consider voltaic cell at 25 deg celsius in which the following reaction takes place 3O2 + 4NO +2H2O yields 4NO3- + 4H+ calculate E not

Answer 1

Cathode (reduction): O₂ + 4H⁺ + 4e- ==> 2H₂0 Eº = +1.229 V (reduction potential)

Anode (oxidation): NO + 2H₂O ==> NO₃^- + 4H⁺ + 3e- Eº = +0.958 V (reduction potential)

Multiply reduction rxn by 3 and oxidation rxn by 4 and cancel like items on opposites sides to get...

3O₂ + 4NO + 2H₂O ==> 4NO₃^- + 4H⁺ Eº = 1.229 V - 0.958 V = 0.271 V