Question

What is the potential of an electrode which originally contained 0.1 m no3no3- and 0.4 m h+ and which has been treated by 80% of the cadmium necessary to reduce all the no3no3- to no(g) at 1 bar? Given : no3+4h++3eno+2h2o; e=0.96 v ; log 2 = 0.3?

Answer 1

Explanation:

reduce all the NO_(3)^(-) to NO(g) at 1 bar ? Given : NO_(3)^(-)+4H^(+)+ 3e^(-)rarr NO+2H_(2)O, E^(@)=0.96 , log 2 = 0.3.