Question

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 1

`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`