Correct Answer - `a. 0.887V`,`-0.046V`
The standard reduction potential for the half cell.
`NO_(3)^(c-)(aq)+2H^(o+)(aq)+e^(-) rarr NO_(2)(g)+H_(2)O(g)` is `0.78V`.
By Nerst equation for the above half cell reaction.
`a.` `E_(cell)=E^(c-)._(cell)-(0.0591)/(n)log.([NO_(2)][H_(2)O])/([NO_(3)^(c-)][H^(o+)]^(2))`
`=0.78-(0.591)/(1)log.(1xx1)/(1xx(8)^(2))`
`=0.78+0.059log64`
`=0.887V`
`b.` In neutral solution `(H^(o+))=10^(-7)M`
So`E=0.78-(0.0591)/(1) log.(1xx1)/(1xx(10^(-7))^(2))`
or `=0.78+0.059xxlog(1xx10^(-14))`
`=-0.046V`