Potassium chlorate is prepared by the electrolysis of KCl in basic solution 6OH- + Cl- - ClO3- + 3H

Q: Why do we divide by 0.6 instead of multiplying?

When efficiency is less than 100%, it means a portion of the current is wasted on side reactions (like heating or electrolyzing water). To get the *required* amount of product, you must supply *more* total charge. Dividing by 0.6 increases the required charge, which correctly models this physical reality.

Q: Where did the 6 Faradays come from?

Look at the balanced half-reaction provided in the question: $6\text{OH}^- + \text{Cl}^- \rightarrow \text{ClO}_3^- + 3\text{H}_2\text{O} + 6\text{e}^-$. The $6\text{e}^-$ tells us that 6 moles of electrons are transferred for every 1 mole of $\text{ClO}_3^-$ produced. One mole of electrons carries one Faraday of charge.

Q: Why do we use the molar mass of $\text{KClO}_3$ and not just $\text{ClO}_3^-$?

The question specifically asks for the time required to produce 10 g of the entire salt, $\text{KClO}_3$. Therefore, we must use the molar mass of the entire compound (122 g/mol) to find the correct number of moles.

Q: Do I need to memorize the molar mass of $\text{KClO}_3$?

Usually, no. In JEE Main, molar masses of complex or less common compounds are often provided in the question, just as it was here ($122 \text{ g mol}^{-1}$). However, knowing atomic masses of common elements (K=39, Cl=35.5, O=16) is highly recommended.

Q: Why did we divide by 3600 at the end?

The formula $Q = I \times t$ uses standard SI units, meaning the time $t$ is calculated in seconds. Since the question explicitly asks for the time rounded to the nearest *hour*, we must convert seconds to hours by dividing by 60 (to get minutes) and then by 60 again (to get hours), which is a total division by 3600.

Question

Potassium chlorate is prepared by the electrolysis of $	ext{KCl}$ in basic solution
$6	ext{OH}^- + 	ext{Cl}^- ightarrow 	ext{ClO}_3^- + 3	ext{H}_2	ext{O} + 6	ext{e}^-$. If only $60\%$ of the current is utilised in the reaction, the time (rounded to the nearest hour) required to produce $10	ext{ g}$ of $	ext{KClO}_3$ using a current of $2	ext{ A}$ is ........ .
(Given : $F = 96,500	ext{ C mol}^{-1}$; molar mass of $	ext{KClO}_3 = 122	ext{ g mol}^{-1}$)

Sigma Percentile · Accepted Answer

### The Setup
Imagine an industrial electrolytic cell where we are passing a steady current of $2	ext{ A}$ through a basic potassium chloride solution. Our goal is to produce exactly $10	ext{ g}$ of potassium chlorate ($	ext{KClO}_3$).

The core chemical transformation happening at the anode is given by the balanced half-reaction:
$$6	ext{OH}^- + 	ext{Cl}^- ightarrow 	ext{ClO}_3^- + 3	ext{H}_2	ext{O} + 6	ext{e}^-$$

By looking closely at this equation, we can extract a vital piece of stoichiometric information: to produce exactly $1	ext{ mole}$ of chlorate ions ($	ext{ClO}_3^-$), the reaction must transfer $6	ext{ moles}$ of electrons. In the language of electrochemistry, this means we need a theoretical charge of $6	ext{ Faradays}$ ($6	ext{ F}$) per mole of product.

### The Efficiency Catch
Here is where many students make a critical error. The problem states that the current efficiency is only $60\%$. This means that out of all the electrical energy pumped into the cell, only $60\%$ is actually driving our desired reaction. The rest is lost to side reactions or heat.

Because the process is inefficient, we must supply **more** total charge to get the same amount of product. To find the actual charge required per mole, we divide the theoretical charge by the efficiency factor ($0.6$):
$$Q_{	ext{act}} = \frac{6	ext{ F}}{0.6} = 10	ext{ F}$$

So, in reality, we need to pump $10	ext{ Faradays}$ of charge into the cell for every mole of $	ext{KClO}_3$ we want to harvest.

### The Master Calculation
Now, let's figure out how many moles of $	ext{KClO}_3$ we actually need. We are asked to produce $10	ext{ g}$, and the molar mass is given as $122	ext{ g mol}^{-1}$.
$$n = \frac{10}{122}	ext{ mol}$$

The total charge ($Q$) required for our specific batch is simply the number of moles multiplied by the actual charge per mole:
$$Q = \frac{10}{122} 	imes 10	ext{ F} = \frac{100}{122}	ext{ F}$$

According to Faraday's Law, total charge is the product of current and time ($Q = I 	imes t$). Remembering that $1	ext{ F} = 96500	ext{ C}$, we can set up our master equation:
$$\frac{100}{122} 	imes 96500 = 2 	imes t$$

### The Final Countdown
Isolating time ($t$), we get:
$$t = \frac{100 	imes 96500}{122 	imes 2}	ext{ seconds}$$

Since the question explicitly asks for the time in hours, we must divide our result by $3600$ (the number of seconds in an hour):
$$t = \frac{100 	imes 96500}{122 	imes 2 	imes 3600}	ext{ hours}$$

Calculating this fraction yields approximately $10.98	ext{ hours}$. Rounding this off to the nearest integer, we arrive at our final answer:
**$11	ext{ hours}$**.

Enter Numerical Value:

Visualized Solution

$The Electrolytic Reaction$

$Theoretical Charge Requirement$

$Actual Charge with Efficiency$

$Moles of Product$

$Total Charge Required$

$Applying Faraday’s Law$

$Solving for Time (in seconds)$

$Converting to Hours$

$Final Answer$

$The Way Forward$

The Sigma Insight: Electrolytic Conduction

The Setup Imagine an industrial electrolytic cell where we are passing a steady current of $2 A$ through a basic potassium chloride solution

The Efficiency Catch Here is where many students make a critical error

The Master Calculation Now, let's figure out how many moles of $KClO_{3}$ we actually need

The Final Countdown

Similar Questions

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Electrolytic Reaction

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Theoretical Charge Requirement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Actual Charge with Efficiency

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles of Product

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Charge Required

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying Faraday’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Time (in seconds)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Converting to Hours

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward