Given `:`
`NO_(3)^(c-) rarrNO_(2)(` acidic medium `), " "E^(c-)=0.8V`
`NO_(3)^(c-) rarr NO_(3)OH(` acidic medium`)" "E^(c-)=0.74V`
At what `pH` the above two half reactions will have same `EMF` values ? Assume the concentration of all the species to be unity. `(` Take `0.059~~0.06)`

Explanation

Balance both equations in acidic medium
`NO_(3)^(c-)+e^(-)+2H^(o+)rarrNO_(2)+H_(2)O " "(i)`
NO_(3)+6e^(c-)+7H^(o+)rarr NH_(2)OH+2H_(2)O" "(ii)`
`E_(1)=E_(1)^(c-) -0.06 log .([NO_(2)])/([NO_(3)^(c-)][H^(o+)]^(2)) [[NO_(2)]=[NO_(3)^(c-)]1M]`
`=0.8-0.06log[H^(o+)]^(2)=0.8-0.06xx[-logH^(o+)]`
`=0.08-0.06xx2pH`
`E_(1)=0.8-0.12pH" "(iii)`
Similarly, `E_(2)` for reaction `(ii),`
`E_(2)=E_(2)^(c-)-(0.06)/(6)log.([NH_(2)OH])/([NO_(3)^(c-)][H^(o+)]^(7))[[NH_(2)OH]=[NO_(3)^(c-)]=1M]`
`=0.74-0.01log [H^(o+)]^(-7)`
`=0.74-0.01xx7(-log H^(o+))`
`=0.74-0.07pH " "(iv)`
Since `E_(1)=E_(2),[`equate Eqs. `(iii)` and `(iv)]`
`:.0.8-0.12pH=0.74-0.07pH`
`pH=(0.06)/(0.05)=1.2`