Balance each redoc reaction occuring in basic solution

Question

NO2^−(aq)+Al(s)→NH3(g)+AlO2^−(aq)

Al(s)+MnO4^−(aq)→MnO2(s)+Al(OH)4^−(aq)

Answer 1

Again, I'll do one since it is in basic solution. You should follow the procedure to do the others.

NO₂^- ==> NH3

NO₂^- ==> NH₃ + 2H₂O .. to balance the O atoms

NO₂^- + 7H⁺ ==> NH₃ + 2H₂O .. to balance the H (but since we are supposed to use basic soln, see next step)

NO₂^- + 7H⁺ + 7OH^- ==> NH₃ + 2H₂O + 7OH^- (to cancel out the added H+ and add OH-)

NO₂^- + 7H⁺ + 7OH^- + 6e- ==> NH₃ + 2H₂O + 7OH^- .. to balance the charge & this is the balanced 1/2 reaction

Al ==> AlO₂

Al + 2H₂O==> AlO₂ .. to balance O atoms

Al + 2H₂O==> AlO₂ + 4H⁺ .. to balance H atoms (but since in basic soln, see next step)

Al + 2H₂O==> AlO₂ + 4H⁺

Al + 2H₂O + 4OH^-==> AlO₂ + 4H⁺ + 4OH^- .. (to cancel out the added H⁺ and add OH^-)

Al + 2H₂O + 4OH^-==> AlO₂ + 4H⁺ + 4OH^- + 4e- .. to balance the charge & this is the balanced 1/2 reaction

NO₂^- + 7H⁺ + 7OH^- + 6e- ==> NH₃ + 2H₂O + 7OH^- x 2 to equalize electrons

Al + 2H₂O + 4OH^-==> AlO₂ + 4H⁺ + 4OH^- + 4e- x 3 to equalize electrons

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2NO₂^- + 14H⁺ + 14OH^- + 12e- ==> 2NH₃ + 4H₂O + 14OH^-

3Al + 6H₂O + 12OH^-==> 3AlO₂ + 12H⁺ + 12OH^- + 12e-

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2NO₂^- + 3Al + 4H₂O ==> 2NH₃ + 3AlO₂ + 2OH^- .. balanced overall equation