A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of \(4.00 \mathrm{~A}\). How long will it take to plate out \(99 \%\) of the platinum in \(0.50 \mathrm{~L}\) of a \(0.010 \mathrm{M}\) solution of \(\mathrm{Pt}^{4+}\) ?

First, we need to find the number of moles (\(n\)) of \(\mathrm{Pt}^{4+}\) ions in the solution. We can use the formula: $$n = C \times V$$ We can then find the mass of the platinum using the formula: $$m = n \times M$$

Next, we need to calculate the charge required to plate out \(99 \%\) of the platinum. We can use Faraday's law of electrolysis: $$Q = n \times F \times z$$ Here, \(z\) represents the charge number of the \(\mathrm{Pt}^{4+}\) ions, which is \(4\).

Finally, we can use the current (\(I\)) to find the time required for the electrolysis process. We can use the formula: $$t = \frac{Q}{I}$$ Now, let's plug in the values and find the time required.

1. Calculate the moles and mass of the platinum $$n = (0.010 \mathrm{M})(0.50 \mathrm{~L}) = 0.005 \mathrm{mol}$$ $$m = (0.005 \mathrm{mol})(195.08 \mathrm{g/mol}) = 0.976 \mathrm{g}$$ 2. Calculate the charge required for the electrolysis process Since we want to plate \(99 \%\) of the platinum, we'll use \(0.99m\) to calculate the charge required. $$Q = (0.005 \mathrm{mol})(96485 \mathrm{C/mol})(4) = 1930.14 \times 0.99 = 1910.84 \mathrm{C}$$ 3. Calculate the time required for the electrolysis process $$t = \frac{1910.84 \mathrm{C}}{4.00 \mathrm{~ A}} = 477.71 \mathrm{s}$$ The time required to plate out \(99 \%\) of the platinum in the \(0.50 \mathrm{~L}\) of \(0.010 \mathrm{M}\) solution of \(\mathrm{Pt}^{4+}\) is approximately \(477.71 \mathrm{s}\).