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21 May, 15:06

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

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  1. 21 May, 17:29
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    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
  2. 21 May, 18:19
    0
    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
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