Question

1. combine the oxidation and reduction half - reactions for hydrogen peroxide and permanganate ions, respectively, and write the balanced chemical equations for the overall reaction between h₂o₂ and mno₄⁻ in acid solution. hint: the number of electrons transferred must “cancel out.”

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.
$H_2O_2
ightarrow O_2 + 2H^++ 2e^-$

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.
$MnO_4^-+8H^++5e^-
ightarrow Mn^{2 + } + 4H_2O$

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: $5H_2O_2
ightarrow 5O_2 + 10H^++ 10e^-$
Reduction: $2MnO_4^-+16H^++10e^-
ightarrow 2Mn^{2 + } + 8H_2O$

Step4: Combine half - reactions

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

Answer:

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