Chapter 20, Problem 55A

(a)

Interpretation Introduction

Interpretation: The chemical equation that is formed by the heating of tungsten oxide with hydrogen, $W{O}_3\left(aq\right)+H_2\to W\left(s\right)+3H_{2}O\left(aq\right)$ 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 55A

The balanced equation is, $W{O}_3\left(aq\right)+3H_2\to W\left(s\right)+3H_{2}O\left(aq\right)$ .

Explanation of Solution

The given equation is a redox reaction, $W{O}_3\left(aq\right)+H_2\to W\left(s\right)+3H_{2}O\left(aq\right)$

The half-reactions involved in the reaction are,

Oxidation: $H_2\to H_{2}O+3e^{-}$

Reduction: $W^{6+}+6e^{-}\to W$

The electron count can be balanced by multiplying the oxidation equation by 6 and the reduction equation by 2. Therefore the equation becomes,

  $2W{O}_3\left(aq\right)+6H_2\to 2W\left(s\right)+6H_{2}O\left(aq\right)$

The equation can again be simplified to,

  $W{O}_3\left(aq\right)+3H_2\to W\left(s\right)+3H_{2}O\left(aq\right)$

(b)

Interpretation Introduction

Interpretation: The reducing agent in the reaction, c is to be determined.

Concept Introduction: The reagents that undergo oxidation is identified as reducing agent and the reagent that undergo reduction is identified as oxidizing agent.

Answer to Problem 55A

The reducing agent is $H_2$ .

Explanation of Solution

In the reaction,

  $W{O}_3\left(aq\right)+H_2\to W\left(s\right)+3H_{2}O\left(aq\right)$

The half-reactions are,

Oxidation: $H_2\to H_{2}O+3e^{-}$

Reduction: $W^{6+}+6e^{-}\to W$

So, the reducing agent is $H_2$ of which oxidation is happening.

(c)

Interpretation Introduction

Interpretation: The element that is undergoing an increase in oxidation in the given reaction, $W{O}_3\left(aq\right)+H_2\to W\left(s\right)+3H_{2}O\left(aq\right)$ is to be determined.

Concept Introduction: The oxidation number of chemical species is found to increase only when it undergoes an oxidation process.

Answer to Problem 55A

The oxidation number is increasing for $H$ .

Explanation of Solution

In the reaction,

  $W{O}_3\left(aq\right)+H_2\to W\left(s\right)+3H_{2}O\left(aq\right)$

The half-reactions are,

Oxidation: $H_2\to H_{2}O+3e^{-}$

Reduction: $W^{6+}+6e^{-}\to W$

Since oxidation is happening to $H_2$ , the oxidation number is increasing from zero to +1 by the loss of electrons.

So, the oxidation number is increasing in $H$ .