Oxidation-Reduction Reactions Chemistry Worksheet Page 9
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6.2 Oxidation Numbers
Example 6.1 shows how we can use our new tools.
e
6.1 - Oxidation Numbers and Redox Reactions
xample
O
3
The following equations represent the reactions that lead to the formation of
bjeCtive
O
4
ammonium phosphate for fertilizers. Determine the oxidation number for each atom
bjeCtive
O
5
in the formulas. Decide whether each reaction is a redox reaction, and if it is, identify
bjeCtive
O
6
what element is oxidized, what is reduced, what the oxidizing agent is, and what the
bjeCtive
reducing agent is.
+ 6SiO
+ 10C → P
+ 10CO + 6CaSiO
a. 2Ca
(PO
)
3
4
2
2
4
3
+ 5O
+ 6H
O → 4H
b. P
PO
4
2
2
3
4
+ NH
→ (NH
c. H
PO
)
PO
3
4
3
4
3
4
Solution
a. The first step is to determine the oxidation number for each atom in the
reaction. Let’s consider the first equation above:
Pure elements,
Monatomic ion,
oxidation number
oxidation number
Combined oxygen,
zero
equal to its charge (+2)
oxidation number −2
+ 6SiO
+ 10C
+ 10CO + 6CaSiO
2Ca
(PO
)
P
3
4
2
2
4
3
Monatomic ion,
Combined oxygen,
oxidation number
oxidation number −2
equal to its charge (+2)
Because the sum of the oxidation numbers for the atoms in an uncharged
molecule is zero, the oxidation number of the carbon atom in CO is +2:
(ox # C) + (ox # O) = 0
(ox # C) + −2 = 0
(ox # C) = +2
Using a similar process, we can assign a +4 oxidation number to the silicon
atom in SiO
:
2
(ox # Si) + 2(ox # O) = 0
(ox # Si) + 2(−2) = 0
(ox # Si) = +4
Calcium phosphate, Ca
(PO
)
, is an ionic compound that contains
3
4
2
2+
3‒
monatomic calcium ions, Ca
, and polyatomic phosphate ions, PO
. The
4
oxidation number of each phosphorus atom can be determined in two ways.
The following shows how it can be done considering the whole formula.
3(ox # Ca) + 2(ox # P) + 8(ox # O) = 0
3(+2) + 2(ox # P) + 8(−2) = 0
(ox # P) = +5
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