A metal bucket is be electroplated by using `ZnCI_(2)` as an electrolyte. How many moles of zinc are deposited in 20 min by a constant of
10A?

Options

A. 0.01
B. 0.03
C. 0.06
D. `0.10`

Answer

To solve the problem of how many moles of zinc are deposited in 20 minutes by a constant current of 10 A using ZnCl₂ as an electrolyte, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Current (I) = 10 A (Amperes)
- Time (t) = 20 minutes

2. **Convert Time to Seconds:**
- Since current is in Amperes (Coulombs per second), we need to convert the time from minutes to seconds.
- \( t = 20 \text{ minutes} \times 60 \text{ seconds/minute} = 1200 \text{ seconds} \)

3. **Calculate Total Charge (Q):**
- Use the formula \( Q = I \times t \)
- \( Q = 10 \text{ A} \times 1200 \text{ seconds} = 12000 \text{ Coulombs} \)

4. **Identify the Electrochemical Reaction:**
- The electroplating reaction for zinc from zinc chloride can be represented as:
\[ \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn (s)} \]
- This indicates that 2 moles of electrons (2e⁻) are required to deposit 1 mole of zinc.

5. **Determine the Number of Faradays (Z):**
- The number of moles of electrons involved in the reaction is represented by Z. Here, Z = 2 (since 2 moles of electrons are needed to deposit 1 mole of zinc).

6. **Use Faraday's Constant (F):**
- Faraday's constant \( F = 96500 \text{ C/mol} \) (Coulombs per mole of electrons).

7. **Calculate the Number of Moles of Zinc (n):**
- Using the formula \( n = \frac{Q}{Z \times F} \)
- Substitute the values:
\[ n = \frac{12000 \text{ C}}{2 \times 96500 \text{ C/mol}} = \frac{12000}{193000} \approx 0.0621 \text{ moles} \]
- Rounding to two decimal places, we find \( n \approx 0.06 \text{ moles} \).

### Final Answer:
The number of moles of zinc deposited is approximately **0.06 moles**.