Chapter 8, Problem 8.96QA

Interpretation Introduction

To write:

The complete and balanced redox reaction of

a) (basic solution)

b) (acidic solution)

c) $I_{2\left(s\right)}+S_2{O_3^{2-}}_{\left(aq\right)} \to I_{\left(aq\right)}^{-}+ S_4{O_6^{2-}}_{\left(aq\right)}$ (neutral solution)

Answer to Problem 8.96QA

Solution:

 Complete and balanced redox reaction is

a) $2M{n}_{\left(aq\right)}^{2+}+O_{2\left(g\right)}+4O{H}_{\left(aq\right)}^{-} \to 2 Mn{O}_{2\left(s\right)}+2H_2{O}_{\left(l\right)}$

b) $Mn{O}_{2\left(s\right) }+{2I}_{\left(aq\right)}^{-}+4H_{\left(aq\right)}^{+}\to M{n}_{\left(aq\right)}^{2+}+I_{2\left(g\right)}+2H_2{O}_{\left(l\right)}$

c) $I_{2\left(s\right)}+{2S}_2{O_3^{2-}}_{\left(aq\right)} \to 2I_{\left(aq\right)}^{-}+ S_4{O_6^{2-}}_{\left(aq\right)}$

Explanation of Solution

1) Concept:

To balance a redox reaction, first we have to assign oxidation numbers to each of the elements in the compound and decide which element is going to be reduced and oxidized. Oxidation is loss of electrons or increase in oxidation number while reduction is gain of electrons or decrease in oxidation number.

2) Given:

i) $M{n}_{\left(aq\right)}^{2+}+O_{2\left(g\right)} \to Mn{O}_{2\left(s\right)}$ (basic solution)

ii) (acidic solution)

iii) $I_{2\left(s\right)}+S_2{O_3^{2-}}_{\left(aq\right)} \to I_{\left(aq\right)}^{-}+ S_4{O_6^{2-}}_{\left(aq\right)}$ (neutral solution)

3) Calculations:

a) $M{n}_{\left(aq\right)}^{2+}+O_{2\left(g\right)} \to Mn{O}_{2\left(s\right)}$

i) The following preliminary expression relates the known reactant and products:

$M{n}_{\left(aq\right)}^{2+}+O_{2\left(g\right)} \to Mn{O}_{2\left(s\right)}$

There are equal number of $Mn$ and $O_2$ on both product and reactant sides, so it is already balanced.

ii) We perform an $∆O.N.$ analysis for manganese and oxygen.

$M{n}_{\left(aq\right)}^{2+}+O_{2\left(g\right)} \to Mn{O}_{2\left(s\right)}$

As $O$ changes oxidation number from $0 to-2$ in the reaction, $∆O.N.$ for $O$ is $2\left(-2\right)=-4$

$Mn$ changes oxidation number from $+2 to+4$ in the reaction, so $∆O.N.$ for $Mn$ is $+2$.

iii) To balance these $∆O.N$ values, we need a coefficient of 2 in front of $M{n}_{\left(aq\right)}^{2+}$. This also requires a 2 in front of $Mn{O}_{2\left(s\right)}$ to balance the number of $Mn$ atoms.

$2 M{n}_{\left(aq\right)}^{2+}+O_{2\left(g\right)} \to 2 Mn{O}_{2\left(s\right)}$

iv) The ionic charge on the reactant is $+4,$ and on the product, it is zero. The reaction is carried out in a basic solution, we need to add $4O{H}^{-}$ on the reactant side to balance the charge on both sides.

$2 M{n}_{\left(aq\right)}^{2+}+O_{2\left(g\right)}+4O{H}^{-} \to 2Mn{O}_{2\left(s\right)}$

v) Taking an inventory of atoms on both sides of the reaction arrow, we find that the left side has four more $H$ atoms and two more $O$ atoms. We correct this imbalance by adding two more molecules of $H_{2}O$ to the right side of the reaction:

$2M{n}_{\left(aq\right)}^{2+}+O_{2\left(g\right)}+4O{H}_{\left(aq\right)}^{-} \to 2Mn{O}_{2\left(s\right)}+2H_2{O}_{\left(l\right)}$

Now the reaction is balanced.

b) $Mn{O}_{2\left(s\right)}+I_{\left(aq\right)}^{-} \to M{n}_{\left(aq\right)}^{2+}+I_{2\left(s\right)}$

i) The following preliminary expression relates the known reactant and products:

$Mn{O}_{2\left(s\right) }+I_{\left(aq\right)}^{-}\to M{n}_{\left(aq\right)}^{2+}+I_{2\left(g\right)}$

There are equal numbers of $Mn$on both product and reactant sides, but there are unequal numbers of $I$ in the reaction, so, to balance it, we have to add the coefficient 2 to the $I_{aq}^{-}$ present in the reactant side.

