Chapter 24, Problem 24.64P

Interpretation Introduction

Interpretation:

The given reaction has to be balanced in acidic solution and also concentration of the original $mo{O}_4^{2-}$ solution has to be calculated.

    $Mn{O}_4^{-}+M{o}^{3+}\to M{n}^{2+}+Mo{O}_2^{2+}$

Concept introduction:

Procedure for balancing the redox reaction in acidic and basic solution is given below.

1. 1) The oxidation half reaction equation and a reduction half reaction equation of the given reaction have to be separated.
2. 2) The half reaction equation has to be balanced with respect to all atoms other than oxygen and hydrogen.
3. 3) Each half reaction equation has to be balanced with respect to oxygen atoms by adding the suitable number of $H_{2}O$ molecules to the side where oxygen atom is deficient.  (For metal hydroxides, directly add $O{H}^{-}$ ions and skip steps $3–4.)$
4. 4) In Acidic Solution: Each half reaction equation has to be balanced with respect to hydrogen atoms by adding the suitable number of $H^{+}$ ions to the side deficient in hydrogen atoms.

   In Basic Solution: Each half reaction has to be balanced with respect to hydrogen atoms by adding a number of $H_{2}O$ molecules equal to the number of excess hydrogen atoms to the side deficient in hydrogen atoms and an equal number of $O{H}^{-}$ ions to the side opposite to the added $H_{2}O$ molecules.

5. 5) Each half reaction equation has to be balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge.
6. 6) Each half reaction equation has to be multiplied by an integer that makes the number of electrons given by the oxidation half reaction equation equal to the number of electrons taken by the reduction half reaction equation.
7. 7) The complete balanced equation is obtained by adding the two half reaction equations and canceling or combining any like terms.

Answer to Problem 24.64P

The concentration of original $Mo{O}_4^{2-}$ solution is $0.103M.$

Explanation of Solution

The given can be rewritten as follows,

    $Mn{O}_4^{-}+M{o}^{3+}\to M{n}^{2+}+Mo{O}_2^{2+}$

The given reaction can be written as two half reactions.

The oxidation half reaction is $M{o}^{3+}\to Mo{O}_2^{2+}.$

The oxidation half reaction can be balanced using above procedure.

The balanced oxidation half reaction with respect to all atoms other than oxygen and hydrogen is given below.

    $M{o}^{3+}\to Mo{O}_2^{2+}$

The given reaction is balanced with respect to oxygen atom by adding suitable number of $H_{2}O$ molecules to the side where oxygen atom is deficient as follows,

    $M{o}^{3+}+2H_{2}O\to Mo{O}_2^{2+}$

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of $H^{+}$ ions to the side deficient in hydrogen atoms as below.

    $M{o}^{3+}+2H_{2}O\to Mo{O}_2^{2+}+4H^{+}$

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

    $M{o}^{3+}+2H_{2}O\to Mo{O}_2^{2+}+4H^{+}+3e^{-}$

The reduction half reaction is $Mn{O}_4^{-}\to M{n}^{2+}.$

The reduction half reaction can be balanced using above procedure.

The balanced reduction half reaction with respect to all atoms other than oxygen and hydrogen is given below.

    $Mn{O}_4^{-}\to M{n}^{2+}$

The given reaction is balanced with respect to oxygen atom by adding suitable number of $H_{2}O$ molecules to the side where oxygen atom is deficient as follows,

    $Mn{O}_4^{-}\to M{n}^{2+}+4H_{2}O$

The above reaction is balanced with respect to hydrogen atom by adding the suitable number of $H^{+}$ ions to the side deficient in hydrogen atoms as below.

    $Mn{O}_4^{-}+8H^{+}\to M{n}^{2+}+4H_{2}O$

The reaction equation is balanced with respect to charge by adding the suitable number of electrons to the side with the excess positive charge as follows,

    $Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$

The overall reaction can be balanced as follows,

    $\frac{\begin{array}{l}M{o}^{3+}+2H_{2}O\to Mo{O}_2^{2+}+4H^{+}+3e^{-}]\times 5\\ \\ Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O]\times 3\end{array}}{3Mn{O}_4^{-}+4H^{+}+5M{o}^{3+}\to 3M{n}^{2+}+2H_{2}O+5Mo{O}_2^{2+}}$

The balanced reaction is $\text{3}{\text{MnO}}_{\text{4}}^{\text{-}}\text{+}\text{4}{\text{H}}^{\text{+}}\text{+}\text{5}{\text{Mo}}^{\text{3+}}\to \text{3}{\text{Mn}}^{\text{2+}}\text{+}\text{2}{\text{H}}_{\text{2}}\text{O+}\text{5}{\text{MoO}}_{\text{2}}^{\text{2+}}\text{.}$

It is given that the filtrate required $20.85mL\left(0.02085L\right)of0.0955MKMn{O}_4$ for the reaction.

Therefore number of moles of $KMn{O}_4$ reacted can be calculated as follows,

    $\begin{array}{c}NumberofmolesofKMn{O}_{4}reacted=Molarity\times Volume\\ \\ =0.0955M\times 0.02085L\\ \\ =0.00199mole\end{array}$

According to the reaction three mole of $KMn{O}_4$ react with five mole of $M{o}^{3+}.$

Therefore one mole of  $KMn{O}_4$ react with $\frac{5}{3}moleofM{o}^{3+}.$

Hence $0.00199mole$ of $KMn{O}_4$ will be reacting with $0.003316moleofM{o}^{3+}.$

All the $M{o}^{3+}$ is formed from $Mo{O}_4^{2-}.$

Therefore $0.003316moleofM{o}^{3+}$ is formed from $0.003316moleofMo{O}_4^{2-}.$

Also given that the volume of $Mo{O}_4^{2-}$ solution is $32.15mL\left(0.03215L\right).$

Hence the concentration of original $Mo{O}_4^{2-}$ solution can be calculated as follows,

    $\begin{array}{c}MolarityofMo{O}_4^{2-}solution=\frac{numberofmoles}{Volumeofsolution\left(L\right)}\\ \\ =\frac{0.003316mole}{0.03215L}\\ \\ =0.103M.\end{array}$

Molarity of $Mo{O}_4^{2-}$ solution is $0.103M.$