Calculate the current in amperes required to liberate 10g of iodine electrolytically in one hour from KI solution .

Answer

To calculate the current required to liberate 10 grams of iodine electrolytically from a potassium iodide (KI) solution, we can follow these steps:

### Step 1: Determine the molar mass of iodine
The atomic weight of iodine (I) is given as 127 g/mol.

### Step 2: Calculate the number of moles of iodine
To find the number of moles of iodine in 10 grams, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
\[ \text{Number of moles of I} = \frac{10 \, \text{g}}{127 \, \text{g/mol}} \approx 0.0787 \, \text{mol} \]

### Step 3: Determine the charge required to liberate iodine
Since 1 mole of iodine requires 1 Faraday (96500 coulombs) to be liberated, we can calculate the total charge (Q) required for 0.0787 moles:
\[ Q = \text{Number of moles} \times \text{Faraday's constant} \]
\[ Q = 0.0787 \, \text{mol} \times 96500 \, \text{C/mol} \approx 7596.55 \, \text{C} \]

### Step 4: Convert time from hours to seconds
The time given is 1 hour, which we need to convert into seconds:
\[ \text{Time (s)} = 1 \, \text{hour} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 3600 \, \text{s} \]

### Step 5: Calculate the current (I)
Using the relationship between charge, current, and time:
\[ I = \frac{Q}{t} \]
Substituting the values we found:
\[ I = \frac{7596.55 \, \text{C}}{3600 \, \text{s}} \approx 2.11 \, \text{A} \]

### Final Answer
The current required to liberate 10 grams of iodine from KI solution in one hour is approximately **2.11 amperes**.

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