Asked by rose
I'm having trouble balancing these two oxidation-reduction reactions that occur in a basic solution. Can anyone help? Please? The numbers after the formulas are subscripts and any with a minus sign means they are negatively charged (-1). Thanks!!
Al(s)+ MnO4-(aq)---> MnO2(s)+Al(OH)4-(aq)
NO2-(aq)+Al(s)-->NH3(g)+AlO2- (aq)
Al(s)+ MnO4-(aq)---> MnO2(s)+Al(OH)4-(aq)
NO2-(aq)+Al(s)-->NH3(g)+AlO2- (aq)
Answers
Answered by
DrBob222
Here are the two half equations (unbalanced) for the first one.
Al + 4OH^- ==> Al(OH)4^- + 3e
MnO4^- + 2H2O + 3e ==> MnO2 + 4OH^-
Add the two equations to obtain the balanced equation.
Al + 4OH^- ==> Al(OH)4^- + 3e
MnO4^- + 2H2O + 3e ==> MnO2 + 4OH^-
Add the two equations to obtain the balanced equation.
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