Chapter 20, Problem 34A

(a)

Interpretation Introduction

Interpretation: The given reaction, $Ba\left(s\right)+O_2\left(g\right)\to BaO\left(s\right)$ is to be balanced whether the first substance is being oxidized or reduced is to be identified.

Concept Introduction: The reaction that involves the loss of electrons or the addition of oxygen is oxidation. While that involves the gain of electrons or removal of oxygen is reduction.

Answer to Problem 34A

The balanced chemical equation is,

  $2Ba\left(s\right)+O_2\left(g\right)\to 2BaO\left(s\right)$

And, the first element, $Ba$ is undergoing oxidation.

Explanation of Solution

The given reaction,

  $Ba\left(s\right)+O_2\left(g\right)\to BaO\left(s\right)$

It is a redox reaction that involves the oxidation and reduction process. It can be represented as,

  $Ba\to B{a}^{2+}\left(g\right)+2e^{-}$

  $O_2+2e^{-}\to 2O^{-}$

The reaction is not balanced since the number of oxygen is not balanced on the product side. Oxygen can be balanced by adding two to $BaO$ . So it is necessary to add two to $Ba$ too.

Therefore the balanced equation is,

  $2Ba\left(s\right)+O_2\left(g\right)\to 2BaO\left(s\right)$

The first substance in the reaction is barium, $Ba$ , and is oxidized.

(b)

Interpretation Introduction

Interpretation: The given reaction, $CuO\left(s\right)+H_2\left(g\right)\to Cu\left(s\right)+H_{2}O\left(l\right)$ is to be balanced and whether the first substance is being oxidized or reduced is to be identified.

Concept Introduction: The reaction that involves the loss of electrons or the addition of oxygen is oxidation. While that involves the gain of electrons or removal of oxygen is reduction.

Answer to Problem 34A

The balanced chemical equation is $CuO\left(s\right)+H_2\left(g\right)\to Cu\left(s\right)+H_{2}O\left(l\right)$

And, the first substance $CuO$ is undergoing reduction.

Explanation of Solution

The given reaction, $CuO\left(s\right)+H_2\left(g\right)\to Cu\left(s\right)+H_{2}O\left(l\right)$

It is a redox reaction that involves both oxidation and reduction. It can be represented as,

  $C{u}^{2+}+2e^{-}\to Cu$

  $H_2\to 2H^{+}+2e^{-}$

The reaction is a balanced equation since the number of copper, oxygen, and hydrogen is equal on both sides.

The first substance, $C{u}^{2+}$ in $CuO$ is undergoing reduction since it involves the loss of electrons.

(c)

Interpretation Introduction

Interpretation: The given reaction, $C_2{H}_4\left(g\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(l\right)$ is to be balanced and to identify whether the first substance is being oxidized or reduced.

Concept Introduction: The reaction that involves the loss of electrons or the addition of oxygen is oxidation. While that involves the gain of electrons or removal of oxygen is reduction.

Answer to Problem 34A

The balanced chemical equation is, $C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(l\right)$ .

And the first substance $C_2{H}_4$ underwent oxidation.

Explanation of Solution

The given reaction,

  $C_2{H}_4\left(g\right)+O_2\left(g\right)\to C{O}_2\left(g\right)+H_{2}O\left(l\right)$

By following the oxidation number method the reduced and oxidized can be found. The oxidation number of carbon in $C_2{H}_4$ is $-2$ . And in $C{O}_2$ is +4. So carbon is said to be oxidized.

The oxidation number of $O_2$ is zero while in $H_{2}O$ is -2 so it is said to be reduced by accepting the two electrons donated by $C_2{H}_4$ . The difference in electron count is six upon changing the oxidation number from $-2$ to +4. Now, the electron count can be balanced by adding three to $O_2$ and two to $C{O}_2$ and $H_{2}O$ .

So, the balanced equation is,

  $C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(l\right)$

The first species, $C^{2-}$ in $C_2{H}_4$ is undergoing oxidation by the loss of electrons.

(d)

Interpretation Introduction

Interpretation: The given reaction, $CaO\left(s\right)+Al\left(s\right)\to A{l}_2{O}_3\left(s\right)+Ca\left(s\right)$ is to be balanced and whether the first substance is being oxidized or reduced is to be identified.

Concept Introduction: The reaction that involves the loss of electrons or the addition of oxygen is oxidation. While that involves the gain of electrons or removal of oxygen is reduction.

Answer to Problem 34A

The balanced chemical equation is $3CaO\left(s\right)+2Al\left(s\right)\to A{l}_2{O}_3\left(s\right)+\text{ 3}Ca\left(s\right)$ .

And, the first substance, $CaO$ undergo reduction.

Explanation of Solution

The given reaction,

  $CaO\left(s\right)+Al\left(s\right)\to A{l}_2{O}_3\left(s\right)+Ca\left(s\right)$

It is an example of a redox reaction. The oxidation number of aluminum is increasing from zero to +3 by the loss of three electrons. While the oxidation number of Ca is decreasing from +2 to 0 by accepting electrons. The electron count in the reaction can be balanced by adding two to aluminum and three to calcium.

The balanced half-reactions are,

  $2Al\to 2A{l}^{3+}+6e^{-}$

  $3C{a}^{2+}+6e^{-}\to 3Ca$

So, the overall balanced equation is, a $3CaO\left(s\right)+2Al\left(s\right)\to A{l}_2{O}_3\left(s\right)+\text{ 3}Ca\left(s\right)$