Chapter 24, Problem 24.31P

(a)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

    $MnO+Pb{O}_2\to Mn{O}_4^{-}+P{b}^{2+}$

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.

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is $MnO\to Mn{O}_4^{-}.$

The reduction half reaction is $Pb{O}_2\to P{b}^{2+}.$

Each 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.

    $MnO\to Mn{O}_4^{-}$

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,

    $MnO+3H_{2}O\to Mn{O}_4^{-}$

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.

    $MnO+3H_{2}O\to Mn{O}_4^{-}+6H^{+}$

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,

    $MnO+3H_{2}O\to Mn{O}_4^{-}+6H^{+}+5e^{-}$

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

    $Pb{O}_2\to P{b}^{2+}$

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,

    $Pb{O}_2\to P{b}^{2+}+2H_{2}O$

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.

    $Pb{O}_2+4H^{+}\to P{b}^{2+}+2H_{2}O$

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,

    $Pb{O}_2+4H^{+}+2e^{-}\to P{b}^{2+}+2H_{2}O$

The overall reaction can be balanced as follows,

    $\frac{\begin{array}{l}MnO+3H_{2}O\to Mn{O}_4^{-}+6H^{+}+5e^{-}]\times 2\\ \\ Pb{O}_2+4H^{+}+2e^{-}\to P{b}^{2+}+2H_{2}O]\times 5\end{array}}{2MnO+5Pb{O}_2+8H^{+}\to 2Mn{O}_4^{-}+5P{b}^{2+}+4H_{2}O}$

The balanced reaction is $2MnO+5Pb{O}_2+8H^{+}\to 2Mn{O}_4^{-}+5P{b}^{2+}+4H_{2}O.$

(b)

Interpretation Introduction

Interpretation:

The following reaction that occurs in acidic solution has to be balanced.

    $A{s}_2{S}_5+N{O}_3^{-}\to H_{3}As{O}_4+HS{O}_4^{-}+N{O}_2$

Concept introduction:

Refer to part (a).

Explanation of Solution

The given reaction can be written as two half reactions as follows,

The oxidation half reaction is $A{s}_2{S}_5\to H_{3}As{O}_4+HS{O}_4^{-}.$

The reduction half reaction is $N{O}_3^{-}\to N{O}_2.$

Each 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.

    $A{s}_2{S}_5\to 2H_{3}As{O}_4+5HS{O}_4^{-}.$

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,

    $A{s}_2{S}_5+28H_{2}O\to 2H_{3}As{O}_4+5HS{O}_4^{-}$

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.

    $A{s}_2{S}_5+28H_{2}O\to 2H_{3}As{O}_4+5HS{O}_4^{-}+45H^{+}$

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,

    $A{s}_2{S}_5+28H_{2}O\to 2H_{3}As{O}_4+5HS{O}_4^{-}+45H^{+}+40e^{-}$

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

    $N{O}_3^{-}\to N{O}_2$

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,

    $N{O}_3^{-}\to N{O}_2+H_{2}O$

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.

    $2H^{+}+N{O}_3^{-}\to N{O}_2+H_{2}O$

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,

    $2H^{+}+N{O}_3^{-}+1e^{-}\to N{O}_2+H_{2}O$

The overall reaction can be balanced as follows,

    $\frac{\begin{array}{l}A{s}_2{S}_5+28H_{2}O\to 2H_{3}As{O}_4+5HS{O}_4^{-}+45H^{+}+40e^{-}\\ \\ 2H^{+}+N{O}_3^{-}+1e^{-}\to N{O}_2+H_{2}O]\times 40\end{array}}{35H^{+}+40N{O}_3^{-}+A{s}_2{S}_5\to 40N{O}_2+2H_{3}As{O}_4+12H_{2}O+5HS{O}_4^{-}}$

The balanced reaction is $35H^{+}+40N{O}_3^{-}+A{s}_2{S}_5\to 40N{O}_2+2H_{3}As{O}_4+12H_{2}O+5HS{O}_4^{-}.$