What is the potential of an electrode which originally contained 0.1 MNO_3^-) and 0.4MH^+ and which has been treated by 80% of the cadmium necessary to reduce all the NO_3^- to NO(g) at 1 bar? Give: NO_3^(-)+4H^++3e^-)to NO+2H_2O,E^(@)=0.96V, log2=0.3

What is the potential of an electrode which originally contained 0.1 M

1.	What is the potential of an electrode which originally contained 0.1 MNO_3^-) and 0.4MH^+ and which has been treated by 80% of the cadmium necessary to reduce all the NO_3^- to NO(g) at 1 bar? Give: NO_3^(-)+4H^++3e^-)to NO+2H_2O,E^(@)=0.96V, log2=0.3
Answer» 0.84V 1.08V 1.23V 1.36V Answer :A