A current of 0.965 A is passed through 500 ml of 0.2 M solution of ZnSO4 for 10 min. The molarity of Zn2+ after deposition of Zn is
1. 0.1M
2. 0.5M
3. 0.8M
4. 0.194M
To determine the molarity of Zn2+ after the deposition of Zn, you can use Faraday's Law. Faraday's Law relates the amount of substance deposited or liberated during electrolysis to the current and the equivalent weight of the substance.
The equation for Faraday's Law is:
moles of substance = (current × time) / (n × F)
Where:
- moles of substance is the amount of substance deposited or liberated
- current is the electric current in amperes (A)
- time is the duration of the electrolysis in seconds (s)
- n is the number of electrons involved in the transfer of one mole of the substance
- F is Faraday's constant (96,485 C/mol)
In this case, ZnSO4 dissociates into Zn2+ and SO4 2-. The oxidation state of Zn is +2, and since there are no electrons involved in the redox reaction, n = 0. Thus, the moles of substance can be simplified to:
moles of Zn2+ = (current × time) / F
Now, calculate the moles of Zn2+ using the given values:
current = 0.965 A
time = 10 min = 600 s
moles of Zn2+ = (0.965 A × 600 s) / 96485 C/mol
moles of Zn2+ ≈ 0.060 mol
Next, calculate the new volume of the solution after the deposition of Zn. The Zn2+ cations are removed from the solution, so the volume is reduced. The initial volume was 500 ml, but 1 mole of Zn2+ occupies 2 moles of ions. Therefore, the new volume can be calculated as:
new volume = initial volume - (moles of Zn2+ × 2)
new volume = 500 ml - (0.060 mol × 2)
new volume = 500 ml - 0.120 ml
new volume ≈ 499.880 ml
Finally, calculate the new molarity of Zn2+ by dividing the moles of Zn2+ by the new volume in liters:
new molarity = moles of Zn2+ / (new volume / 1000)
new molarity = 0.060 mol / (499.880 ml / 1000)
new molarity ≈ 0.120 M
Therefore, the molarity of Zn2+ after the deposition of Zn is approximately 0.120 M, which is not one of the given answer choices.
To find the molarity of Zn2+ after the deposition of Zn, we need to calculate the amount of Zn deposited.
First, let's calculate the number of moles of Zn2+ initially present in the solution:
molarity = moles/volume
0.2 M = moles/0.5 L
moles = 0.2 M x 0.5 L
moles = 0.1
Since the stoichiometry of the reaction is 1:1 between Zn2+ and Zn, the number of moles of Zn which will be deposited will also be 0.1 moles.
To find the mass of Zn deposited, we need to use Faraday's law of electrolysis:
mass = (current x time) / (n x F)
Where:
current = 0.965 A
time = 10 min = 600 sec (converted to seconds)
n = number of moles
F = Faraday's constant = 96,485 C/mol
mass = (0.965 A x 600 sec) / (0.1 mol x 96,485 C/mol)
mass = 0.05958 g
Now, we can calculate the new volume and molarity of the solution after the deposition of Zn.
The volume after deposition can be calculated by subtracting the volume of Zn deposited from the initial volume:
volume after deposition = initial volume - volume of Zn deposited
volume after deposition = 500 ml - (0.05958 g / (density of Zn))
Since the density of Zn is 7.14 g/cm³, we need to convert the volume from ml to cm³:
1 ml = 1 cm³
volume after deposition = 500 cm³ - (0.05958 g / 7.14 g/cm³)
volume after deposition = 500 cm³ - 8.35 cm³
volume after deposition = 491.65 cm³
Now, we can calculate the molarity of Zn2+ after the deposition of Zn:
molarity = moles/volume
molarity = 0.1 mol / (491.65 cm³ / 1000)
molarity = 0.203 M
The molarity of Zn2+ after the deposition of Zn is approximately 0.203 M.
So, the correct answer is not provided in the options given.
amperes x seconds = coulombs
coulombs/96,485 = Faraday's
1 Faraday will deposit 1 equivalent weight of Zn^+2 or 1/2 mole Zn^+2.
I get 0.965 x 10 min x (60 sec/min) = ?? coulombs.
That divided by 96,485 = 0.006 Faraday's and that will deposit 0.003 mole Zn^+2.
How many moles Zn^+2 in 500 mL 0.2 M ZnSO4. M x L = 0.2 x 0.50 = 0.1 mol
0.003 is deposited. How much is left. What is the molarity of that solution?