Chapter 20, Problem 70A

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation of the redox reaction between rubidium and iodine is to be found and the oxidizing agent is to be identified.

Concept Introduction: The reaction which involves both oxidation and reduction are redox reaction. The chemical species that undergo reduction and gain of electrons are oxidizing agents.

Answer to Problem 70A

The balanced equation is,

  $2Rb+I_2\to 2RbI$

And, $I_2$ is the oxidizing agent.

Explanation of Solution

The reaction of rubidium, $Rb$ , and iodine, $I_2$ results in the formation of rubidium iodide, $RbI$ . The chemical equation can therefore be written as,

  $2Rb+I_2\to 2RbI$

So, here rubidium is giving the electrons present on them to iodine.

  $Rb\to R{b}^{+}+1e^{-}$

And $I_2$ accepts the electron as,

  $I_2+e^{-}\to 2I^{-}$

So, the reduction is happening to $I_2$ and so it is the oxidizing agent.

(b)

Interpretation Introduction

Interpretation: The balanced chemical equation of the redox reaction between barium and water is to be found and the oxidizing agent is to be identified.

Concept Introduction: The reaction which involves both oxidation and reduction are redox reaction. The chemical species that undergo reduction and gain of electrons are oxidizing agents.

Answer to Problem 70A

The balanced equation is,

  $Ba+2H_{2}O\to Ba{\left(OH\right)}_2+H_2$

And, $H_{2}O$ is the oxidizing agent.

Explanation of Solution

The reaction of barium, $Ba$ , and water, $H_{2}O$ results in the formation of Barium hydroxide, $Ba{\left(OH\right)}_2$ , and hydrogen gas. The chemical equation can therefore be written as,

  $Ba+2H_{2}O\to Ba{\left(OH\right)}_2+H_2$

So here barium is giving the electrons present on them to H.

  $Ba\to B{a}^{2+}+2e^{-}$

And, $H_{2}O$ accepts the electron as,

  $H^{+}+e^{-}\to H_2$

So reduction is happening to $H^{+}$ in water and so $H_{2}O$ is the oxidizing agent.

(c)

Interpretation Introduction

Interpretation: The balanced chemical equation of the redox reaction between aluminum and iron (II) sulphate is to be found and the oxidizing agent is to be identified.

Concept Introduction: The reaction which involves both oxidation and reduction are redox reaction. The chemical species that undergo the reduction and gain of electrons are oxidizing agents.

Answer to Problem 70A

The balanced equation is,

  $2Al+3FeS{O}_4\to A{l}_2{\left(S{O}_4\right)}_3+3Fe$

And, $FeS{O}_4$ is the oxidizing agent.

Explanation of Solution

The reaction of aluminum, Al, and iron (III) sulphate, $FeS{O}_4$ results in the formation of aluminum sulphate, $A{l}_2{\left(S{O}_4\right)}_3$ . The chemical equation can therefore be written as,

  $2Al+3FeS{O}_4\to A{l}_2{\left(S{O}_4\right)}_3+3Fe$

So here, aluminum is giving the electrons present on them to $F{e}^{2+}$ .

  $Al\to A{l}^{3+}+3e^{-}$

And $F{e}^{2+}$ accepts the electron as,

  $F{e}^{2+}+e^{-}\to Fe$

So, the reduction is happening to $F{e}^{2+}$ in $FeS{O}_4$ and so, it is the oxidizing agent.

(d)

Interpretation Introduction

Interpretation: The balanced chemical equation of the redox reaction between butene and oxygen is to be found and the oxidizing agent is to be identified.

Concept Introduction: The reaction which involves both oxidation and reduction are redox reaction. The chemical species that undergo the reduction and gain of electrons are oxidizing agents.

Answer to Problem 70A

The balanced equation is,

  $C_4{H}_8+6O_2\to 4C{O}_2+4H_{2}O$

And, $O_2$ is the oxidizing agent.

Explanation of Solution

The reaction of butene, $C_4{H}_8$ , and oxygen, $O_2$ results in the formation of carbon dioxide and water. The chemical equation can therefore be written as,

  $C_4{H}_8+6O_2\to 4C{O}_2+4H_{2}O$

So here, butene is giving the electrons present on them to oxygen.

  $C_4{H}_8\to C^{4+}+4e^{-}$

And, $O_2$ accepts the electron as,

  $O_2+2e^{-}\to O^{2-}$

So, the reduction is happening to $O_2$ and so, it is the oxidizing agent.

(e)

Interpretation Introduction

Interpretation: The balanced chemical equation of the redox reaction between zinc and hydrobromic acid is to be found and the oxidizing agent is to be identified.

Concept Introduction: The reaction which involves both oxidation and reduction are redox reaction. The chemical species that undergo the reduction and gain of electrons are oxidizing agents.

Answer to Problem 70A

The balanced equation is,

  $Zn+2HBr\to ZnB{r}_2+H_2$

And, $HBr$ is the oxidizing agent.

Explanation of Solution

The reaction of zinc, Zn, and hydrobromic acid, $HBr$ results in the formation of zinc bromide, $ZnB{r}_2$ . The chemical equation can therefore be written as,

  $Zn+2HBr\to ZnB{r}_2+H_2$

So here, zinc is giving the electrons present on them to $H^{+}$ in $HBr$ .

  $Zn\to Z{n}^{2+}+2e^{-}$

And, $H^{+}$ accepts the electron as,

  $2H^{+}+2e^{-}\to H_2$

So, the reduction is happening to $H$ and so, it is the oxidizing agent.

(f)

Interpretation Introduction

Interpretation: The balanced chemical equation of the redox reaction between magnesium and bromine is to be found and the oxidizing agent is to be identified.

Concept Introduction: The reaction which involves both oxidation and reduction are redox reaction. The chemical species that undergo the reduction and gain of electrons are oxidizing agents.

Answer to Problem 70A

The balanced equation is,

  $2Mg+B{r}_2\to 2MgBr$

And, $B{r}_2$ is the oxidizing agent.

Explanation of Solution

The reaction of Magnesium, Mg, and bromine, $B{r}_2$ results in the formation of magnesium bromide, $MgBr$ . The chemical equation can therefore be written as,

  $2Mg+B{r}_2\to 2MgBr$

So here, magnesium is giving the electrons present in them to bromine.

  $Mg\to M{g}^{2+}+2e^{-}$

And, $B{r}_2$ accepts the electron as,

  $B{r}_2+2e^{-}\to 2B{r}^{-}$

So, the reduction is happening to $B{r}_2$ and so, it is the oxidizing agent.