Chapter 7.3, Problem 7.20QAP

Interpretation Introduction

Interpretation:

To identify each of the following as a oxidation or a reduction.

a. ${\text{O}}_{\text{2}}{\text{(g) + 4e}}^{\text{-}}\to {\text{2O}}^{\text{-2}}\text{(aq)}$

b. $\text{Ag(s) }\to {\text{Ag}}^{\text{+}}{\text{(aq) + e}}^{\text{-}}$

c. ${\text{Fe}}^{\text{3+}}{\text{(aq) + e}}^{\text{-}}\to {\text{Fe}}^{\text{2+}}\text{(aq)}$

d. ${\text{2Br}}^{\text{-}}\text{(aq)}\to {\text{Br}}_{\text{2}}{\text{(l) + 2e}}^{\text{-}}$

Concept introduction:

Oxidation occurs when an atom’s oxidation state increases during a reaction. Or loss of electrons

Reduction occurs when an atom’s oxidation state decrease during a reaction. Or gain of electrons.

Oxidation rules:

1. Oxidation state of an element (in its pure form) is 0.

2. The oxidation state of representatives is their ionic charges.

3. Oxidation state of oxygen is always -2.The peroxide ion is always -1.

4. H with nonmetals (covalent compounds) is given +1oxidation state. Usually in the front of the formula.

5. The sum of the oxidation numbers must be zero for compounds and whatever the charge of the ionic species is.

_· Oxidation state of “C” in CH₄

$\begin{array}{l}\text{X + 4(+1) = 0}\\ \text{X = -4}\end{array}$

• Oxidation state of diatomic atoms is zero.

Therefore, O₂ oxidation state is zero.

• Oxidation state of “C” in CO₂.

$\begin{array}{l}\text{X + 2(-2) = 0}\\ \text{X = +4}\end{array}$

• Oxidation state of “O” in H₂O.

$\begin{array}{l}\text{2(+1)+ X = 0}\\ \text{X = -2}\end{array}$