Write balanced half-reactions for the following redox reaction: 5cl2(g)+2mn+2(aq)+8h2o(l)→ 10cl−(aq)+2mno−4(aq)+16h+(aq)a. reduction: . oxidation:

a. 5Cl₂ + 10e⁻  → 10Cl⁻

b. 4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺

Explanation:

5Cl2(g)+2Mn+2(aq)+8H2O(l)→ 10Cl−(aq)+2MnO4-(aq)+16H+(aq)

First of all, think in all the oxidation numbers for each compound. That's the way, you can notice the half reaction.

When the oxidation number, decrease, you have reduction. Compound is wining electrons.

When the oxidation number increase, you have oxidation. Compound is losing electrons.

Cl2(g) - Atoms in ground state, has 0 as oxidation number. In products side, you have the anion chloride which act with -1, so chlorine has been reduced.

Mn2+ - The manganese ion is already telling you with 2, that is its oxidation number. On the side products, the element was transformed into the permanganate anion; how oxygen acts with -2 and there are 4 atoms, it has -8 as oxidation state but since the general charge is -1 the Mn is acting with +7.  From + 2 it went to +7, which means that it increased, so it has oxidized.

5Cl₂ + 10e⁻  → 10Cl⁻  - REDUCTION

The chlorine had to gain 1 electron to go from 0 to -1, but being a diatomic molecule were 2 electrons, but finally so that the charges are balanced and because there are 5 moles, it ends up gaining 10 electrons.

Mn²⁺ → MnO₄⁻ - OXIDATION

Look that in main reaction we have H⁺, that is the clue to notice us, that we are in acidic medium. So if we have 4 O, in MnO₄⁻, we have to complete with 4H₂O in the other side of O. And to ballance the protons, we have to add 8H⁺ in product side

4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺ OXIDATION

How do u finish this?. You have to multipply .2, the oxidation one to balance the e⁻ so, they can be cancelled.

5Cl₂ + 10e⁻  → 10Cl⁻

(4H₂O + Mn²⁺ → MnO₄⁻ + 5e⁻ + 8H⁺ ).2

8H₂O + 2Mn²⁺ → 2MnO₄⁻ + 10e⁻ + 16H⁺

5Cl₂ + 10e⁻ + 8H₂O + 2Mn²⁺ → 10Cl⁻ + 2MnO₄⁻ + 10e⁻ + 16H⁺

5Cl₂  + 8H₂O + 2Mn²⁺ → 10Cl⁻ + 2MnO₄⁻  + 16H⁺

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