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Balance each redox reaction occurring in basic solution.(a) NO2? (aq) + Al(s) ? NH3(g) +
Chapter 16, Problem 70P(choose chapter or problem)
Balance each redox reaction occurring in basic solution.
(a) \(\mathrm{NO}_{2}^{~-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{~-}(a q) \)
(b) \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{~-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{~-}(a q)\)
Equation Transcription:
Text Transcription:
NO_2^-(aq)+Al(s) rightarrow NH_3(g)+AlO_2^-(aq)
Al(s)+MnO_4^-(aq) rightarrow MnO_2(s)+Al(OH)_4^-(aq)
Questions & Answers
QUESTION:
Balance each redox reaction occurring in basic solution.
(a) \(\mathrm{NO}_{2}^{~-}(a q)+\mathrm{Al}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{AlO}_{2}^{~-}(a q) \)
(b) \(\mathrm{Al}(s)+\mathrm{MnO}_{4}^{~-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{Al}(\mathrm{OH})_{4}^{~-}(a q)\)
Equation Transcription:
Text Transcription:
NO_2^-(aq)+Al(s) rightarrow NH_3(g)+AlO_2^-(aq)
Al(s)+MnO_4^-(aq) rightarrow MnO_2(s)+Al(OH)_4^-(aq)
ANSWER:
Solution: Here, we are going to balance the given redox reactions.
Step1:
- NO2-(aq) + Al(s) → NH3(g) + AlO2-(aq)
The two half reactions are:
Oxidation: Al(s) → AlO2-(aq) [ Al oxidized from 0 to +3]
Reduction: NO2- (aq) → NH3 (g) [ N reduced from +3 to -3 ]
Step2:
To balance the O atoms in the reduction half reaction, we add two water molecule on the right:
NO2- (aq) → NH3 (g) + 2H2O
To balance the H atoms, we add seven H+ ions on the left:
NO2- (aq) + 7H+ → NH3 (g) + 2H2O
As the reaction takes place in a basic solution, therefore, for seven H+ ions, we add seven OH– ions to both sides of the equation:
NO2- (aq) + 7H+ + 7OH- → NH3 (g) + 2H2O + 7OH-
Replacing the H+ and OH– ions with water, the resultant equation is:
NO2-