Calculate the equivalent weight of the underlined species in the following unbalanced reaction:
`underline(Br)_2+overset(ɵ)(O)toBr^(ɵ)+BrO_2^(ɵ)` (basic medium)
(b). `underline(HCuCl)_2toCu+Cu^(2+)+4Cl^(ɵ)+2H^(o+)` (acidic medium)
(c). `underline(MnO_4^(2-))toMnO_2+MnO_4^(ɵ)` (acidic medium)
(d). `KCiO_3overset(Delta)toKCl+O_2`

Explanation

(a). `2e^(-)+Br_2to2Br^(ɵ)`
`Br_2to2BrO_2^(ɵ)+6e^(-)`
`(Ew of Br_2=((Mw)/(2)+(Mw)/(6))=((160)/(2)+(160)/(6))`
`=80+26.66=106.66`
(b). `HCuCl_2toCu+Cu^(2+)`
`e^(-)+Cu^(o+)toCu`
`Cu^(o+)toCu^(2+)+e^(-)`
Mw of `HCuCl_2=((Mw)/(1)+(Mw)/(2))=135.5+135.5=271`
(c). `2e^(-)+MnO_4^(2-)toMnO_2`
`MnO_4^(2-)toMnO_4^(ɵ)+e^(-)`
(Mw of `MnO_4^(ɵ)=55+4xx16=119)`
Ew of `MnO_4^(ɵ)=((Mw)/(2)+(Mw)/(1))=((119)/(2)+(119)/(2))`
`=59.5+119=178.5`
(d).
Mw of `KCiO_3=39+35.5xx16xx3=122.5`
Ew of `KClO_3=(Mw)/(6)=(122.5)/(6)=20.42`