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# Tricky question - Redox systems in acidic and basic conditions Watch

1. 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?

Attachment 545149545151
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2. (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?

Attachment 545149545151
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.
3. (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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