In the following reactions:
a. `3MnO_(2)+4Alrarr3Mn+2Al_(2)O_(3)`
b. `2MnO_(4)^(ө)+16H^(o+)+10Cl^(ө)rarr2Mn^(2+)+5Cl_(2)+8H_(2)O`
Which species is reduced and which is oxidised?

Answer

A

Explanation

a. Change in oxidation numbers:
`I =? -1` to `0, Cl implies 0` to `-1`, and `Himplies+1` to `+1` ( no change)
Ioding in `HI` is oxidised and chlorine is reduced.
b. Change in oxidation number:
`Mn implies -4` to, `Al implies 0` to `3`, and `O implies -2` to `-2`
Manganese in `MnO_(2)` is reduced and aluminium is oxidised.
c. Change in oxidation number:
`Mnimplies + 7` to `2, Oimplies -2` to `-2` and `Cl implies -1` to `0`
Manganese in `MnO_(4)^(ө)` is reduced and `Cl^(ө)` is oxidised
iv. `Cu^(2+)` is reduced beacuse its oxidation number is lowered to `+1` in `CuBr` whereas `SO_(2)` is oxidised beacuse the oxidation number of `S(+4)` is raised to `(+6)` in `SO_(4)^(2-)`.