The standard reduction potential for the half-cell, `NO_(3(aq.))^(-)+2H_((aq.))^(+)+e^(-)rarrNO_(2(g))+2H_(2)O` is `0.78V`.
(i) Calculate the reduction potential in 8M `H^(+)`.
(ii) What will be the reduction potential of the half-cell in a neutral solution. Assume all the other species to be at unit concentration

Explanation

(i) The redox reaction is :
`NO_(3)^(-)(aq)+2H^(+)(aq)+e^(-) to NO_(2)(g)+H_(2)O(l)`
According to Nernst equation :
`E=E^(@)-(0.0591)/(n)"log"([NO_(2)(g)][H_(2)O(l)])/([NO_(3)^(-)(aq)][H^(+)(aq)]^(2))`
According to available data,
`[NO_(2)(g)]=1,[H_(2)O(l)]=1, [NO_(3)^(-)(aq)]=1,n=1,E^(@)=0.78" V", H^(+)=8M`
`:. " " E=0.78-(0.0591)/(1)"log"(1)/((8)^(2))=0.78+0.0591 log 8^(2)=0.78+0.0591(2xx"log "8)`
`=0.78+0.0591xx2xx0.9031=0.78+0.1067=0.8867" V"`
(ii) For neutral solution `(H^(+)]=10^(-7)M`
`E=0.78-(0.0591)/(1)"log"(1)/((10^(-7))^(2))=0.78-0.0591(log 10^(14))`
`=0.78-0.0591(14" log "10)=0.78-0.0591xx14" "(because "log" 10=1)`
`=0.78-0.8274=-0.0474" V"`.