`Br_2` undergoes disproportionation reaction in basic medium to give `Br^(ɵ)` ion and `BrO_3^(ɵ)` (bromate) ion in reduction and oxidation reaction.

Explanation

`Br_(2)+2e^(-) to 2Br^(-) ` (n=2) (reduction)
`Br_(2)+12O^(-) H to 2BrO_(3)^(-)+6H_(2)O+10e^(-) (n=10)`
`Ew(Br_(2))=Mw(Br_(2))/(2)+Mw(Br_(2))/(10)=(80 xx 2)/(2)+(80 xx2)/(10)=80+16=96g`
Alternatively, `Ew(Br_(2))`
`=Mw(Br_(2)) (1/2+1/10) rArr Mw (Br_(2)) xx 6/10`
`=(80 xx 2 xx 6)/(10)=96g`
n-factor of `Br_(2)=(10)/(6)`
`Ew(Br_(2))=Mw(Br_(2))/("mean n-factor")=Mw(Br_(2))/(10//6)`
`=(6 xx Mw (Br_(2)))/(10)`