(a) `NaBr` on reaction with conc `H_(2)SO_(4)` gives `SO_(2),HBr`, and `Br_(2)`, whereas `NaCl` with conc `H_(2)SO_(4)` gives `HCl` but no `Cl_(2)` or `SO_(2)` is produced. Explain ?
(b) Which of the following reactions occur ?
(i) `Br_2+2NaCltoCl_2+NaBr`
(ii) `Cl_2+NaBrtoBr_2+2NaCl`

Explanation

(a) Since the reduction potential of `F_(2) gt Cl_(2) gt Br_(2) gt I_(2),F_(2)` can oxidise `Cl^(ɵ),Br^(ɵ)`, and `I^(ɵ)` but not vice versa. Similarly, `Cl_(2)` can oxidise `I^(ɵ)` only `I^(ɵ)` but not vice vera. `Br^(ɵ)` is more easily oxidised than `Cl^(ɵ)`. So `NaBr(Br^(ɵ))` on reaction with conc `H_(2)SO_(4)`, undergoes oxidation to `Br_(2)` whereas `h_(2)SO_(4)(SO_(4)^(2-))` is reduced to `SO_(2)` So the reaction of `NaBr` with conc `H_(2)SO_(4)` is as follows :
`Br^(ɵ) + H_(2)SO_(4) ("conc") rarr NBr + HSO_(4)^(ɵ)`
`2Br^(ɵ) + 3H_(2) SO_(4) ("conc") rarr Br_(2)+SO_(2) + 2H_(2) O + 2HSO_(4)^(ɵ)`
Reaction of `NaCl` with conc `H_(2)SO_(4)` ltbr. `Cl^(ɵ) +H_(2)SO_(4)("conc") rarr HCl + HSO_(4)^(ɵ)`
(b) From the above equation it is clear that `Br^(ɵ)` is more easily oxidised than `Cl^(ɵ)`
Hence, `Cl_2` will oxidise `Br^(ɵ)` but not vice versa. So reaction (ii) is feasible.
`Br^(ɵ)+Cl_2toBr_2+2Cl^(ɵ)`