Two possible reactions of ${\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}$with${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$to produce${\mathbf{O}}_{\mathbf{2}}$andare${\mathbf{Mn}}^{\mathbf{+}}$

Scheme:$$\begin{gathered} {\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\to }{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}} \\ {\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\to }{\mathbf{O}}_{\mathbf{2}} \end{gathered}$$

Scheme:$$\begin{gathered} {\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\to }{\mathbf{O}}_{\mathbf{2}}\mathbf{+}{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}} \\ {\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

(a) Complete the half reactions for both schemes by adding ${\mathbf{e}}^{\mathbf{+}}$and ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$and ${\mathbf{H}}^{\mathbf{+}}$write a balanced net equation for each scheme.

(b) Sodium peroxyborate tetrahydrate, ${\mathbf{NaBO}}_{\mathbf{3}}\mathbf{\cdot }\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{(}\mathbf{FM}\mathbf{153}\mathbf{.}\mathbf{86}\mathbf{)}$produces ${\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}$when dissolved in acid ${\mathbf{BO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\to }{\mathbf{H}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}{\mathbf{BO}}_{\mathbf{3}}^{\mathbf{-}}$. To decide whether Scheme 1 or 2 Schemeoccurs student at the U.S. Naval academy weighed $\mathbf{0}\mathbf{.}\mathbf{123}{\mathbf{gNaBO}}_{\mathbf{3}}\mathbf{.}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$into a 100mLvolumetric flask added 20mLof $\mathbf{1}\mathbf{M}\mathbf{}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}$and diluted to the mark with ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$. Then they titratedof this solution with$\mathbf{0}\mathbf{.}\mathbf{01046}{\mathbf{MKMnO}}_{\mathbf{4}}$until the first pale pink color persisted. How may mL of${\mathbf{KMnO}}_{\mathbf{4}}$are required in Scheme 1and 2 Scheme?

(The Scheme 1stoichiometry was observed).

A redox titration happens when the analyte and the titrant undergo an oxidation–reduction process. The endpoint is frequently detected using an indicator, much as it is in acid–base titrations. Oxalic acid titrated against potassium permanganate in acid medium is an example of redox titration.

(a) To balance the equation we have to write two half reactions so we can determine the ratio of moles.

In scheme 1,${\mathrm{KMnO}}_4$ is reduced to$M{n}^{2+}$ and$H_2{O}_2$ is oxidized to $O_2$.

The half reactions are

Reduction:$Mn{O}_4^{-}+5e^{-}+8H^{+}\rightleftharpoons M{n}^{2+}+4H_{2}O\hspace{1em}/\cdot 2$

Oxidation: $H_2{O}_2\rightleftharpoons O_2+2H^{+}+2e^{-}/\cdot 5$

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

Reduction:$2Mn{O}_4^{-}+10e^{-}+16H^{+}\rightleftharpoons 2M{n}^{2+}+8H_{2}O$

Oxidation: $5H_2{O}_2\rightleftharpoons 5O_2+10H^{+}+10e^{-}$

In Scheme $2{\mathrm{KMnO}}_4$is both oxidized and reduced to $M{n}^{2+}$and $O_2$while $H_2{O}_2$is reduced to water.

The half reactions are:

Reduction:$H_2{O}_2+2H^{+}+2e^{-}\rightleftharpoons 2H_{2}O/\cdot 3$

Oxidation:$Mn{O}_4^{-}\rightleftharpoons M{n}^{2+}+2O_2+3e^{-}\hspace{1em}/\cdot 2$

To balance the electrons on the left and right side, we multiply the first reaction with 3 and the second reaction with 2 and we get:

Reduction:$3H_2{O}_2+6H^{+}+6e^{-}\rightleftharpoons 6H_{2}O$

Oxidation: $2Mn{O}_4^{-}\rightleftharpoons 2M{n}^{2+}+4O_2+6e^{-}$

Hence the half reactions for Scheme 1 and Scheme 2 are

Reduction:$Mn{O}_4^{-}+5e^{-}+8H^{+}\rightleftharpoons M{n}^{2+}+4H_{2}O\hspace{1em}/\cdot 2$

Oxidation:$H_2{O}_2\rightleftharpoons O_2+2H^{+}+2e^{-}/\cdot 5$

and

Reduction:$H_2{O}_2+2H^{+}+2e^{-}\rightleftharpoons 2H_{2}O/\cdot 3$

Oxidation:$Mn{O}_4^{-}\rightleftharpoons M{n}^{2+}+2O_2+3e^{-}\hspace{1em}/\cdot 2$

respectively.