$Mn{O}_{2\left(s\right) }+{2I}_{\left(aq\right)}^{-}\to M{n}_{\left(aq\right)}^{2+}+I_{2\left(g\right)}$

ii) We perform an $∆O.N.$ analysis for manganese and oxygen.

$Mn{O}_{2\left(s\right) }+{2 I}_{\left(aq\right)}^{-}\to M{n}_{\left(aq\right)}^{2+}+I_{2\left(g\right)}$

The $I$ changes oxidation number from $-1 to 0$ in the reaction, and there are two $I$ in the reaction, so $∆O.N.$ for $I$ is $+2$.

$Mn$ changes the oxidation number from $+4 to+2$ in the reaction, so $∆O.N.$ for $Mn$ is $2-4= -2$.

The change in the oxidation number is the same for oxidation and reduction.

iii) The ionic charge on the reactant is $-2,$ and on the product, it is $+2.$ As the reaction is carried out in an acidic solution, we need to add $4H^{+}$ on the reactant side to balance the charge on both sides.

$Mn{O}_{2\left(s\right) }+{2I}_{\left(aq\right)}^{-}+4H_{\left(aq\right)}^{+}\to M{n}_{\left(aq\right)}^{2+}+I_{2\left(g\right)}$

iv) Taking an inventory of atoms on both sides of the reaction arrow, we find that the left side has four more $H$ atoms and two more $O$ atoms. We correct this imbalance by adding two more molecules of $H_{2}O$ to the right side of the reaction:

$Mn{O}_{2\left(s\right) }+{2I}_{\left(aq\right)}^{-}+4H_{\left(aq\right)}^{+}\to M{n}_{\left(aq\right)}^{2+}+I_{2\left(g\right)}+2H_2{O}_{\left(l\right)}$

Now the reaction is balanced

c) $I_{2\left(s\right)}+S_2{O_3^{2-}}_{\left(aq\right)} \to I_{\left(aq\right)}^{-}+ S_4{O_6^{2-}}_{\left(aq\right)}$

i) The following preliminary expression relates the known reactant and products:

$I_{2\left(s\right)}+S_2{O_3^{2-}}_{\left(aq\right)} \to I_{\left(aq\right)}^{-}+ S_4{O_6^{2-}}_{\left(aq\right)}$

There are unequal numbers of $I$, $O,$ and $S$ in the reaction, so, to balance it, we have to add coefficient 2 to the $I_{aq}^{-}$ present in the product side and add coefficient 2 to the $S_2{O_3^{2-}}_{\left(aq\right)}$ on reactant side, so the chemical reaction is

$I_{2\left(s\right)}+{2S}_2{O_3^{2-}}_{\left(aq\right)} \to 2I_{\left(aq\right)}^{-}+ S_4{O_6^{2-}}_{\left(aq\right)}$

ii) We perform an $∆O.N.$ analysis for iodine and sulphur.

$I_{2\left(s\right)}+{2 S}_2{O_3^{2-}}_{\left(aq\right)} \to 2I_{\left(aq\right)}^{-}+ S_4{O_6^{2-}}_{\left(aq\right)}$

The $I$ changes oxidation number from $0 to-2$ in the reaction, so $∆O.N.$ for $I$ is $-2-0= -2$.

The $S$ changes oxidation number from $+2 to+2.5$ in the reaction, and there are four $S$ in the reaction, so $∆O.N.$ for $S$ is $10-8= +2$.

The change in the oxidation number is the same for oxidation and reduction.

Also there are same number of charges on the reactant and product sides, so the reaction is balanced.

$I_{2\left(s\right)}+{2 S}_2{O_3^{2-}}_{\left(aq\right)} \to 2I_{\left(aq\right)}^{-}+ S_4{O_6^{2-}}_{\left(aq\right)}$

Conclusion:

In a redox reaction, substances either gain electrons or thereby undergo reduction, or they lose electrons and undergo oxidation. The given reaction is redox as the oxidation number of elements in the reactants changes during the reaction.