Chapter 20, Problem 46A

(a)

Interpretation Introduction

Interpretation: The reaction $CuS\left(s\right)+N{O}_3{}^{-}\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+N{O}_2\left(g\right)+S{O}_2\left(g\right)$ is to be balanced.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.

Answer to Problem 46A

The balanced chemical equation of the reaction is, $CuS\left(s\right)+8N{O}_3{}^{-}\left(aq\right){\text{+8H}}^{+}\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+6N{O}_2\left(g\right)+S{O}_2\left(g\right)+4H_{2}O$ .

Explanation of Solution

In the given reaction,

  $CuS\left(s\right)+N{O}_3{}^{-}\left(aq\right)\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+N{O}_2\left(g\right)+S{O}_2\left(g\right)$

Sulfur, $S^{2-}$ in $CuS$ is undergoing oxidation by the loss of six electrons present on it and forms $S{O}_2$ . And N in $N{O}_3{}^{-}$ is undergoing reduction by the acceptance of electrons and forms NO2. The half-reaction can be represented as,

Oxidation: $S^{2-}\to S^{4+}+6e^{-}$

Reduction: $N^{5+}+1e^{-}\to N^{4+}$

The number of oxygen can be balanced by the addition of 4 water molecules. Now the number of H in the reactant side can be balanced by the addition of ${\text{8H}}^{+}$ . The reaction can be balanced by the addition of 8 to $N{O}_3{}^{-}$ . Then the number of nitrogen in the product side can be balanced by adding 6 to $\left(N{O}_2\right)$ . Thus, the entire chemical equation becomes balanced.

The balanced equation is,

  $CuS\left(s\right)+8N{O}_3{}^{-}\left(aq\right){\text{+8H}}^{+}\to Cu{\left(N{O}_3\right)}_2\left(aq\right)+6N{O}_2\left(g\right)+S{O}_2\left(g\right)+4H_{2}O$

(b)

Interpretation Introduction

Interpretation: The reaction $I^{-}\left(aq\right)+N{O}_3{}^{-}\left(aq\right)\to I_2\left(aq\right)+NO\left(g\right)$ is to be balanced.

Concept Introduction: The chemical reaction that involves both oxidation and reduction is the redox reaction. Redox reactions can be balanced by following the half-reaction method. That is,

• First, the given redox reaction is divided into half reactions one oxidation and the other reduction.
• Based on mass and charge the half-reactions are balanced.
• The electrons on both sides of the reaction are equalized.
• Then add the half-reactions together and the common one is eliminated to get the balanced equation.

Answer to Problem 46A

The balanced chemical equation of the reaction is, $6I^{-}\left(aq\right)+3N{O}_3{}^{-}\left(aq\right){\text{ +12H}}^{+}+12e^{-}\to 3I_2\left(aq\right)+3NO\left(g\right)+6H_{2}O$

Explanation of Solution

In the given reaction,

  $I^{-}\left(aq\right)+N{O}_3{}^{-}\left(aq\right)\to I_2\left(aq\right)+NO\left(g\right)$

Copper, $I^{-}$ changes its oxidation state from -1 to zero in $I_2$ by the loss of electrons. And the $N^{5+}$ in $N{O}_3{}^{-}$ changes its oxidation state from +5 to +2 in NO. The half-reaction can be represented as,

Oxidation: $2I^{-}\to I^{2+}+2e^{-}$

Reduction: $N^{5+}+3e^{-}\to N^{2+}$

Multiplying the oxidation equation by 3 and the reduction equation by 2 will make the electron count balanced. Nitrogen count is also balanced. The oxygen count can be balanced by the addition of 6 molecules of water on the product side. And in the reactant side corresponding $H^{+}$ and electron must be added to make the reaction balanced.

Therefore the balanced chemical equation is,

  $6I^{-}\left(aq\right)+3N{O}_3{}^{-}\left(aq\right){\text{ +12H}}^{+}+12e^{-}\to 3I_2\left(aq\right)+3NO\left(g\right)+6H_{2}O$