How many hours does it require to reduce 3 mol of `Fe^(3+)` to `Fe^(2+)` by passing 2.00 A current ?

Explanation

To solve the problem of how many hours it requires to reduce 3 moles of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \) by passing a current of 2.00 A, we can follow these steps:

### Step 1: Determine the number of moles of electrons required
To reduce \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \), each mole of \( \text{Fe}^{3+} \) requires 1 mole of electrons. Therefore, for 3 moles of \( \text{Fe}^{3+} \), we need:
\[ \text{Moles of electrons} = 3 \text{ moles} \]

### Step 2: Calculate the total charge required
Using Faraday's constant, which is approximately \( 96500 \, \text{C/mol} \), we can calculate the total charge (\( Q \)) required for 3 moles of electrons:
\[ Q = \text{Moles of electrons} \times \text{Faraday's constant} = 3 \, \text{mol} \times 96500 \, \text{C/mol} = 289500 \, \text{C} \]

### Step 3: Use the current to find the time
The relationship between charge, current, and time is given by:
\[ Q = I \times t \]
Where:
- \( Q \) is the total charge in coulombs
- \( I \) is the current in amperes (A)
- \( t \) is the time in seconds (s)

Rearranging this equation to solve for time (\( t \)):
\[ t = \frac{Q}{I} = \frac{289500 \, \text{C}}{2.00 \, \text{A}} = 144750 \, \text{s} \]

### Step 4: Convert time from seconds to hours
To convert seconds into hours, we divide by the number of seconds in an hour (3600 s/h):
\[ t_{\text{hours}} = \frac{144750 \, \text{s}}{3600 \, \text{s/h}} \approx 40.21 \, \text{hours} \]

### Final Answer
It requires approximately **40.21 hours** to reduce 3 moles of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \) by passing a current of 2.00 A.

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