Answer:
NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-
There is 1 hydroxide ion, on the reactant side
Explanation:
NO2−(aq) + Al(s) → NH3(aq) + Al(OH)4−(aq)
Step 1: The half reactions
NO2- (aq) → NH3(g)
Al(s) → Al(OH)4-
<u>Step 2: </u>Balancing electrons
NO2- → NH3
On the left side N has an oxidation number of +3 and on the right side -3.
NO2- +6e-→ NH3
Al(s) → Al(OH)4-
On the left side, Al has an oxidation number of 0 and on the right side +3.
Al(s) → Al(OH)4- +3e-
To have the same amount of electrons transfered, we have to multiply the second reaction by 2
NO2- +6e-→ NH3
2(Al(s) → Al(OH)4- +6e-)
<u>Step 3:</u> Balance with OH/H2O
NO2- +6e +5H2O → NH3 +7OH-
2Al +8OH- → 2Al(OH)4- + 6e-
<u>Step 4:</u> The netto reaction
NO2- + 5H2O + 2Al + 8OH- → NH3 +7OH- + 2Al(OH)4-
NO2- + 5H2O + 2Al + OH- → NH3 + 2Al(OH)4-
There is 1 hydroxide ion, on the reactant side
