Chapter 19, Problem 67AP

The concentration of a hydrogen peroxide solution can be conveniently determined by titration against a standardised potassium permanganate solution in an acidic medium according to the following unbalanced equation:

${\text{MnO}}_{\text{4}}+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\underset{}{\to }{\text{O}}_{\text{2}}+{\text{Mn}}^{2+}$

(a) Balance this equation. (b) If 36.44 mL of a $\text{0}\text{.01652 }M{\text{ KMnO}}_{\text{4}}$ solution is required to completely oxidize $\text{25}{\text{.00 mL of an H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution, calculate the molarity of the ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution.

Interpretation Introduction

Interpretation:

The balance equation for the given reaction and the molarity of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution are to be determined.

Concept introduction:

Oxidation is the addition of the electronegative element and the removal of the electropositive element in a chemical reaction. Reduction is the addition of the electropositive element and the removal of the electronegative element in a chemical reaction.

The chemical reaction in which oxidation and reduction take place simultaneously or the gain and loss of electrons take place in the chemical reaction is called a redox reaction.

A galvanic cell is made up of two half-cells that are cathodic and anodic. This cell converts chemical energy into electrical energy.

When the cell constant is positive and free energy is negative, then the reaction will occur spontaneously.

Answer to Problem 67AP

Solution:

a)

The overall balanced equation is as follows:

$2{\text{MnO}}_4^{-}\left(aq\right)+5{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{O}}_{\text{2}}\left(g\right)$

b)

$0.0602\text{M}$

Explanation of Solution

a) Balance the equation

${\text{MnO}}_4^{-}\left(aq\right)+{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)\to {\text{Mn}}^{2+}\left(aq\right)+{\text{O}}_{\text{2}}\left(g\right)$

Consider that the galvanic cellis made up of ${\text{MnO}}_4^{-}$ half-cells and ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ half-cells.

The cell diagram is as follows:

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\underset{\text{anode}}{\left(aq\right){\text{|O}}_2}\left(g\right)\underset{\text{salt bridge}}{\text{|| }}\underset{\text{cathode}}{{\text{MnO}}_4^{-}\left(aq\right){\text{|Mn}}^{2+}}\left(aq\right)$

${\text{MnO}}_4^{-}$ undergoes half-reaction as reduction at the cathode as:

${\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)$.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ undergoes half-reaction as oxidation at the anode as:

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(g\right)\to {\text{O}}_{\text{2}}\left(g\right){\text{+ 2e}}^{-}+2{\text{H}}^{+}$.

The number of electrons that takes part in the two half-cell reactions should be the same.

${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ half-cell reaction is multiplied by $5$ to make the number of electrons that take part in the reaction.

$5\times \left({\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(g\right)\to {\text{O}}_{\text{2}}\left(g\right){\text{+ 2e}}^{-}+2{\text{H}}^{+}\right)$

${\text{MnO}}_4^{-}$ half-cell reaction is multiplied by $2$ to make the number of electrons that take part in the reaction.

$2\times \left({\text{MnO}}_4^{-}\left(aq\right)+8{\text{H}}^{+}\left(aq\right)+5e^{-}\to {\text{Mn}}^{2+}\left(aq\right)+4{\text{H}}_2\text{O}\left(l\right)\right)$

Adding the two half-cell reactions gives the overall cell reaction, which is as follows:

$\begin{array}{l}5{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(g\right)\to 5{\text{O}}_{\text{2}}\left(g\right)+10{\text{H}}^{+}+10e^{-}\\ 2{\text{MnO}}_4^{-}\left(aq\right)+10e^{-}+16{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)\\ \overline{5{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+2{\text{MnO}}_4^{-}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\rightleftharpoons 5{\text{O}}_{\text{2}}\left(g\right)+2{\text{Mn}}^{2+}\left(aq\right)}+8{\text{H}}_{\text{2}}\text{O}\left(l\right)\end{array}$

Hence, the overall balanced equation is as follows:

$2{\text{MnO}}_4^{-}\left(aq\right)+5{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{O}}_{\text{2}}\left(g\right)$.

b) Molality of the ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution

The volume of ${\text{KMnO}}_{\text{4}}$ is $36.44\text{mL}$, the concentration of ${\text{KMnO}}_{\text{4}}$ is $0.01652\text{M}$, and the volume of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution is $25\text{mL}$.

The number of moles of ${\text{KMnO}}_4$ is calculated as follows:

$\text{Moles}\text{of}{\text{KMnO}}_{\text{4}}=\text{Concentration}\text{of}{\text{KMnO}}_{\text{4}}\times \text{Volume}\text{of}{\text{KMnO}}_{\text{4}}$.

Substitute the value of the concentration and the volume in the above expression as follows:

$\begin{array}{l}\text{Moles}\text{of}{\text{KMnO}}_{\text{4}}=\left(36.44\text{mL}\right)\times \left(\frac{0.01652\text{mol}}{1000\text{mL}\text{sol}}\right)\\ =6.020\times {10}^{-4}\text{mol}{\text{KMnO}}_{\text{4}}\end{array}$

The overall balanced equation is as follows:

$2{\text{MnO}}_4^{-}\left(aq\right)+5{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Mn}}^{2+}\left(aq\right)+8{\text{H}}_{\text{2}}\text{O}\left(l\right)+5{\text{O}}_{\text{2}}\left(g\right)$

From the balance equation, 2 moles of ${\text{MnO}}_4^{-}$ are stoichiometrically equivalent to 5 moles of hydrogen peroxide.

The number of moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ oxidized is calculated as follows:

$\begin{array}{c}\text{Moles}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}=\left(6.020\times {10}^{-4}\text{mol}{\text{MnO}}_4^{-}\right)\times \left(\frac{5\text{mol}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}}{2\text{mol}{\text{MnO}}_4^{-}}\right)\\ =1.505\times {10}^{-3}\text{mol}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\end{array}$

The molarity of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is calculated as follows:

$\text{Molarity}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}=\frac{\text{Moles}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}}{\text{Volume}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}}$.

Substitute the value of the number of moles and the volume in the above expression as follows:

$\begin{array}{c}\text{Molarity}\text{of}{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}=\left(\frac{1.505\times {10}^{-3}\text{mol}}{25\times {10}^{-3}\text{L}}\right)\\ =0.0602\text{mol}/\text{L}\\ \text{=}\text{0}\text{.0602}\text{M}\end{array}$

Therefore, the molarity of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is $0.0602\text{M}$.