A constant current flowed for 2 hours through a potassium iodide solution oxidising the iodide ion to iodine `(2I^(-) rarr I_(2) + 2e^(-))`.
At the end of the experiment, the iodine was titrated with 21.75 mL of 0.0831 M soldium thiosulphate solution.
`(I_(2) + 2S_(2)O_(3)^(2-) rarr 2I^(-) +S_(4) O_(6)^(2-))`
What was the average rate of current flow in apmeres ?

Answer

To find the average rate of current flow in amperes during the electrolysis of potassium iodide solution, we can follow these steps:

### Step 1: Calculate the moles of sodium thiosulfate used
The volume of sodium thiosulfate solution used is given as 21.75 mL, and its molarity is 0.0831 M.

\[ \text{Moles of } Na_2S_2O_3 = \text{Volume (L)} \times \text{Molarity} = \frac{21.75 \, \text{mL}}{1000} \times 0.0831 \, \text{mol/L} \]

Calculating this gives:

\[ \text{Moles of } Na_2S_2O_3 = 0.02175 \, \text{L} \times 0.0831 \, \text{mol/L} = 0.001806825 \, \text{mol} \]

### Step 2: Relate moles of iodine to moles of thiosulfate
From the balanced equation, we know that 1 mole of \( I_2 \) reacts with 2 moles of \( S_2O_3^{2-} \). Therefore, the moles of iodine produced can be calculated as:

\[ \text{Moles of } I_2 = \frac{1}{2} \times \text{Moles of } Na_2S_2O_3 = \frac{1}{2} \times 0.001806825 \, \text{mol} = 0.0009034125 \, \text{mol} \]

### Step 3: Calculate the total moles of iodide ions oxidized
Since 1 mole of \( I_2 \) is produced from 2 moles of \( I^- \), the moles of iodide ions oxidized are:

\[ \text{Moles of } I^- = 2 \times \text{Moles of } I_2 = 2 \times 0.0009034125 \, \text{mol} = 0.001806825 \, \text{mol} \]

### Step 4: Calculate the total charge flow
Using Faraday's constant (\( F = 96500 \, \text{C/mol} \)), we can calculate the total charge (\( Q \)) that has flowed through the circuit:

\[ Q = \text{Moles of } I^- \times F = 0.001806825 \, \text{mol} \times 96500 \, \text{C/mol} = 174.5 \, \text{C} \]

### Step 5: Calculate the total time in seconds
The time is given as 2 hours. We need to convert this into seconds:

\[ \text{Time} = 2 \, \text{hours} \times 3600 \, \text{s/hour} = 7200 \, \text{s} \]

### Step 6: Calculate the average current
Using the formula for current (\( I = \frac{Q}{t} \)):

\[ I = \frac{Q}{t} = \frac{174.5 \, \text{C}}{7200 \, \text{s}} \approx 0.0242 \, \text{A} \]

Thus, the average rate of current flow is approximately **0.0242 A**.

### Summary of Steps:
1. Calculate moles of sodium thiosulfate used.
2. Relate moles of iodine to moles of thiosulfate.
3. Calculate total moles of iodide ions oxidized.
4. Calculate total charge flow using Faraday's constant.
5. Convert time from hours to seconds.
6. Calculate average current using the charge and time.