Add the two reactions we got in the previous step.$3H_2{O}_2+6H^{+}+6e^{-}+2Mn{O}_4^{-}\rightleftharpoons 6H_{2}O+2M{n}^{2+}+4O_2+6e^{-}$

The net reaction is$3H_2{O}_2+6H^{+}+6e^{-}+2Mn{O}_4^{-}\rightleftharpoons 6H_{2}O+2M{n}^{2+}+4O_2$

(b) First we should determine the moles of${\mathrm{NaBO}}_3$ that were used in the titration with ${\mathrm{KMnO}}_4$.

$$\begin{gathered} \mathrm{n}({\mathrm{NaBO}}_3\cdot 4{\mathrm{H}}_2\mathrm{O})=\frac{\mathrm{m}({\mathrm{NaBO}}_3\cdot 4{\mathrm{H}}_2\mathrm{O})}{\mathrm{M}({\mathrm{NaBO}}_3\cdot 4{\mathrm{H}}_2\mathrm{O})} \\ \mathrm{n}({\mathrm{NaBO}}_3\cdot 4{\mathrm{H}}_2\mathrm{O})=\frac{1.023\mathrm{g}}{153.86\mathrm{g}/\mathrm{mol}} \\ \mathrm{n}({\mathrm{NaBO}}_3\cdot 4{\mathrm{H}}_2\mathrm{O})=6.649\mathrm{mmol} \end{gathered}$$

Since 10mL of the 100mL solution of peroxyborate was used we have to divide by 10

$\mathrm{n}({\mathrm{NaBO}}_3\cdot 4{\mathrm{H}}_2\mathrm{O})=6.649\mathrm{mmol}:10=0.6649\mathrm{mmol}$

The ratio of moles is

$$\begin{gathered} n\left(H_2{O}_2\right)=n(NaB{O}_3\cdot 4H_{2}O) \\ n\left(H_2{O}_2\right)=0.6649mmol \end{gathered}$$

Volume of required in Scheme is

$$\begin{gathered} \mathrm{n}\left({\mathrm{KMnO}}_4\right)=\frac{2}{5}\mathrm{n}\left({\mathrm{H}}_2{\mathrm{O}}_2\right) \\ \mathrm{c}\left({\mathrm{KMnO}}_4\right)\cdot \mathrm{V}\left({\mathrm{KMnO}}_4\right)=\frac{2}{5}\mathrm{n}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)\hspace{1em}/:\mathrm{c}\left({\mathrm{KMnO}}_4\right) \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=\frac{2\cdot \mathrm{n}\left({\mathrm{H}}_2{\mathrm{O}}_2\right)}{5\cdot \mathrm{c}\left({\mathrm{KMnO}}_4\right)} \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=\frac{2\cdot 0.6649\mathrm{mmol}}{5\cdot 0.01046\hspace{0.33em}\mathrm{M}} \\ \mathrm{V}\left({\mathrm{KMnO}}_4\right)=25.43\mathrm{mL} \end{gathered}$$

Hence 25.43mL of${\mathrm{KMnO}}_4$ is used in Scheme 1.

Volume of$KMn{O}_4$required in Scheme2 is

$$\begin{gathered} n\left(KMn{O}_4\right)=\frac{2}{3}n\left(H_2{O}_2\right) \\ c\left(KMn{O}_4\right)\cdot V\left(KMn{O}_4\right)=\frac{2}{3}n\left(H_2{O}_2\right)/:c\left(KMn{O}_4\right) \\ V\left(KMn{O}_4\right)=\frac{2\cdot n\left(H_2{O}_2\right)}{3\cdot c\left(KMn{O}_4\right)} \\ V\left(KMn{O}_4\right)=\frac{2\cdot 0.6649\hspace{0.33em}mmol}{3\cdot 0.01046\hspace{0.33em}M} \\ V\left(KMn{O}_4\right)=42.38\hspace{0.33em}mL \end{gathered}$$

Hence 42.38mL of $KMn{O}_4$is required in Scheme2.