# 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.