© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Required Electrochemical Reaction
Potassium chlorate (KClO3) is produced during the electrolysis of a potassium chloride (KCl) solution in basic conditions via the net reaction:
6\,OH^- + Cl^- \to ClO_3^- + 3\,H_2O + 6\,e^-
From this, we see that to form one mole of ClO_3^- (which corresponds to one mole of KClO3 in terms of the chlorate ion), 6 moles of electrons are required; equivalently, 6 Faradays (F) of charge are needed per mole of ClO_3^- .
Step 2: Calculate the Number of Moles of KClO3 to be Produced
We want to produce 10 g of KClO3. The molar mass of KClO3 is given as 122 g mol–1.
Number of moles of KClO3 =
\displaystyle \frac{10\,\text{g}}{122\,\text{g mol}^{-1}}
Step 3: Find the Total Charge Required
For 1 mole of KClO3, the charge required is:
\displaystyle 6\,F = 6 \times (96500\,\text{C mol}^{-1})
Therefore, to produce
\displaystyle \frac{10}{122}
moles of KClO3, the total charge needed in coulombs is:
\displaystyle \frac{10}{122} \times 6\,F \;=\; \frac{10}{122} \times 6 \times 96500\,\text{C}
Step 4: Account for the Current Efficiency
Only 60% of the current is effectively utilized. If Q_{\text{req}} is the calculated charge, the actual charge needed considering 60% efficiency is:
\displaystyle Q_{\text{actual}} = \frac{Q_{\text{req}}}{0.60}
Step 5: Relate the Actual Charge to the Current and Time
If I = 2\,\text{A} (amperes) is the current used, and t is the time in seconds, the relationship is given by:
\displaystyle Q_{\text{actual}} = I \times t
Since we want t in hours, recall that 1\,\text{hour} = 3600\,\text{seconds} . Hence,
\displaystyle t(\text{seconds}) = Q_{\text{actual}} \,\big/\, I.
Once we calculate t(\text{seconds}) , we can convert it to hours by dividing by 3600.
Step 6: Perform the Calculation
Compute the moles of KClO3 to be formed:
\displaystyle \frac{10}{122}\,\text{mol} \approx 0.081967\,\text{mol}.
Charge required ( Q_{\text{req}} ) for these moles:
\displaystyle 0.081967 \times 6 \times 96500 \,\text{C} \approx 0.081967 \times 579000 \,\text{C} \approx 47500\,\text{C} \; (\text{roughly, for illustration}).
Considering 60% efficiency, actual charge needed:
\displaystyle Q_{\text{actual}} = \frac{47500}{0.60} \approx 79166.7\,\text{C}.
Since the current is 2\,\text{A} :
\displaystyle t(\text{seconds}) = \frac{79166.7}{2} \approx 39583.35\,\text{seconds}.
Converting to hours:
\displaystyle t(\text{hours}) = \frac{39583.35}{3600} \approx 11\,\text{hours}.
Step 7: Final Answer
The time required to produce 10 g of KClO3 under the given conditions (with 60% current efficiency) and a current of 2 A is about 11 hours.
