50 ml of ZnSO4 solution was transferred to a mercury cathode and enough solid potassium nitrate is added to make the solution 0.1 M in KNO3. The electrolysis of Zn2+ is carried to completion of -1.3 V vs. SCE with the passage of 241 C of electricity. calculate the initial concentration of zinc ion ?
To calculate the initial concentration of zinc ions (Zn2+), we need to use the information given about the electrolysis process and the passage of electricity. Let's go step by step.
Step 1: Calculate the number of moles of electrons passed through the cell.
We know that the passage of 241 C (Coulombs) of electricity corresponds to the transfer of 1 mole of electrons. Therefore, the number of moles of electrons (n) can be calculated using Faraday's Law:
n = 241 C / (Faraday's Constant)
The Faraday's Constant is equal to 96485 C/mol.
Substituting the values:
n = 241 / 96485
n ≈ 0.002498 moles of electrons
Step 2: Calculate the number of moles of zinc (Zn) deposited.
In the electrolysis process, 2 moles of electrons are required to deposit 1 mole of zinc (Zn) ions. Therefore, the number of moles of Zn deposited can be calculated as:
Moles of Zn = 0.002498 moles of electrons / 2
Moles of Zn ≈ 0.001249 moles of Zn
Step 3: Calculate the initial concentration of Zn2+ ions.
Now, we need to consider the balanced equation for the electrolysis of Zn2+ ions:
Zn2+ (aq) + 2e- -> Zn (s)
From the equation, we can see that 1 mole of Zn2+ ions requires 2 moles of electrons for reduction.
Moles of Zn2+ = 2 * Moles of Zn
Moles of Zn2+ ≈ 2 * 0.001249 ≈ 0.002498 moles of Zn2+
Finally, we calculate the initial concentration (C) using the given volume (V) of the ZnSO4 solution:
C = Moles of Zn2+ / Volume of ZnSO4 solution
Volume of ZnSO4 solution = 50 mL = 0.05 L
C = 0.002498 moles of Zn2+ / 0.05 L
C = 0.04996 M
Therefore, the initial concentration of zinc ions (Zn2+) is approximately 0.04996 M.