2. combine the oxidation and reduction half - reactions for hydrogen peroxide and permanganate ions…

Explanation:

Step1: Write oxidation half - reaction

The oxidation of hydrogen peroxide ($H_2O_2$) in acid solution gives oxygen gas ($O_2$). The oxidation state of oxygen in $H_2O_2$ is - 1 and in $O_2$ is 0. $H_2O_2\rightarrow O_2 + 2H^++ 2e^-$

Step2: Write reduction half - reaction

The permanganate ion ($MnO_4^-$) is reduced in acid solution to $Mn^{2 + }$. The oxidation state of Mn in $MnO_4^-$ is + 7 and in $Mn^{2+}$ is + 2. $MnO_4^-+8H^++5e^-\rightarrow Mn^{2 + } + 4H_2O$

Step3: Multiply half - reactions to balance electrons

To cancel out the electrons, we multiply the oxidation half - reaction by 5 and the reduction half - reaction by 2. Oxidation: $5H_2O_2\rightarrow 5O_2 + 10H^++ 10e^-$ Reduction: $2MnO_4^-+16H^++10e^-\rightarrow 2Mn^{2 + } + 8H_2O$

Step4: Combine half - reactions

Adding the two half - reactions together gives the overall balanced chemical equation: $2MnO_4^-+5H_2O_2 + 6H^+\rightarrow 2Mn^{2 + }+5O_2 + 8H_2O$

Answer:

$2MnO_4^-+5H_2O_2 + 6H^+\rightarrow 2Mn^{2 + }+5O_2 + 8H_2O$