Potassium chlorate is prepared by the electrolusis of `KCl` in basic medium as:
`Cl^(-) + 6OH^(-) rarr ClO_(3)^(-) + 3H_(2)O + 6e`
If only `60%` of current is utilised in the reaction, the time to produce `10 g` of `KClO_(3)` using current of `2` ampere : (`mol. wt.` of `KClO_(3) = 122.5`)

Options

A. `1.94 hr`
B. `10.94 hr`
C. `19.09 hr`
D. `12.09 hr`

Explanation

To solve the problem of calculating the time required to produce 10 g of KClO₃ through the electrolysis of KCl, we will follow these steps:

### Step 1: Calculate the number of moles of KClO₃ produced
To find the number of moles of KClO₃, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of KClO₃ = 10 g
- Molar mass of KClO₃ = 122.5 g/mol

Calculating the number of moles:
\[ \text{Number of moles of KClO₃} = \frac{10 \, \text{g}}{122.5 \, \text{g/mol}} \approx 0.0816 \, \text{mol} \]

### Step 2: Determine the number of electrons required
From the given reaction:
\[ \text{Cl}^- + 6 \text{OH}^- \rightarrow \text{ClO}_3^- + 3 \text{H}_2\text{O} + 6e^- \]
We see that 6 moles of electrons are required to produce 1 mole of KClO₃. Therefore, for 0.0816 moles of KClO₃:
\[ \text{Number of moles of electrons} = 0.0816 \, \text{mol} \times 6 = 0.4896 \, \text{mol} \]

### Step 3: Convert moles of electrons to coulombs
Using Faraday's constant (approximately 96500 C/mol), we can convert moles of electrons to total charge (Q):
\[ Q = \text{Number of moles of electrons} \times \text{Faraday's constant} \]
\[ Q = 0.4896 \, \text{mol} \times 96500 \, \text{C/mol} \approx 47200 \, \text{C} \]

### Step 4: Calculate the effective current used
Since only 60% of the current is utilized, we need to calculate the effective current:
\[ \text{Effective current} = 0.60 \times 2 \, \text{A} = 1.2 \, \text{A} \]

### Step 5: Calculate the time required
Using the formula \( Q = I \times t \), we can rearrange it to find time (t):
\[ t = \frac{Q}{I} \]
Substituting the values:
\[ t = \frac{47200 \, \text{C}}{1.2 \, \text{A}} \approx 39333.33 \, \text{s} \]

### Step 6: Convert time from seconds to hours
To convert seconds to hours:
\[ t \text{ (in hours)} = \frac{39333.33 \, \text{s}}{3600 \, \text{s/hour}} \approx 10.93 \, \text{hours} \]

### Final Answer
The time required to produce 10 g of KClO₃ using a current of 2 A is approximately **10.93 hours**.

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