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

Options

A. 1.0mol
B. 0.20mol
C. 0.4mol
D. 0.80mol

Explanation

To solve the problem of how many moles of platinum (Pt) may be deposited on the cathode when 0.80 F of electricity is passed through a 1.0 M solution of Pt^(4+), we can follow these steps:

### Step 1: Understand the reaction
The reduction reaction for platinum ions (Pt^(4+)) at the cathode can be represented as:
\[ \text{Pt}^{4+} + 4e^- \rightarrow \text{Pt} \]
This indicates that one mole of Pt^(4+) requires 4 moles of electrons (4 Faradays) to be reduced to solid platinum (Pt).

### Step 2: Calculate the number of moles of electrons
We know that 1 Faraday (F) corresponds to the charge of 1 mole of electrons, which is approximately 96485 coulombs. In this case, we have 0.80 F of electricity passed through the solution. Therefore, the number of moles of electrons (n) can be calculated using the formula:
\[ n = \frac{\text{Charge (in Farads)}}{\text{Number of electrons required per mole of substance}} \]
For Pt^(4+), we need 4 electrons per mole.

### Step 3: Calculate the moles of Pt deposited
Using the amount of electricity passed (0.80 F), we can calculate the moles of Pt deposited:
1. Calculate the moles of electrons:
\[ \text{Moles of electrons} = 0.80 \, \text{F} \, \text{(since 1 F = 1 mole of electrons)} \]
Therefore, we have 0.80 moles of electrons.

2. Since 4 moles of electrons are required to deposit 1 mole of Pt, we can find the moles of Pt deposited:
\[ \text{Moles of Pt} = \frac{\text{Moles of electrons}}{4} = \frac{0.80}{4} = 0.20 \, \text{moles of Pt} \]

### Final Answer:
Thus, the number of moles of platinum (Pt) that may be deposited on the cathode is **0.20 moles**.

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