What volume of `0.1N FeSO_(4)` can be oxidized by a current of 2 ampere-hours ?

Options

A. `0.746 L`
B. `7.46L`
C. `1.482L`
D. `0.373L`

Explanation

To solve the problem of determining what volume of `0.1N FeSO4` can be oxidized by a current of 2 ampere-hours, we can follow these steps:

### Step 1: Calculate the total charge in coulombs
We know that 1 ampere-hour (Ah) is equivalent to 3600 coulombs. Therefore, for 2 ampere-hours:
\[ \text{Total charge (Q)} = 2 \, \text{A} \times 3600 \, \text{s} = 7200 \, \text{C} \]

### Step 2: Calculate the number of equivalents of FeSO4
The number of equivalents can be calculated using Faraday's law, where 1 equivalent corresponds to the charge required to oxidize or reduce 1 mole of electrons (which is approximately 96500 coulombs).
\[ \text{Number of equivalents} = \frac{\text{Total charge (Q)}}{96500 \, \text{C/equiv}} = \frac{7200 \, \text{C}}{96500 \, \text{C/equiv}} \approx 0.0746 \, \text{equiv} \]

### Step 3: Use the relationship between normality, volume, and equivalents
The relationship between normality (N), volume (V), and number of equivalents (n) is given by:
\[ \text{Normality} \times \text{Volume} = \text{Number of equivalents} \]
Rearranging this gives us:
\[ \text{Volume} = \frac{\text{Number of equivalents}}{\text{Normality}} \]
Substituting the values we have:
\[ \text{Volume} = \frac{0.0746 \, \text{equiv}}{0.1 \, \text{N}} = 0.746 \, \text{L} \]

### Conclusion
The volume of `0.1N FeSO4` that can be oxidized by a current of 2 ampere-hours is **0.746 liters**.

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