Balancing chemistry please?

Question

Balance the following equations, using the oxidation number method, and enter the coefficients in the boxes. The coefficients must be the smallest possible integers.
All boxes must contain an entry.
NO3- + Al + H2O + OH- NH3 + Al(OH)4- + H2O + OH-

Note that neither H2O nor OH- should appear on more than one side of the equation. Thus, some of the coefficients above must be 0.

Tries 0/5
PH3 + CrO42- + H2O + OH- P + Cr(OH)4- + OH- + H2O

Note that neither H2O nor OH- should appear on more than one side of the equation. Thus, some of the coefficients above must be 0




I kept trying to do this until I just couldn't. I can't get it please help me in anyway.

Answer 1

To balance the redox equation, please remember to always start from half reaction equation, such as the equations of EMF involving electrons. Please pay attention to what I am doing here. For the reaction:
NO3(-) + Al ---> NH3 + Al(OH)4(-)
with appropriately added OH- and/ or H2O. Here are the half reactions (with e- as the electrons):
Al + 4OH- ---> Al(OH)4(-) + 3e-................(1)
where Al is oxidized from Ox state of "0" to "3", thus 3e- is at the oxidized side.
NO3(-) + 8e- + 6H2O ---> NH3 + 9OH-.....(2)
where N is reduced from Ox state of "5" to "-3", thus 8e- is at the oxidized side.
8(1)+3(2) to eliminate electrons, we get:
8Al + 3NO3(-) + 18 H2O + 32OH- ---> 8Al(OH)4(-) + 3NH3 + 27OH-
cancel 27 OH- from both sides we get the final equation:
8Al + 3NO3(-) + 18 H2O + 5OH- ---> 8Al(OH)4(-) + 3NH3

For the reaction:
PH3 + CrO4(2-) ---> P + Cr(OH)4(-)
with appropriately inserted OH- and/ or H2O. Here are the half reactions (with e- as the electrons):
PH3 + 3OH- ---> P + 3H2O + 3e-
CrO4(2-) + 4H2O + 3e- ---> Cr(OH)4(-) + 4OH-
Adding both half equations up, we get:
CrO4(2-) + 4H2O + PH3 + 3OH- ---> Cr(OH)4(-) + 4OH- + P + 3H2O
Cancel 3H2O and 3OH- from both sides, we get the final equation:
CrO4(2-) + H2O + PH3 ---> Cr(OH)4(-) + P + OH-

Do you get it? Can you do the same in the future?