How many moles of Pt may be deposited on the cathode when 0.80F of electricity is passed through 1.0M solution of `Pt^(4+)`?

Options

A. 0.1 mol
B. 0.2 mol
C. 0.4 mol
D. 0.6 mol

Explanation

To solve the problem of how many moles of platinum (Pt) can be deposited at the cathode when 0.80 Faraday of electricity is passed through a 1.0 M solution of \( \text{Pt}^{4+} \), we can follow these steps:

### Step 1: Understand the Reaction
The reduction reaction for platinum ions can be represented as:
\[ \text{Pt}^{4+} + 4e^- \rightarrow \text{Pt} \]
This equation tells us that 1 mole of \( \text{Pt} \) is deposited when 4 moles of electrons (or 4 Faradays of charge) are used.

### Step 2: Apply Faraday's First Law of Electrolysis
According to Faraday's first law, the amount of substance deposited (in moles) at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. The formula is:
\[ W = Z \cdot it \]
Where:
- \( W \) is the mass of the substance deposited,
- \( Z \) is the electrochemical equivalent,
- \( i \) is the current,
- \( t \) is the time.

However, for our purpose, we can simplify this to find the number of moles deposited using the relationship between Faraday's and moles of electrons.

### Step 3: Calculate Moles of Platinum Deposited
From the stoichiometry of the reaction, we know that:
- 4 Faradays are required to deposit 1 mole of platinum.

Now, if 0.80 Faraday of electricity is passed:
\[ \text{Moles of Pt deposited} = \frac{\text{Faradays passed}}{\text{Faradays required for 1 mole of Pt}} = \frac{0.80 \, \text{F}}{4 \, \text{F/mole}} = 0.20 \, \text{moles} \]

### Conclusion
Thus, when 0.80 Faraday of electricity is passed through a 1.0 M solution of \( \text{Pt}^{4+} \), 0.20 moles of platinum will be deposited on the cathode.

### Final Answer
**0.20 moles of platinum may be deposited on the cathode.**

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