Chapter 20, Problem 81A

(a)

Interpretation Introduction

Interpretation: The given half-reactions, $H{g}^{2+}+2e^{-}\to Hg$ , $Al\to A{l}^{3+}+3e^{-}$ is to be combined into a balanced one.

Concept Introduction: The reactions that involve oxidation and reduction are separated with the help of half-reactions. The combination and cancellation of common substances from half reactions will make a balanced chemical equation.

Answer to Problem 81A

The balanced equation is $3HgO+2Al\to A{l}_2{O}_3+3Hg$ .

Explanation of Solution

The half-reactions,

  $H{g}^{2+}+2e^{-}\to Hg$

  $Al\to A{l}^{3+}+3e^{-}$

The $H{g}^{2+}$ is $HgO$ and $A{l}^{3+}$ is present in $A{l}_2{O}_3$ . Multiplying the reduction equation of $H{g}^{2+}$ by 3 and the oxidation equation of $A{l}^{3+}$ by 2 will balance the electron count. Then the common species cancellation and balancing the oxygen count will give the reaction as,

  $3HgO+2Al\to A{l}_2{O}_3+3Hg$

(b)

Interpretation Introduction

Interpretation: The given half-reactions, $Mn{O}_2{}^{+}+4H^{+}+2e^{-}\to M{n}^{2+}+2H_{2}O$ , $Fe\to F{e}^{2+}+2e^{-}$ is to be combined into a balanced one.

Concept Introduction: The reactions that involve oxidation and reduction are separated with the help of half-reactions. The combination and cancellation of common substances from half reactions will make a balanced chemical equation.

Answer to Problem 81A

The balanced equation is $Mn{O}_2+Fe\to MnO+FeO$ .

Explanation of Solution

The half-reactions,

  $Mn{O}_2{}^{+}+4H^{+}+2e^{-}\to M{n}^{2+}+2H_{2}O$

  $Fe\to F{e}^{2+}+2e^{-}$

The $F{e}^{2+}$ is $FeO$ and $M{n}^{2+}$ is $MnO$ . The electron count is balanced. Then the common species cancellation and balancing the oxygen count by adding H2O will give the reaction as,

  $Mn{O}_2+Fe\to MnO+FeO$

(c)

Interpretation Introduction

Interpretation: The given half-reactions, $F{e}^{3+}+e^{-}\to F{e}^{2+}$ , $Cd\to C{d}^{2+}+2e^{-}$ is to be combined into a balanced one.

Concept Introduction: The reactions that involve oxidation and reduction are separated with the help of half-reactions. The combination and cancellation of common substances from half reactions will make a balanced chemical equation.

Answer to Problem 81A

The balanced equation is $F{e}_2{O}_3+Cd\to CdO+2FeO$ .

Explanation of Solution

The half-reactions,

  $F{e}^{3+}+e^{-}\to F{e}^{2+}$

  $Cd\to C{d}^{2+}+2e^{-}$

  $F{e}^{3+}$ in the given reaction is $F{e}_2{O}_3$ , $C{d}^{2+}$ is CdO and $F{e}^{2+}$ is FeO. The equation can be balanced by adding 2 to FeO on the product side. The balanced chemical equation is,

  $F{e}_2{O}_3+Cd\to CdO+2FeO$