Chapter 20, Problem 56A

(a)

Interpretation Introduction

Interpretation: Whether silver in the reaction is oxidized or reduced is to be determined.

Concept Introduction: The species that involves the loss of electrons are said to be oxidized while the one that involves the gain of electrons is said to be reduced.

Answer to Problem 56A

Silver is oxidized in the reaction.

Explanation of Solution

In the reaction,

  $Ag+H_{2}S\to A{g}_{2}S+H_2$

The half-reactions involved are,

Oxidation: $Ag\to A{g}_{2}S+1e^{-}$

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

The removal of electrons is taking place in $Ag$ , silver. So, $Ag$ is said to be oxidized.

(b)

Interpretation Introduction

Interpretation: The oxidizing and reducing agents in the given reaction, $Ag+H_{2}S\to A{g}_{2}S+H_2$ is to be determined.

Concept Introduction: The species that undergo reduction is identified as an oxidizing agent while the one that undergoes oxidation is identified as a reducing agent.

Answer to Problem 56A

Reducing agent is $Ag$ and oxidizing agent is $H_{2}S$ .

Explanation of Solution

In the reaction,

  $Ag+H_{2}S\to A{g}_{2}S+H_2$

The half-reactions involved are,

Oxidation: $Ag\to A{g}_{2}S+1e^{-}$

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

The removal of electrons is taking place in $Ag$ , silver. So $Ag$ is said to be oxidized while $H^{+}$ in the $H_{2}O$ accepts electron and form $H_2$ by the reduction process.

So, the reducing agent is $Ag$ and oxidizing agent is $H_{2}S$ .

(c)

Interpretation Introduction

Interpretation: The given chemical equation, $Ag+H_{2}S\to A{g}_{2}S+H_2$ 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 56A

The balanced chemical equation is,

  $2Ag+H_{2}S\to A{g}_{2}S+H_2$

Explanation of Solution

In the reaction,

  $Ag+H_{2}S\to A{g}_{2}S+H_2$

The half-reactions involved are,

Oxidation: $Ag\to A{g}_{2}S+1e^{-}$

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

The half-reaction can be balanced first by multiplying 2 by Ag.

So the equation becomes,

  $2Ag+H_{2}S\to A{g}_{2}S+H_2$

All the other chemical species are equal. So the balanced equation is,

  $2Ag+H_{2}S\to A{g}_{2}S+H_2$