Animated Solution for Chemistry - Electrochemistry: Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation $6{\text{OH}}^{-}+{\text{Cl}}^{-}⟶{\text{ClO}}_3^{-}+3{\text{H}}_2\text{O}+6e^{-}$ A current of $x\text{A}$ has to be passed for $10\text{ h}$ to produce $10.0\text{ g}$ of potassium chlorate. The value of $x$ is …… . (Nearest integer) (Molar mass of ${\text{KClO}}_3=122.6{\text{ g mol}}^{-1}$, $F=96500\text{C}$)

A Chemical Symphony

Imagine an electrolytic cell where a solution of potassium chloride is undergoing electrolysis in a basic medium. This isn't just a random mixture; it's a carefully orchestrated environment where electrical energy is converted into chemical energy. The basic medium provides the necessary hydroxide ions ($O{H}^{-}$) that participate directly in the anodic oxidation process.

Our goal here is to determine the exact amount of electrical current required to produce a specific mass of potassium chlorate ($KCl{O}_3$). To do this, we must bridge the gap between the macroscopic world of grams and hours, and the microscopic world of electrons and ions.

The Bridge Between Charge and Mass

To find the required current, we rely on a fundamental principle of electrochemistry: Faraday's First Law of Electrolysis. This law elegantly states that the mass ($W$) of a substance deposited or produced at an electrode is directly proportional to the total electrical charge ($Q$) passed through the solution.

Mathematically, this is expressed as:

Since charge ($Q$) is the product of current ($I$) and time ($t$), we can rewrite this as:

Here, $Z$ is the electrochemical equivalent, which is defined as the molar mass ($M$) divided by the product of the n-factor ($n$) and Faraday's constant ($F$). Substituting this gives us our master equation:

Decoding the Electron Transfer

Before we can plug numbers into our master equation, we need to determine the n-factor. This is where the provided chemical equation becomes our most valuable tool:

Look closely at the stoichiometry. For every one mole of chlorate ion ($Cl{O}_3^{-}$) produced, exactly six moles of electrons ($e^{-}$) are released. This direct relationship tells us that our n-factor ($n$) is exactly $6$. This is a crucial step; misidentifying the n-factor is a common pitfall that leads to incorrect results.

Bringing It All Together

Now, we are ready to substitute our known values into the master equation. We are given: - Mass ($W$) = $10.0\text{ g}$ - Molar mass ($M$) = $122.6{\text{ g mol}}^{-1}$ - n-factor ($n$) = $6$ - Faraday's constant ($F$) = $96500{\text{ C mol}}^{-1}$ - Time ($t$) = $10\text{ hours}$

A critical warning: Time must always be in seconds when using Faraday's law because one Ampere is defined as one Coulomb per second. Therefore, we must convert the 10 hours into seconds by multiplying by $3600$ ($60\text{ minutes}\times 60\text{ seconds}$).

Substituting these values yields:

Rearranging the equation to solve for the current ($x$):

Upon calculating this expression, we find:

The question specifically asks for the nearest integer. Since $1.311$ is closer to $1$ than to $2$, we round down.

Final Answer: The required current is $1\text{ A}$.