Question

for the reaction
mathrm{kclo}_{3} longrightarrow mathrm{kcl}+\frac{3}{2} mathrm{o}_{2}
assign oxidation numbers to each element on each side of the equation.

k in kclo₃:
cl in kclo₃:
o in kclo₃:

k in kcl:
cl in kcl:
o in o₂:

which element is oxidized?
o cl
o o
o k

which element is reduced?
o o
o k
o cl

Explanation:

Step1: Determine oxidation number of K in KClO₃

In compounds, group - 1 metals like K have an oxidation number of + 1. So, the oxidation number of K in KClO₃ is + 1.

Step2: Determine oxidation number of K in KCl

Again, K has an oxidation number of + 1 in KCl.

Step3: Determine oxidation number of Cl in KClO₃

Let the oxidation number of Cl be \(x\). In KClO₃, K = + 1 and O = - 2. The sum of oxidation numbers in a neutral compound is 0. So, \(+1+x + 3\times(-2)=0\), which gives \(x=+5\).

Step4: Determine oxidation number of Cl in KCl

In KCl, K = + 1. Let the oxidation number of Cl be \(y\). Then \(+1 + y=0\), so \(y=-1\).

Step5: Determine oxidation number of O in KClO₃

In most compounds, oxygen has an oxidation number of - 2. So, the oxidation number of O in KClO₃ is - 2.

Step6: Determine oxidation number of O in O₂

In elemental form, the oxidation number of an element is 0. So, the oxidation number of O in O₂ is 0.

Step7: Identify oxidized and reduced elements

Oxidation is an increase in oxidation number. Oxygen goes from - 2 in KClO₃ to 0 in O₂, so oxygen is oxidized. Reduction is a decrease in oxidation number. Chlorine goes from + 5 in KClO₃ to - 1 in KCl, so chlorine is reduced.

Answer:

K in KClO₃: + 1
K in KCl: + 1
Cl in KClO₃: + 5
Cl in KCl: - 1
O in KClO₃: - 2
O in O₂: 0
Which element is oxidized? O
Which element is reduced? Cl