3. Use the changes in oxidation numbers to identify which atoms are oxidized and which are reduced in each reaction:
a. 2H2 (g) + O2(g) →2H2O(l)
b. 2KNO3(s)→2KNO2(s)+ O2(g)
c. NH4NO2(s) →N2(g) + 2H2O(g)
Solution:
(a): 2H2(g) + O2(g) → 2H2O(l)
Oxidation half-reaction:
H20 − 2e− → 2H+
Reduction half-reaction:
O20 + 4e− → 2O−2
Therefore,
Hydrogen (H) atoms are oxidized
Oxygen (O) atoms are reduced
(b): 2KNO3(s) → 2KNO2(s) + O2(g)
Oxidation half-reaction:
2O−2 − 4e− → O20
Reduction half-reaction:
N+5 + 2e− → N+3
Therefore,
Oxygen (O) atoms are oxidized
Nitrogen (N) atoms are reduced
(c): NH4NO2(s) → N2(g) + 2H2O(g)
Oxidation half-reaction:
2N−3 − 6e− → N20
Reduction half-reaction:
2N+3 + 6e− → N20
Therefore,
Nitrogen (N) atoms in NH4+ are oxidized
Nitrogen (N) atoms in NO2− are reduced
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