91Ó°ÊÓ

• Redox reactions are oxidation-reduction chemical reactions in which the oxidation states of the reactants change. The term redox refers to the reduction-oxidation process.
• All redox reactions can be divided down into two types of reactions: reduction and oxidation.
• In a redox reaction, or Oxidation-Reduction process, the oxidation and reduction reactions always happen at the same time.

To balance the equation we have to write two half-reactions so we can determine the ratio of moles. In strongly acidic conditions, ${\mathrm{KMnO}}_4$ is reduced to ${\mathrm{Mn}}^{2+}$ and ${\mathrm{H}}_2{\mathrm{O}}_2$ oxidized do oxygen as we can see in table 16-3. The half- reactions are:

Reduction :${\mathrm{MnO}}_4^{-}+5{\mathrm{e}}^{-}+8{\mathrm{H}}^{+}⇌{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}×2$

Oxidation: ${\mathrm{H}}_2{\mathrm{O}}_2⇌{\mathrm{O}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}×5$

To balance the number of electrons on the left and right side, we multiply the first reaction with 2 and the second reaction with 5 and we get:

$$\begin{gathered} \mathrm{Reduction}âˆ\P 2{\mathrm{MnO}}_4^{-}+10{\mathrm{e}}^{-}+16{\mathrm{H}}^{+}⇌2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Oxidation}âˆ\P 5{\mathrm{H}}_2{\mathrm{O}}_2⇌5{\mathrm{O}}_2+10{\mathrm{H}}^{+}+10{\mathrm{e}}^{-} \end{gathered}$$

Net reaction : $2{\mathrm{MnO}}_4^{-}+5{\mathrm{H}}_2{\mathrm{O}}_2+6{\mathrm{H}}^{+}⇌2{\mathrm{Mn}}^{2+}+5{\mathrm{O}}_2+8{\mathrm{H}}_2\mathrm{O}$

To find the molarity of ${\mathrm{H}}_2{\mathrm{O}}_2$, we have to calculate the molarity of ${\mathrm{KMnO}}_4$and put it in ratio with ${\mathrm{H}}_2{\mathrm{O}}_2$. The data that is given:

localid="1667555758794" $$\begin{gathered} \mathrm{V}(\mathrm{titrant})=27.66\mathrm{mL} \\ \mathrm{V}\left(\mathrm{blank}\right)=0.04\mathrm{mL} \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)=0.02123\mathrm{M} \\ \mathrm{V}\left(\mathrm{analyte}\right)=\mathrm{V}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=25.00\mathrm{mL} \end{gathered}$$

We can calculate the volume of${\mathrm{KMnO}}_4$by subtracting the volume of titrant with the volume of blank:

localid="1667555773710" $$\begin{gathered} \mathrm{V}\left({\mathrm{KMnO}}_4\right)=\mathrm{V}\left(\mathrm{titrant}\right)-\mathrm{V}\left(\mathrm{blank}\right) \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=(27.66-0.04)\mathrm{mL} \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=27.62\mathrm{mL} \end{gathered}$$

Now we can put moles of localid="1667555782210" ${\mathrm{H}}_2{\mathrm{O}}_2$and ${\mathrm{H}}_2{\mathrm{O}}_2$in ratio. The ratio of moles is 5 : 2 so we can write:

localid="1667555790634" $$\begin{gathered} \mathrm{n}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=\frac{5}{2}\mathrm{n}\left({\mathrm{KMnO}}_4\right) \\ \mathrm{c}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)â‹\dots \mathrm{V}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=\frac{5}{2}×\mathrm{c}\left({\mathrm{KMnO}}_4\right)×\mathrm{V}\left({\mathrm{KMnO}}_4\right)/:\mathrm{V} \\ \mathrm{c}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=\frac{5×\mathrm{c}\left({\mathrm{KMnO}}_4\right)×\mathrm{V}\left({\mathrm{KMnO}}_4\right)}{2×\mathrm{V}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)} \\ \mathrm{c}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=\frac{5×0.02123\mathrm{M}×27.62\mathrm{ml}}{2×25\mathrm{ml}} \\ \mathrm{c}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=0.0586\mathrm{M} \end{gathered}$$

Since 25mL of ${\mathrm{H}}_2{\mathrm{O}}_2$was diluted in 250mL volumetric flask, the original solution has 10 times larger concentration

localid="1667555804423" $$\begin{gathered} \mathrm{c}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=0.0586×10\mathrm{M} \\ \mathrm{c}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=0.586\mathrm{M} \end{gathered}$$

Hence, the molarity of the commercial ${\mathrm{H}}_2{\mathrm{O}}_2$is$\mathrm{c}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)=0.586\mathrm{M}$