Problem 28 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Potassium chlorate is prepared by the electrolysis of \(\mathrm{KCl}\) in basic solution \(6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}\) If only \(60 \%\) of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce \(10 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\) using a current of \(2 \mathrm{~A}\) is (Given : \(\mathrm{F}=\overline{96,500 \mathrm{C}} \mathrm{mol}^{-1} ;\) molar mass of \(\mathrm{KClO}_{3}=122 \mathrm{~g} \mathrm{~mol}^{-1}\) )

Step-by-Step Solution

Answer

It takes 11 hours to produce 10 g of KClO3.

Step 1: Calculate moles of KClO3 needed

We need to find out how many moles of \( \mathrm{KClO}_3 \) correspond to 10 grams. Use the formula: \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]. Given the molar mass of \( \mathrm{KClO}_3 \) is 122 g/mol, the moles are: \[ \text{moles} = \frac{10}{122} \approx 0.08197 \text{ moles} \].

Step 2: Determine the charge required

Each mole of \( \mathrm{KClO}_3 \) requires 6 moles of electrons, as per the reaction. Therefore, 0.08197 moles will require \( 0.08197 \times 6 = 0.49182 \) moles of electrons. Using Faraday's constant (\( F = 96,500 \mathrm{~C/mol} \)), the total charge \( Q \) is: \[ Q = 0.49182 \times 96500 \approx 47440.39 \text{ C} \].

Step 3: Adjust charge for efficiency

Since only 60% of the current is used efficiently, the effective charge \( Q_{\text{effective}} \) becomes: \[ Q_{\text{effective}} = \frac{47440.39}{0.60} = 79067.32 \text{ C} \].

Step 4: Calculate time required

Using the formula for current \( I = \frac{Q}{t} \), rearrange to \( t = \frac{Q}{I} \). Given the current \( I = 2 \mathrm{~A} \), the time \( t \) is: \[ t = \frac{79067.32}{2} = 39533.66 \text{ seconds} \].

Step 5: Convert time to hours

Convert seconds to hours by dividing by 3600 (since there are 3600 seconds in an hour): \[ \text{hours} = \frac{39533.66}{3600} \approx 10.98 \text{ hours} \]. Round this to the nearest hour to get 11 hours.

Key Concepts

Potassium Chlorate

Potassium chlorate, represented as \( \text{KClO}_3 \), is a chemical compound that finds use in various applications. Its preparation often involves the process of electrolysis, specifically by electrolyzing potassium chloride \( \text{KCl} \) in a basic solution. This chemical reaction yields potassium chlorate alongside other by-products such as water and releases electrons, as expressed in the reaction:

• \[ 6 \text{OH}^{-} + \text{Cl}^{-} \rightarrow \text{ClO}_{3}^{-} + 3\text{H}_2\text{O} + 6e^{-} \]

Potassium chlorate is noted for its role as a strong oxidizing agent. In everyday uses, it might be found in products ranging from safety matches to fireworks, demonstrating its significance beyond purely academic interest. Understanding its preparation through electrolysis is crucial for comprehending the intricacies of industrial chemical processes.

Faraday's Law of Electrolysis

In the context of electrolysis, Faraday's Law provides a quintessential framework for understanding how electric charge is involved in chemical reactions. Named after the scientist Michael Faraday, this law helps us calculate the amount of substance produced at an electrode during electrolysis. Fundamentally, the law states:

• The mass of a substance altered at an electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte.
• This is usually expressed mathematically as \( m = \frac{Q}{F} \times \frac{M}{n} \), where \( m \) is the mass, \( Q \) is the total electric charge, \( F \) represents Faraday's constant (approximately \( 96,500 \text{ C/mol} \)), \( M \) is the molar mass of the substance, and \( n \) is the number of moles of electrons required.

Faraday's law is instrumental in the electrolysis process, especially when calculating how much potassium chlorate can be produced. By applying this law, we can link the charge required to produce a certain amount of chemical products, ensuring accurate industrial applications.

Electrochemical Reactions

Electrochemical reactions form the backbone of electrolysis processes. These involve chemical reactions driven by or producing electrical energy. When electricity is passed through an electrolyte solution, it prompts the movement of ions, facilitating reactions like the conversion of \( \text{KCl} \) to potassium chlorate in a basic solution.Understanding electrochemical reactions involves focusing on the transfer of electrons. Electrons move from one species to another in a methodical dance, which is key in the reaction:

• The reduction at the cathode and the oxidation at the anode define these reactions.
• In potassium chlorate preparation, electrons released play a vital role in forming \( \text{ClO}_3^{-} \).

Electrochemical reactions are also closely linked with energy changes, which are central to powering processes in batteries and electrolysis systems. Mastering these reactions opens doors to understanding larger applications in fields like energy storage and metallurgy.