The element indium is to be obtained by electrolysis of a molten halide of the element. Passage of a current of 3.20 A for a period of 40.0 min results in formation of 3.05 g of In. what is the oxidation state of indium in the halide melt? (Atomic mass of In=114.8)

Options

A. 3
B. 2
C. 5
D. 1

Answer

To determine the oxidation state of indium in the molten halide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Current (I) = 3.20 A
- Time (t) = 40.0 min
- Mass of indium deposited (m) = 3.05 g
- Atomic mass of indium (M) = 114.8 g/mol

2. **Convert Time from Minutes to Seconds:**
\[ t = 40.0 \text{ min} \times 60 \text{ s/min} = 2400 \text{ s} \]

3. **Calculate the Total Charge (Q) Passed:**
Using the formula \( Q = I \times t \):
\[ Q = 3.20 \text{ A} \times 2400 \text{ s} = 7680 \text{ C} \]

4. **Calculate the Number of Moles of Indium Deposited:**
Using the formula:
\[ \text{Moles of In} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.05 \text{ g}}{114.8 \text{ g/mol}} \approx 0.0266 \text{ mol} \]

5. **Calculate the Total Charge Required to Deposit One Mole of Indium:**
According to Faraday's laws, the charge required to deposit one mole of a substance is given by:
\[ Q = n \times F \]
where \( n \) is the number of moles of electrons transferred and \( F \) (Faraday's constant) = 96500 C/mol.

6. **Calculate the Charge per Mole of Indium:**
Rearranging the formula gives:
\[ n = \frac{Q}{F} = \frac{7680 \text{ C}}{96500 \text{ C/mol}} \approx 0.0797 \text{ mol of electrons} \]

7. **Determine the Oxidation State (n) of Indium:**
Since we have calculated the moles of indium deposited (0.0266 mol), we can find the oxidation state (n) using the relationship:
\[ n = \frac{\text{moles of electrons}}{\text{moles of indium}} = \frac{0.0797}{0.0266} \approx 3 \]

8. **Conclusion:**
Therefore, the oxidation state of indium in the molten halide is +3.

### Final Answer:
The oxidation state of indium in the halide melt is +3.

---