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Newbie

Joined: 18 Jul 2018
Posts: 2

# Reduction potential problem

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.

Distinguished Member

Joined: 27 Sep 2017
Posts: 269
Location: Berlin, Germany
 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.

Newbie

Joined: 18 Jul 2018
Posts: 2

 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?
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