equation
F is Faraday's constant (96485 C/mol)

For silver (Ag), the molar mass (M) is 107.87 g/mol, and the number of electrons transferred (z) is 1.

Therefore, Q = (2g / 107.87 g/mol) * 1 * 96485 C/mol

Now, we can equate the two equations for Q:

I * 3.25 = (2g / 107.87 g/mol) * 1 * 96485 C/mol

Simplifying and solving for I:

I = (2g / 107.87 g/mol) * 1 * 96485 C/mol / 3.25 seconds

I ≈ 553.063 A

Now, we can calculate the quantity of electricity that passed through the second cell using the same method:

Q = I * 3.25 seconds

And then calculate the mass of copper (Cu) deposited in the second cell using the equation:

m = (Q / (z * F)) * M

For copper (Cu), the molar mass (M) is 63.55 g/mol, and the number of electrons transferred (z) is 2, as indicated by the balanced redox equation.

Therefore, m = (Q / (2 * 96485 C/mol)) * 63.55 g/mol

Substituting the value of Q from the previous calculation:

m ≈ ((I * 3.25 seconds) / (2 * 96485 C/mol)) * 63.55 g/mol

m ≈ (553.063 A * 3.25 seconds) / (2 * 96485 C/mol) * 63.55 g/mol

m ≈ 0.0679 g

Therefore, approximately 0.0679 grams of copper are deposited on the second cell.