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Enter the oxidation number of one atom of each element in each reactant and product.

CH4(g)+2O2(g)⟶CO2(g)+2H2O(g)
C in CH4 :
H in CH4 :
O in O2 :
C in CO2 :
O in CO2 :
H in H2O :
O in H2O :
Which atom is reduced?
Which atom is oxidized?

Respuesta :

Answer:

1. -4

2. +1

3. 0

4. +4

5. -2

6. +1

7. -2

reduced = H

oxidized = O

Explanation:

Know oxidation rules.

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The oxidation numbers of the atoms of the specified elements in each of the given atoms are;

1) -4

1) -42) +1

1) -42) +13) 0

1) -42) +13) 04) +4

1) -42) +13) 04) +45) -2

1) -42) +13) 04) +45) -26) +1

1) -42) +13) 04) +45) -26) +17) -2

1) -42) +13) 04) +45) -26) +17) -2Atom oxidized = C

1) -42) +13) 04) +45) -26) +17) -2Atom oxidized = CAtom reduced = O

1) C in CH4

To get the oxidation number of C;

Oxidation state of hydrogen atom is +1 and so if the oxidation state of C is x, then we have;

x + 4(+1) = 0

x + 4 = 0

x = -4

2) H in CH4

Oxidation state on Carbon atom in this case is -4. Thus;

-4 + 4x = 0

4x = 4

x = +1

3) O in O2

This is oxygen gas that exists in it's free state and as such oxidation number is 0.

4) C in CO2

Oxidation state of O here is -2. Thus;

x + 2(-2) = 0

x - 4 = 0

x = +4

5) O in CO2

Oxidation state of C is +4 here. Thus;

4 + 2x = 0

2x = -4

x = -4/2

x = -2

6) H in H2O

Oxidation state of oxygen here is -2. Thus;

2x - 2 = 0

2x = 2

x = 2/2

x = +1

7) O in H2O

Oxidation state of hydrogen here is +1. Thus;

2(1) + x = 0

x = -2

Finally, oxidation number of carbon increased, then it is the atom that was oxidized while the atom reduced is the Oxygen atom.

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