Balance the following unbalanced redox reaction (assume acidic solution if necessary) : Cr2O72 - + Cl - → Cl2 + Cr3 + Indicate the coefficient that will be used for Cl2 (g) in this reaction

Equation of the reaction

14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O

The coefficient that will be used for Cl₂ in this reaction is 3

Explanation:

We use the method of electron-ion to the balance.

By assumption, the redox reaction is happening at acidic medium.

Cr₂O₇²⁻ + Cl⁻ → Cl₂ + Cr³⁺

Chloride is raising the oxidation state from - 1 in the chloride, to 0 in the chloride dyatomic. This is the half reaction of oxidation

2Cl⁻ → Cl₂ + 2e⁻ Oxidation

In the dichromate anion, chromium acts with + 6 in oxidation state, and we have 2 Cr, so the global charge of the element is + 12. To change to Cr³⁺ it has release 3 electrons, but we have 2 Cr, so it finally released 6 e-. The oxidation state was decreased, so this is the reduction half reaction.

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O Reduction

As we have 7 O in the product side, we add 7 water, to the opposite place. In order to balance the H (protons) we, add the amount of them, in the opposite side, again.

(2Cl⁻ → Cl₂ + 2e⁻) ₓ3

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) ₓ1

By multiplying the half reactions, in order to remove the electrons and we then add, the equations as thus:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6Cl⁻ → 3Cl₂ + 6e⁻ + 2Cr³⁺ + 7H₂O

The balanced redox reaction after obtaining thesame amount of electrons is this: therefore cancelling can now take place.

14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O

14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O

The coefficient that will be used for Cl₂ in this reaction is 3

Explanation:

We use the method of electron-ion to the balance.

We assume that the redox reaction is happening at acidic medium.

Cr₂O₇²⁻ + Cl⁻ → Cl₂ + Cr³⁺

Chloride is raising the oxidation state from - 1 in the chloride, to 0 in the chloride dyatomic. This is the half reaction of oxidation

2Cl⁻ → Cl₂ + 2e⁻ Oxidation

In the dichromate anion, chromium acts with + 6 in oxidation state, and we have 2 Cr, so the global charge of the element is + 12. To change to Cr³⁺ it has release 3 electrons, but we have 2 Cr, so it finally released 6 e-. The oxidation state was decreased, so this is the reduction half reaction.

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O Reduction

As we have 7 O in the product side, we add 7 water, to the opposite place. In order to balance the H (protons) we, add the amount of them, in the opposite side, again.

(2Cl⁻ → Cl₂ + 2e⁻) ₓ3

(14 H⁺ + Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ + 7H₂O) ₓ1

We multiply the half reactions, in order to remove the electrons and we sum, the equations:

14 H⁺ + Cr₂O₇²⁻ + 6e⁻ + 6Cl⁻ → 3Cl₂ + 6e⁻ + 2Cr³⁺ + 7H₂O

Now, that we have the same amount of electrons, they can cancelled, so the balanced redox reaction is:

14 H⁺ + Cr₂O₇²⁻ + 6Cl⁻ → 3Cl₂ + 2Cr³⁺ + 7H₂O