The standard reduction potential of the following reaction at `25^(@)C,` is `0.78V, NO_(3)^(-)(aq)+2H^(+)(aq)+e rarr NO_(2)+H_(2)O.`
(i) If the concentration of `H^(+)` is 8(M), what will be the standard reduction potential of the half - cell ? (ii) What will be the reduction potential of the half - cell in neutral solution? Suppose all other constituents are at unit concentration.

Explanation

(i) `E=E^(@)-(0.059)/(1)log.(1)/([H^(+)]^(2))`
or, `E=0.78-0.059log.(1)/((8)^(2))`
`therefore E=.8867V.`
(ii) In neutral solution, `[H^(+)]=10^(-7)(M)`
`E=0.78-0.059log.(1)/((10^(-7))^(2))=-0.046V`