A pollutant in smog is nitrogen dioxide, NO2. The gas has a reddish brown color and is responsible for the red-brown color associated with this type of air pollution. Nitrogen dioxide is also a contributor to acid rain because when rain passes through air contaminated with NO2, it dissolves and undergoes the following reaction:
3NO2(g) + H2O → NO(g) + 2H+(aq) + 2NO3-(aq).
In this reaction, which element is reduced and which is oxidized?
Which element is the oxidizing agent and which is the reducing agent?
3NO2(g) + H2O → NO(g) + 2H+(aq) + 2NO3-(aq).
In this reaction, which element is reduced and which is oxidized?
Which element is the oxidizing agent and which is the reducing agent?
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For oxdn state purposes O is virtually always O^2- (as it is here) Also H is +1.
3NO2(g) + H2O → NO(g) + 2H+(aq) + 2NO3-(aq).
NO2 = [N] - 4(2O^2-) = 0 so it is N(IV) in NO2
NO N is N(II); and in NO3^- we have [N] - 6(3O^2-) = -1 hence N(V)
N(IV) in NO2 → NO N(II) N has gained two e⁻s so NO2 has been reduced
N(IV) in NO2 → N(V) in [NO3]^- N has lost one e⁻ therefore NO2 has been oxidized
Conclusion: NO2 has been both reduced and oxidized and consequently it is both
the oxidizing and reducing agent.
3NO2(g) + H2O → NO(g) + 2H+(aq) + 2NO3-(aq).
NO2 = [N] - 4(2O^2-) = 0 so it is N(IV) in NO2
NO N is N(II); and in NO3^- we have [N] - 6(3O^2-) = -1 hence N(V)
N(IV) in NO2 → NO N(II) N has gained two e⁻s so NO2 has been reduced
N(IV) in NO2 → N(V) in [NO3]^- N has lost one e⁻ therefore NO2 has been oxidized
Conclusion: NO2 has been both reduced and oxidized and consequently it is both
the oxidizing and reducing agent.