Chemistry Forum

Dear all, I have an issue of what might be the most possible reaction on anode and cathode during electrolysis of my chemical.

These are the reduction potentials that I found on web.

Electrodes: Graphite electrodes
Voltage: 60 V
Current: 5A

Oxidation @ anode:

R₁. H₂O -> 1/2 O₂ + 2H+ + 2e- Eo ox = -1.23 V
R₂. NH₃OH+ -> 1/2 N₂ + H₂O + 2H+ + 2e- Eo ox = 1.87 V

Reduction @ cathode:

R₃. NO₃- + 4H+ + 3e- -> NO + H₂O Eo red= 0.96 V
R₄. NO₃- + 3H+ + 2e- -> HNO₂ + H₂O Eo red= 0.94 V
R₅. NO₃- + 2H+ + 2e- -> NO₂ + H₂O Eo red= 0.80 V
R₆. 2NO₃- + 12H+ + 10e- -> N₂ + 6H₂o Eo red= 1.25 V
R₇. 2H₂O + 2e- -> H₂ + 2OH- Eo red= -0.83 V
R₈. 2NH₃OH+ + H₂ + 2e- -> N₂H₅+ + 2H₂O Eo red = 1.42 V

From my understanding, the more positive is the Eo ox and Eo red, the more possible the reaction will occur.
From the reactions above, it should be R₂ and R₈ that will occur for electrolysis of hydroxylamine nitrate, while for electrolysis of concentrated nitric acid, that will be R₁ and R₃. However, I did notice NO₂ produced during the reaction.

I would say you boil your solution . Compare your potentials and the used voltage and current.

Nitrate decompose to NO, NO₂ under this circumstances

hydroxylammonium will go to hydrogen and hydroxylamin.

ChenBeier wrote:I would say you boil your solution . Compare your potentials and the used voltage and current.

Nitrate decompose to NO, NO₂ under this circumstances

hydroxylammonium will go to hydrogen and hydroxylamin.

How should I compare the potentials with the used current and voltage?
From what I read on electrolysis theory, only the reaction with most positive potential will proceed.

Could you please let me know which reaction shows that the solution is boiling?

The NO₃- should form NO and then proceed to form NO₂ at cathode. But this is only observed in nitric acid not hydroxylamine nitrate.

How does the hydroxylammonium go to hydrogen and hydroxylamine where no equation with reduction potential was found related to hydrogen and hydroxylamine?