Question

Complete and balance the following skeleton equation for a reaction in acidic solution: $H_{2}O_{2}+MnO_{4}^{-}+H^{+} \longrightarrow Mn^{2+}+H_{2}O+O_{2}$

Answer 1

The given reaction is

$MnO_4^- + H_2O_2 + H^+ \rightarrow Mn^{2+} + H_2O +O_2$

We have to balance this equation, so first let us apply reduction method to balance the electrons .

Le the oxidation number for Mn be n.

So,

$n+ 4\times(-2) =-1 \\=> n= 7$

Then,

$Mn^{+7} +5e^-\rightarrow Mn^{+2}$

In the same way, we get $H^{2+} +e^-\rightarrow H^+$

On adding both equations: $MnO_4^- + 5H_2O_2 + aH^+\rightarrow Mn^{2+} + 5H_2O +bO_2$

Now we have to find the value of a and b,

Now by hit and trial we can balance all the hydrogen and oxygen atoms,  

$MnO_4^- + 5H_2O_2 + 10H^+\rightarrow Mn^{2+} + 10H_2O +2O_2$

Answer 2

Explanation:

hope it helps you..................