# Fuel cell potential for the electrolysis of ammonia

I am trying to calculate the fuel cell potential for the electrolysis of ammonia with the following equation:

E = Gibbs free energy for the reaction/(# electrons X Faraday's Constant)


The half reactions are as follows:

$$2NH_3 + 6OH^- -> N_2 + 6H_2O + 6e (Anode) E^0 = -0.77 V$$ $$6H_2O + 6e -> 3H_2 + 6OH^- (Cathode) E^0 = -0.83 V$$

I obtained the potencial values from the following paper and (double checked online as well): https://www.researchgate.net/publication/327875102_Review-Ammonia_Oxidation_Electrocatalysis_for_Hydrogen_Generation_and_Fuel_Cells Equations (4a,4b and 4c)

The overall reaction is then:

$$2NH_3 -> N2 + 3H_2$$ $$E^0 = +0.06 V$$ ** in the paper they seem to imply that the cell voltage is $$+ 0.06 V$$ which I don't understand why, since the oxidation takes place at the anode, therefore for the anode the potential should be $$E^0 = +0.77 V$$ from my understanding.

I have doubled check my standard enthalpies of formation and entropies of formation. However, when I calculate the Gibbs free energy of the reaction I keep getting $$E^0 = -0.06 V$$ for the overall reaction. My Gibbs free energy is positive and the temperature I am using is 298 K, also NH3 is in a gaseous state. I have tried calculating the $$E^0$$ for $$H_2 + 0.5 O_2 -> H_2O$$ and I obtain a cell voltage of $$E^0 = 1.23 V$$ which corresponds to the reaction Gibbs free energy of $$- 237 kJ/mol$$ , I just don't know where I am getting confused in the ammonia reaction.

In the paper they mention that the reaction is thermodynamically favorable, but I don't see how when the Gibbs free energy for the reaction is positive.

• Based on the wiki articles for "Ammonia" and the "Haber process", I think your standard potential value is -0.06 V, as you say. From the second source, equilibrium favors ammonia under standard conditions, though (for kinetic reasons) the reaction doesn't happen at any measurable rate. So the ammonia decomposition reaction would be -0.06 V standard cell potential, as you say. I look forward to seeing a nice answer to this! – Ed V Jun 20 '19 at 13:58