Chapter 16, Problem 16.16P

(a)

Interpretation Introduction

Interpretation:

Two possible reaction of ${\text{MnO}}_4$ with ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ to give ${\text{O}}_2$ and ${\text{Mn}}^{\text{2+}}$ is are given.  Half reaction for both scheme by addition of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{H}}^{\text{+}}$ has to be completed and balanced net equation for both scheme has to be written.

Answer to Problem 16.16P

For scheme 2, the half reactions are,

$\text{2}\left[{\text{8H}}^{\text{+}}+{\text{MnO}}_{\text{4}}^{\text{-}}{\text{+ 5e}}^{-}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}{\text{+ 5H}}_{\text{2}}\text{O}\right]$

$\text{5}\left[{\text{ H}}_{\text{2}}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{O}}_{\text{2}}{\text{+2e}}^{\text{-}}{\text{+2H}}^{\text{+}}\right]$

The net reaction is,

${\text{6H}}^{\text{+}}+2{\text{MnO}}_{\text{4}}^{\text{-}}{\text{+ 5H}}_2{O}_2\stackrel{}{\to }{\text{2Mn}}^{\text{2+}}{\text{+5O}}_2{\text{+ 8H}}_{\text{2}}\text{O}$

For scheme 2, the half reactions are,

$\text{2}\left[{\text{MnO}}_{\text{4}}^{\text{-}}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}{\text{+ 2O}}_{\text{2}}{\text{+3e}}^{\text{-}}\right]$

$\text{3}\left[{\text{ H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{+2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{2H}}_{\text{2}}\text{O}\right]$

The net reaction is,

${\text{6H}}^{\text{+}}+2{\text{MnO}}_{\text{4}}^{\text{-}}{\text{+ 3H}}_2{O}_2\stackrel{}{\to }{\text{2Mn}}^{\text{2+}}{\text{+4O}}_2{\text{+ 6H}}_{\text{2}}\text{O}$

Explanation of Solution

Given

Scheme 1:

${\text{MnO}}_{\text{4}}^{\text{-}}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}$

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{2}\left[{\text{8H}}^{\text{+}}+{\text{MnO}}_{\text{4}}^{\text{-}}{\text{+ 5e}}^{-}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}{\text{+ 5H}}_{\text{2}}\text{O}\right]\\ {}^{+7}{}^{+2}\\ \text{5}\left[{\text{ H}}_{\text{2}}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{O}}_{\text{2}}{\text{+2e}}^{\text{-}}{\text{+2H}}^{\text{+}}\right]\\ {}^{-1}{}^0\end{array}}\\ {\text{6H}}^{\text{+}}+2{\text{MnO}}_{\text{4}}^{\text{-}}{\text{+ 5H}}_2{O}_2\stackrel{}{\to }{\text{2Mn}}^{\text{2+}}{\text{+5O}}_2{\text{+ 8H}}_{\text{2}}\text{O}\end{array}$

Scheme 2:

$\begin{array}{l}\underset{\_}{\begin{array}{l}\text{2}\left[{\text{MnO}}_{\text{4}}^{\text{-}}\stackrel{}{\to }{\text{Mn}}^{\text{2+}}{\text{+ 2O}}_{\text{2}}{\text{+3e}}^{\text{-}}\right]\\ {}^{+7}{}^{-2}{}^{+2}\\ \text{3}\left[{\text{ H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{+2H}}^{\text{+}}{\text{+2e}}^{\text{-}}\stackrel{}{\to }{\text{2H}}_{\text{2}}\text{O}\right]\\ {}^{-1-2}\end{array}}\\ {\text{6H}}^{\text{+}}{\text{+2MnO}}_{\text{4}}^{\text{-}}{\text{+ 3H}}_{\text{2}}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{2Mn}}^{\text{2+}}{\text{+4O}}_{\text{2}}{\text{+ 6H}}_{\text{2}}\text{O}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Volume of potassium permanganate required in scheme 1 and scheme 2 has to be calculated.

Answer to Problem 16.16P

In scheme 1, $25.43mL$ of potassium permanganate is required.

In scheme 2, $42.38mL$ of potassium permanganate is required.

Explanation of Solution

Given

Weight of Sodium peroxyborate tetrahydrate = $\text{1}\text{.023 g}$

Amount of 1 M Sulphuric acid = $\text{20}\text{.0 mL}$

Strength of Potassium permanganate titrated with 1 M Sulphuric acid = $\text{0}\text{.01046}\text{M}$

Amount of Sulphuric acid titrated with potassium permanganate = $\text{10}\text{.00 mL}$

To calculate: amount of ${\text{KMnO}}_{\text{4}}$ required for the titration

$\frac{\text{1}{\text{.023g NaBO}}_{\text{3}}{\text{.4H}}_{\text{2}}\text{O}}{\text{153}\text{.86g/mol}}\text{= 6}{\text{.649 mmol NaBO}}_3$

${\text{1}/\text{10}}^{\text{th}}$ of this amount was titrated.

$\text{= 0}\text{.6649 mmol}$ ${\text{NaBO}}_3$

It producing $\text{0}\text{.6649 mmol}$ hydrogen peroxide by the reaction of ${\text{BO}}_{\text{3}}^{\text{-}}{\text{+ 2H}}_{\text{2}}\text{O}\stackrel{}{\to }{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{BO}}_{\text{3}}^{\text{-}}$.

Scheme 1:

2 moles of ${\text{MnO}}_4^{-}$ react with 5 moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$

Therefore, $\text{0}\text{.6649 mmol}$ hydrogen peroxide requires $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\left(0.6649\right)$= $\text{0}{\text{.2660 mmol MnO}}_{\text{4}}^{\text{-}}$

$\frac{\text{0}{\text{.2660 mmol MnO}}_{\text{4}}^{\text{-}}}{\text{0}{\text{.01046 mmol KMnO}}_{\text{4}}\text{/mL}}\text{= }25.43mLKMn{O}_{\text{4}}$

Scheme 2:

2 moles of ${\text{MnO}}_4^{-}$ react with 3 moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$

Therefore, $\text{0}\text{.6649 mmol}$ hydrogen peroxide requires $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\left(0.6649\right)$= $\text{0}{\text{.4433 mmol MnO}}_{\text{4}}^{\text{-}}$

$\frac{\text{0}{\text{.4433 mmol MnO}}_{\text{4}}^{\text{-}}}{\text{0}{\text{.01046 mmol KMnO}}_{\text{4}}\text{/mL}}\text{= }42.38mLKMn{O}_{\text{4}}$