Question

How do you balance this redox reaction using the oxidation number method ?\[KCl{O_3}(s) \to KCl(s) + {O_2}(g)\]

Answer 1

Hint:Here the given equation is in a unbalanced state hence we need to balance it. Here we need to know that there are two ways to balance a redox reaction which is oxidation number method and half equation method. Selection of method is actually our priority but here the question is to balance using oxidation number method and balancing includes a series of steps which can be followed to balance any skeletal equation. It includes indication of oxidation number, calculating increase or decrease in oxidation number, multiplying the formula to equalize, and final balancing using \[H\] and \[O\] atoms.

Complete answer:
First of all we have to write the skeletal equation given in the question.
 \[KCl{O_3}(s) \to KCl(s) + {O_2}(g)\]
Next, we shall balance the skeletal equation. It is actually easy to oxidize it through specific steps which is as follows,
Step1: First we need to calculate the oxidation number of elements in both the left and right side.
We shall check up on to the left hand side first.
On left hand side , oxidation number of elements are as follows,
\[K = + 1\] , \[O = - 2\] , \[Cl = + 5\]
Next , we shall check up on the right side.
On right hand side, oxidation number of element are as follows,
\[K = + 1\] , \[Cl = - 1\] , \[O = 0\]
Step2: Next step is to check the changes in oxidation number.
As we know , when comparing those left and right sides, there is a difference in oxidation number of elements from reactant stage to product stage. The changes of oxidation number of elements are as follows,
\[O: - 2 \to 0\] ; change in oxidation number is \[ + 2\]
\[Cl: + 5 \to - 1\] ; change in oxidation number is \[ - 6\]
Step3: Change in oxidation number is equalized in this step.
Actually in this step we should know that we need \[3\] atoms of \[O\] for every \[1\] atom of \[Cl\] otherwise \[6\] atoms of \[O\] for every \[2\] \[Cl\] atoms. This makes us understand that we need total changes of \[ + 12\] and \[ - 12\] .
Step4: Inserting coefficients to get the numbers mentioned above.
Here, in this step we need to put out exact coefficients to make them equalize to the above number of total change. Then , the skeletal equation will become as follows,
\[2KCl{O_3} \to 2KCl + 3{O_2}\]
Now we can see every substance on the left and right hand side is having coefficients.
Step5: Check each element one by one as all are balanced or not.
This is the final step to make sure that whether elements on both sides are balanced or not because then only we can assure that the given skeletal equation is balanced without error.
On left hand side: \[2\]\[K\]; \[2\]\[Cl\]; \[6\]\[O\]
On right hand side: \[2\]\[K\]; \[2\]\[Cl\]; \[6\]\[O\]
Now it is sure that the equation is balanced correctly.
Therefore, the correct balanced equation is \[2KCl{O_3} \to 2KCl + 3{O_2}\]
In this way the redox reaction is balanced using the oxidation number method.

Note: Actually while balancing a redox reaction we have to keep in mind that it is a little more complicated as compared to mass balancing. So we should do it with very concentration during calculations like finding oxidation number, analyzing change in oxidation number, multiplying, inserting coefficients etc: Oxidation and reduction process altogether in a chemical reaction is actually termed as redox reaction. Next main point we have to remember is to have a look at the final results and ensure whether it is mass balanced and whether they possess the same total charge on both sides. Working more and more on balancing redox equations will make it get easier and also keep in mind that as long as the atom gets oxidized and reduced , the procedure of balancing will pull out necessary details.