Question

атізнw Question 6 Homework. Unanswered Fill in the Blanks Consider the already balanced chemical reaction shown below: KCLO,(s) +KCl(s) + O2(g) The oxidation state of Kin KCLO2 (s) = The oxidation state of Clin KCLO,(s) = The oxidation state of O in KCLO(s) - The oxidation state of Kin KCLS) The oxidation state of Clin KCl(s) - The oxidation state of O in O2(g): The element being oxidized is (K.CO) The element being reduced is (K.COM

Answer 1

Oxidation number of K = +1

Oxidation number of O = -2

lets the oxidation number of Cl be x

use:

1* oxidation number (K) + 2* oxidation number (O) + 1* oxidation number (Cl) = net charge

1*(+1)+2*(-2)+1* x = 0

-3 + 1 * x = 0

x = 3

So oxidation number of Cl = +3

Oxidation number of each element in KClO2 is

K=+1

O=-2

Cl=+3

Oxidation number of K = +1

lets the oxidation number of Cl be x

use:

1* oxidation number (K) + 1* oxidation number (Cl) = net charge

1*(+1)+1* x = 0

1 + 1 * x = 0

x = -1

So oxidation number of Cl = -1

Oxidation number of each element in KCl is

K=+1

Cl=-1

lets the oxidation number of O be x

use:

2* oxidation number (O) = net charge

2* x = 0

0 + 2 * x = 0

x = 0

So oxidation number of O = 0

Oxidation number of each element in O2 is

O=0

Oxidation state of Cl is changing from +3 to -1

So, Cl is reduced

Oxidation state of O is changing from -2 to 0

So, O is oxidised

Answer:

+1

+3

-2

+1

-1

0

O

Cl