Tricky question - Redox systems in acidic and basic conditions

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?

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 Cl_2, and Cl^- ions are oxidised to Cl₂ because E^ѳ Cl₂/Cl^- (redox system 6) < E^ѳ 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 E^ѳ ClO^-/Cl₂ (redox system 7) _< E^ѳ 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.