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Tricky question - Redox systems in acidic and basic conditions Watch

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    Hey guys - really stuck on this question (7d) - I'm absolutely fine with writing equations in acidic or basic conditions, I just don't understand why for this scenario why this would cause the reverse reactions to occur? Surely the same reactions should still happen? :/

    Would someone mind explaining it to me?

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    (Original post by Serine)
    Hey guys - really stuck on this question (7d) - I'm absolutely fine with writing equations in acidic or basic conditions, I just don't understand why for this scenario why this would cause the reverse reactions to occur? Surely the same reactions should still happen? :/

    Would someone mind explaining it to me?

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    First you have to consider the half equation for 'redox system 7' under acidic and alkaline conditions.

    In acidic conditions:
    ClO-(aq) + 2H+(aq)→ ½Cl2(g) + H2O(l)

    In alkaline conditions:
    ClO-(aq) + H2O(l)→ ½Cl2(g) + 2OH-(aq)


    Therefore under acidic conditions, the position of equilibrium lies to the right so Cl- ions reduce ClO- to Cl2, and Cl- ions are oxidised to Cl2 because Eѳ Cl2/Cl- (redox system 6) < Eѳ ClO-/Cl2 (redox system 7).

    Under alkaline conditions, the position of equilibrium for 'redox system 7' lies well to the left due to the hydroxide ions. Therefore Eѳ ClO-/Cl2 (redox system 7) < Eѳ Cl2/Cl- (redox system 6) Therefore, Cl2 is reduced to ClO- ions by OH- ions. This reaction produces electrons which are then accepted by Cl2 in 'redox system 6' which reduces Cl2 to Cl- ions.

    Its all to do with Le Chatelier's principle and the position of equilibrium, because that's what electrode potentials are, a measure of the position of equilibrium. If the position of equilibrium lies to the left, it has a more negative electrode potential and is a stronger reducing agent; if the position of equilibrium lies to the right, it has a more positive electrode potential and is a stronger oxidising agent.
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    (Original post by B_9710)
    First you have to consider the half equation for 'redox system 7' under acidic and alkaline conditions.

    In acidic conditions:
    ClO-(aq) + 2H+(aq)→ ½Cl2(g) + H2O(l)

    In alkaline conditions:
    ClO-(aq) + H2O(l)→ ½Cl2(g) + 2OH-(aq)


    Therefore under acidic conditions, the position of equilibrium lies to the right so Cl- ions reduce ClO- to Cl2, and Cl- ions are oxidised to Cl2 because Eѳ Cl2/Cl- (redox system 6) < Eѳ ClO-/Cl2 (redox system 7).

    Under alkaline conditions, the position of equilibrium for 'redox system 7' lies well to the left due to the hydroxide ions. Therefore Eѳ ClO-/Cl2 (redox system 7) < Eѳ Cl2/Cl- (redox system 6) Therefore, Cl2 is reduced to ClO- ions by OH- ions. This reaction produces electrons which are then accepted by Cl2 in 'redox system 6' which reduces Cl2 to Cl- ions.

    Its all to do with Le Chatelier's principle and the position of equilibrium, because that's what electrode potentials are, a measure of the position of equilibrium. If the position of equilibrium lies to the left, it has a more negative electrode potential and is a stronger reducing agent; if the position of equilibrium lies to the right, it has a more positive electrode potential and is a stronger oxidising agent.
    Very well explained sir.
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