The following reaction is a single replacement reaction. What happens to the oxidation number of the elements in this reaction? Show proof. Li(s) + CuCl2(s)=LiCl(s) + Cu(s)

1. The oxidation number of Li changes from 0 to +1

2. The oxidation number of Cu changes from +2 to 0.

3. The oxidation number of Cl remains the same in each compound.

Explanation:the equation for the reaction is given below:

Li(s) + CuCl2(s) —> LiCl(s) + Cu(s)

Next, we shall determine the oxidation number of each element in the reaction.

NOTE: 1. Oxidation number of chlorine (Cl) is always –1 in any compound because it is a ground 7 element.

Now, we shall determine the change in oxidation number of each element. This is illustrated below:

1. For Li:

Li = 0 (ground state)

LiCl = 0

Cl = –1

Li + (–1) = 0

Li – 1 = 0

Collect like terms:

Li = 0 + 1

Li = +1

The oxidation number of Li changes from 0 to +1

2. For Cu:

CuCl2 = 0

Cu + 2Cl = 0

Cl = –1

Cu + 2(–1) = 0

Cu – 2 = 0

Collect like terms

Cu = 0 + 2

Cu = +2

Cu = 0 (ground state)

The oxidation number of Cu changes from +2 to 0.

3. For Cl:

The oxidation number of Cl remains the same in each compound.

Summary:

1. The oxidation number of Li changes from 0 to +1

2. The oxidation number of Cu changes from +2 to 0.

3. The oxidation number of Cl remains the same in each compound