Calculate the current (in mA) required to deposite 0.195g of platinum metal in 5.0 hours from a solution of `[PtCl_6^(2-)`:(Atomic mass:Pt=195)

Options

A. 310
B. 31
C. 21.44
D. 5.36

Answer

To calculate the current required to deposit 0.195 g of platinum metal from a solution of \([PtCl_6]^{2-}\) in 5 hours, we can follow these steps:

### Step 1: Determine the oxidation state of platinum in \([PtCl_6]^{2-}\)
The oxidation state of platinum (Pt) can be calculated as follows:
Let the oxidation state of Pt be \(x\). The charge on each chlorine (Cl) is \(-1\), and there are 6 chlorine atoms. Therefore, the equation can be set up as:
\[ x + 6(-1) = -2 \]
This simplifies to:
\[ x - 6 = -2 \implies x = +4 \]
So, the oxidation state of platinum in \([PtCl_6]^{2-}\) is +4.

### Step 2: Determine the change in oxidation state
Platinum is being reduced from +4 to 0 during the deposition process. This means that 4 electrons are involved in the reduction:
\[ Pt^{4+} + 4e^- \rightarrow Pt^0 \]
Thus, the n-factor (number of electrons transferred) is 4.

### Step 3: Use Faraday's law of electrolysis
According to Faraday's law, the mass deposited (W) is given by:
\[ W = Z \cdot I \cdot T \]
Where:
- \(W\) = mass deposited (in grams)
- \(Z\) = electrochemical equivalent (in grams per coulomb)
- \(I\) = current (in amperes)
- \(T\) = time (in seconds)

### Step 4: Calculate the electrochemical equivalent (Z)
The electrochemical equivalent \(Z\) can be calculated using the formula:
\[ Z = \frac{\text{Equivalent weight}}{96500} \]
The equivalent weight is given by:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{195 \text{ g/mol}}{4} = 48.75 \text{ g/equiv} \]
Thus,
\[ Z = \frac{48.75}{96500} \text{ g/C} \]

### Step 5: Convert time from hours to seconds
The time given is 5 hours. We need to convert this to seconds:
\[ T = 5 \text{ hours} \times 60 \text{ minutes/hour} \times 60 \text{ seconds/minute} = 18000 \text{ seconds} \]

### Step 6: Substitute values into Faraday's law
Now we can substitute the values into the equation:
\[ 0.195 = \left(\frac{48.75}{96500}\right) \cdot I \cdot 18000 \]
Rearranging to solve for \(I\):
\[ I = \frac{0.195 \cdot 96500}{48.75 \cdot 18000} \]

### Step 7: Calculate the current
Calculating the current:
\[ I = \frac{0.195 \cdot 96500}{48.75 \cdot 18000} \approx 0.02144 \text{ A} \]
To convert this to milliamperes (mA):
\[ I \approx 21.44 \text{ mA} \]

### Final Answer
The current required to deposit 0.195 g of platinum metal in 5 hours is approximately **21.44 mA**.

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