Answer:
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
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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