Answer:
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺
Explanation:
This is an easy redox reaction:
Br₂ + 2KI → I₂ + 2KBr
We determine the oxidation states.
0 for the elements at ground state.
K does not change the oxidation state during the reaction.
Bromine is reduced to bromide (oxidation state decreases)
and iodide is oxidized to Iodine (oxidation state increases)
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
In oxidation, electrons are released while in reduction, the electrons are gained. To make the ionic equation, we just add K⁺
So we sum both reactions
Br₂ + 2e⁻ + 2I⁻ + 2K⁺ ⇄ 2e⁻ + I₂ + 2Br⁻ + 2K⁺
We cancel the electrons on both sides of the equation:
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺
