Separate this redox reaction into its balanced component half‑reactions. Use the symbol e− for an electron. Cl2+2Li⟶2LiCl

Separation of the redox reaction into its balanced component half-reactions is as follows;

Oxidation half-reaction;

Reduction half-reaction;

By definition;

Oxidation is simply characterized by an increase in oxidation number of a reacting entity.

Reduction is simply characterized by a decrease in oxidation number of a reacting entity.

A reaction in which oxidation and reduction occur simultaneously is termed a Redox reaction as in the given reaction.

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Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation

2Li + 2e⁻ ⟶ 2Li⁺ Reduction

Explanation:

The question requests to split the equation below into half equations;

Cl2+2Li⟶2LiCl

In redox chemistry, splitting into half equations simply means highlighting the reduction and oxidation reactions of the reaction.

Before proceeding, we hav to split the ionic compound; LiCl into it's component ions. So we have;

Cl₂ + 2Li ⟶ 2Li⁺Cl⁻

This leaves us with;

Cl₂ ⟶ 2Cl⁻ i

2Li ⟶ 2Li⁺   ii

Oxidation reactions can be identified by the increase in oxidation number and decrease in the case of reduction.

in reaction i, there is a decrease in oxidation number from 0 to -1. This is the reduction half equation,

in reaction ii, there is an increase in oxidation number from 0 to +1. This is the oxidation half equation

In terms of electrons, we have to even the charge;

Oxidation = Loss of electrons

Reduction = Gain of electrons

The half equations are given as;

Cl₂ ⟶ 2Cl⁻ + 2e⁻ Oxidation

2Li + 2e⁻ ⟶ 2Li⁺ Reduction

it is a good conductor

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