For the reaction NO_(3)^(-)toNO_(2) (acidic medium), E^(@)=0.790V NO_(3)^(-)toNH_(3)OH (acidic medium), E^(@)=0.731V Calculate the pH at which at the above two half reactions will have same E values (Assume the concentrations of all the species to be unity).

For the reaction NO_(3)^(-)toNO_(2) (acidic medium), E^(@)=0.790V NO_

1.	For the reaction NO_(3)^(-)toNO_(2) (acidic medium), E^(@)=0.790V NO_(3)^(-)toNH_(3)OH (acidic medium), E^(@)=0.731V Calculate the pH at which at the above two half reactions will have same E values (Assume the concentrations of all the species to be unity).
Answer» Solution :`NO_(3)^(-)+2H^(+)+E^(-)toNO_(2)+H_(2)O,""E^(@)=0.790V` `NO_(3)^(-)+7H^(+)+6e^(-)toNH_(2)OH+2H_(2)O,E^(@)=0.731V` Since E VALUES for both the reactions are same `E_(NO_(3)^(-)//NO_(2))=E_(NO_(3)^(-)//NH_(3)OH)` `thereforeE_(NO_(3)^(@)//NO_(2))+(0.059)/(1)"LOG"([H^(+)][NO_(3)^(-)])/([NO_(2)])=E_(NO_(3)^(-)//NH_(2)OH)^(@)+(0.059)/(6)"log"([H^(+)]^(7)[NO_(3)^(-)])/([NH_(2)OH])` or `0.790+0.059log[H^(+)]^(2)=0.731+(0.059)/(6)log[H^(+)]` (as concentrations of all SPECIES =1, given) or `0.790+0.118log[H^(+)]=0.731+0.0688log[H^(+)]` or `(0.118-0.0688)log[H^(+)]=0.731-0.790` or `0.0492log[H^(+)]=0.059` or `-log[H^(+)]=(0.059)/(0.0492)=1.1992` or `pH=1.1992`