How many hours are needed to deposit the metal based on the reaction .
`Fe^(2+)(aq) + 2e^(-) to Fe(s)`. using a current strength of 0.02amp.

Answer

To determine how many hours are needed to deposit iron using the given reaction and current strength, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ \text{Fe}^{2+}(aq) + 2e^{-} \rightarrow \text{Fe}(s) \]
This indicates that 1 mole of Fe is deposited by the transfer of 2 moles of electrons.

2. **Determine the Equivalent Weight of Iron**:
The molar mass of iron (Fe) is 56 g/mol. Since iron has a +2 charge in this reaction, the equivalent weight (E) is calculated as:
\[ E = \frac{\text{Molar Mass}}{\text{n}} = \frac{56 \text{ g/mol}}{2} = 28 \text{ g/equiv} \]

3. **Use Faraday's First Law of Electrolysis**:
According to Faraday's first law, the weight of the substance deposited (W) is given by:
\[ W = \frac{E \cdot I \cdot T}{F} \]
where:
- \( W \) = weight of the substance (in grams)
- \( E \) = equivalent weight (in grams/equiv)
- \( I \) = current (in amperes)
- \( T \) = time (in seconds)
- \( F \) = Faraday's constant (approximately 96500 C/equiv)

4. **Rearranging the Formula**:
We need to find the time \( T \). Rearranging the formula gives:
\[ T = \frac{W \cdot F}{E \cdot I} \]

5. **Substituting Values**:
Here, we want to deposit 1 mole of iron, which corresponds to 28 g (the equivalent weight we calculated). Now substituting the values:
\[ T = \frac{28 \text{ g} \cdot 96500 \text{ C/equiv}}{28 \text{ g/equiv} \cdot 0.02 \text{ A}} \]

6. **Simplifying the Equation**:
The 28 g cancels out:
\[ T = \frac{96500}{0.02} \]
\[ T = 4825000 \text{ seconds} \]

7. **Convert Seconds to Hours**:
To convert seconds into hours, divide by 3600 (the number of seconds in an hour):
\[ T_{\text{hours}} = \frac{4825000}{3600} \approx 1340.28 \text{ hours} \]

### Final Answer:
Approximately **1340.28 hours** are needed to deposit the metal.