Question

Balance the following redox reaction by oxidation number method.

MnO2 + Br– → Mn2+ + Br2 + H2O (Acid medium)​

Answer 1

refer the attachment above

Answer 2

Here,

we see that $MnO_{2}$ to $Mn$ is a reduction reaction (gain of electrons)  and $Br^{-}$ to $Br_{2}$ is oxidation (loss of electrons).

To balance molecules of Br, we multiply it by 2

To balance molecules of Mn, we multiply it by 1

∴ Oxidation reaction × 2

  Reduction reaction × 1

$2Br^{-}$ →$Br_{2}$

$MnO_{2}$→$Mn^{+2}$

The equation is,

$2Br^{-} + MnO_{2}$ → $2Br_{2} + Mn^{+2} + H_{2}O$

Now balancing H (added to LHS)  and O (added to RHS) molecules, we get,

$2Br^{-} + MnO_{2} + 4H^{-}$→ $2Br_{2} + Mn^{+2} + 2H_{2}O$.

Hence, this is the balanced equation.

(#SPJ3)