A current of 0.15A is passed through an aqueous solution of K2PtCl4. How long will it take to deposit 1.00g Pt? (M=195.1)
We will first find the charge required to deposit 1g of Pt, and then find the time taken to pass this charge with the given current.
n(Pt) = 1.00 g / M(Pt)
where n(Pt) is the number of moles of Pt, and M(Pt) = 195.1 g/mol is the molar mass of Pt.
n(Pt) = 1.00 / 195.1
n(Pt) ≈ 0.00513 mol
Since K2PtCl4 dissociates into K2PtCl2 and two Cl atoms, we can find out that 2 mole of electrons are needed for each mole of Pt.
n(e-) = n(Pt) * 2
where n(e-) is the number of moles of electrons.
n(e-) = 0.00513 * 2
n(e-) ≈ 0.0103 mol
Now, we can find the total charge required for this process using the Faraday's constant (F).
F = 96485 C/mol is Faraday's constant.
Q = n(e-) * F
Q = 0.0103 * 96485
Q ≈ 993 C (rounded)
Now that we have the required charge (Q) for depositing 1g of Pt, we can find the time (t) it takes with the given current (I).
I = 0.15 A
t = Q / I
t = 993 / 0.15
t ≈ 6620 s (rounded)
So it would take about 6620 seconds to deposit 1g of Pt with a 0.15 A current.
To find out the time it will take to deposit 1.00g of Pt in an aqueous solution of K2PtCl4 with a current of 0.15A, we need to use Faraday's laws of electrolysis.
First, we need to calculate the number of moles of Pt that will be deposited using the formula:
moles = mass / molar mass
Given that the molar mass of Pt is 195.1 g/mol and the mass is 1.00g, we can calculate the number of moles of Pt:
moles = 1.00g / 195.1 g/mol
moles = 0.00512 mol
Next, we use Faraday's law to determine the amount of charge required to deposit this amount of Pt:
charge (Coulombs) = moles * Faraday constant
The Faraday constant is 96,500 C/mol.
charge = 0.00512 mol * 96,500 C/mol
charge = 496.48 C
Lastly, we can calculate the time using the formula:
time (s) = charge (Coulombs) / current (Amperes)
time = 496.48 C / 0.15 A
time = 3309.87 seconds
Therefore, it will take approximately 3309.87 seconds, or about 55 minutes, to deposit 1.00g of Pt with a current of 0.15A.