Chapter U5.17, Problem 1E

Interpretation Introduction

Interpretation: The element being oxidized and reduced on reacting copper and oxygen to form copper (II) oxide to should be determined.

Concept Introduction: The oxidation state is defined as the charge(s) that an atom would have when electron(s) were transferred completely from a molecule or ion.

The element being oxidized is the one whose oxidation increases in the reaction whereas the reduced element is the one whose oxidation number decreases in the reaction.

Answer to Problem 1E

On reacting copper and oxygen to form copper (II) oxide, the copper undergoes oxidation and oxygen undergoes reduction.

Explanation of Solution

The balanced chemical reaction for the formation of copper (II) oxide from copper and oxygen is:

  ${\text{2Cu + O}}_{\text{2}}\to \text{ 2CuO}$

While determining the oxidation state of compound, the element with greater electronegativity is assigned with negative value of oxidation state which is equal to the charge as an anion in ionic compounds and element whose oxidation states are fixed are assigned. For compounds with no charge, the sum of oxidation states is zero.

The oxidation states are determined as:

Since, $\text{Cu}$ and ${\text{O}}_{\text{2}}$ are in its stable elemental state so, the oxidation state of each is zero for $\text{CuO}$ the oxidation is determined as:

The oxidation state of $\text{O}$ is assigned as -2 and the oxidation state of $\text{Cu}$ is assigned as x:

Since, $\text{CuO}$ has no charge so; the sum of oxidation state must be zero.

so:

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

So, the oxidation for each element in the reaction is:

  $\begin{array}{l}{\text{ 2Cu + O}}_{\text{2}}\to \text{ 2CuO }\\ \text{Oxidation states: 0 0 For Cu = +2 }\\ \text{ For O = -2 }\end{array}$

Since, the oxidation state of $\text{Cu}$ increases from 0 to +2 so, it undergoes oxidation and the oxidation state of $\text{O}$ decreases from 0 to -2 so it undergoes reduction.