To calculate the time it takes to deposit 1g of Pt, we need to start by calculating the number of electrons needed to deposit 1g of Pt.
Using the molar mass of Pt (195.1 g/mol), we can convert 1g of Pt to moles:
1g Pt X (1 mol Pt/195.1 g Pt) = 0.00512 mol Pt
We know that the reduction of Pt 4+ to Pt requires 4 electrons:
Pt 4+ + 4e- → Pt
So, the number of electrons needed to deposit 0.00512 mol of Pt is:
0.00512 mol Pt X 4 electrons/mol Pt = 0.0205 mol electrons
Now we use Faraday's constant, which gives us the number of coulombs of charge in one mole of electrons.
One molt of electrons = Faraday's constant (F) = 96,500 C/mol e-
The total charge required to deposit 0.0205 mol electrons is:
0.0205 mol e- X 96,500 C/mol e- = 1,977.25 C
Finally, we can calculate the time (t) it takes to deposit 1g of Pt using the equation:
t = Q/I
where Q is the total charge (1977.25 C) and I is the current (0.15 A).
t = 1977.25 C / 0.15 A = 13,181.67 seconds
Therefore, it would take approximately 3.66 hours to deposit 1g of Pt from an aqueous solution of K2PtCl4 with a current of 0.15A.
A current of 0.15A is passed through an aqueous solution of K2PtCl4. How long will it take to deposit 1g of Pt (M=195.1)
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