Chapter 20, Problem 54A

(a)

Interpretation Introduction

Interpretation: The given reaction of ethanol and dichromate, $C{r}_2{O}_7{}^{-2}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)\to C{r}^{3+}\left(aq\right)+C{O}_2\left(g\right)$ is to be balanced by the half-reaction method.

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.
• In an acidic medium, $H^{+}$ can be used to balance hydrogen count. And to balance oxygen count $H_{2}O$ can be used.

Answer to Problem 54A

The balanced chemical equation is $2C{r}_2{O}_7{}^{-2}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)+16H^{+}\to 4C{r}^{3+}\left(aq\right)+2C{O}_2\left(g\right)+11H_{2}O$

Explanation of Solution

In the given reaction,

  $C{r}_2{O}_7{}^{-2}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)\to C{r}^{3+}\left(aq\right)+C{O}_2\left(g\right)$

The half-reactions involved are,

Oxidation: $C_2{H}_{5}OH\to C{O}_2$

Reduction: $C{r}_2{O}_7{}^{-2}\to C{r}^{3+}$

Now, the other species and then oxygen and hydrogen count in the half-reactions can be balanced by the corresponding addition of them. That is,

  $C{r}_2{O}_7{}^{-2}+14H^{+}\to 2C{r}^{3+}+7H_{2}O$

  $C_2{H}_{5}OH+3H_{2}O\to 2C{O}_2\left(g\right)+12H^{+}$

The electron count in the above equation can be shown as,

  $C{r}_2{O}_7{}^{-2}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O$

The electron count can be balanced by multiplying the reduction equation by 2.

  $C_2{H}_{5}OH+3H_{2}O\to 2C{O}_2\left(g\right)+12H^{+}+12e^{-}$

  $2C{r}_2{O}_7{}^{-2}+28H^{+}+12e^{-}\to 4C{r}^{3+}+14H_{2}O$

The half-reactions can then be added together to get the equation,

  $\begin{array}{c}2C{r}_2{O}_7{}^{-2}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)+3H_{2}O+28H^{+}+12e^{-}\to 4C{r}^{3+}\left(aq\right)+2C{O}_2\left(g\right)+\\ 11H_{2}O+12H^{+}+12e^{-}\end{array}$

Cancellation of common species will simply be the equation as,

  $2C{r}_2{O}_7{}^{-2}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)+16H^{+}\to 4C{r}^{3+}\left(aq\right)+2C{O}_2\left(g\right)+11H_{2}O$

(b)

Interpretation Introduction

Interpretation: To interpret whether dichromium ion is the oxidizing or reducing agent.

Concept Introduction: The chemical species that undergo reduction, the gain of electrons are identified to be oxidizing agents while that undergo oxidation, loss of electrons are identified to be reducing agents.

Answer to Problem 54A

The dichromate ions are oxidizing agents.

Explanation of Solution

In the reaction,

  $C{r}_2{O}_7{}^{-2}\left(aq\right)+C_2{H}_{5}OH\left(aq\right)\to C{r}^{3+}\left(aq\right)+C{O}_2\left(g\right)$

The $C{r}^{6+}$ in $C{r}_2{O}_7{}^{-2}$ gains 3 electrons to form $C{r}^{3+}$ . Since it is undergoing reduction by the gaining of electrons it is an oxidizing agent.