[johnnyjohn993]

NH₂OH + CO₂ =N₂ + CO  (BASIC)

2OH^- +2H⁺ +CO₂ = CO + H₂0  + 2OH^-

2e- + H₂O + CO₂ = CO + 2OH^-


3OH^- +H₂O + 2NH₂OH = N₂ + 5H⁺ +5OH^-

2OH^- +H₂O + 2NH₂OH = N₂ + 5H₂O

2OH^- + 2NH₂OH= N₂ +4H₂O + 2e-
2e- + H₂O + CO₂ =CO + 2OH^-

ANSWER: 2NH₂OH + CO₂ = N₂ + 3H₂O + CO

QUESTIONs :
1)why did we add H₂O to the side of 2NH₂OH of our half reaction instead
placing it on the oxygen and hydrogen deprive N₂ ?(isn't that how we do it? ).

2.)Does its always have to be alternate " that is, adding  your H⁺ to the CO₂ half reaction; and H⁺ to left side of the NH₂OH half reaction ? I got the answer after 30mins T_T it was so confusing though.

3) Do we have cases like we will "add up" our H₂O or OH^-  because we cannot cancel them ?

[Hunter2]

If the process works in alcaline condition then its not allowed to add H⁺

Correct is this:

Oxidation: 2 NH₂OH  + 2OH^- => N₂ + 4 H₂O + 2 e^-

Reduction: CO₂ + H₂O + 2 e^-=> CO + 2 OH^-

Addition gives

2 NH₂OH + CO₂ => N₂ + CO + 3 H₂O

Recipe alcaline condition:

Reduction: add water on edukt side and get OH^- on product side

Oxidation: add OH^- on edukt side and get water on product side



acidic condition:

Reduction : add H⁺ on edukt side and get water on product side

Oxidation : add water on edukt side and get H⁺ on product side

This works every time if oxygen is in the compounds.