[Zed1m]

Hi chemists, for the question above, i got the two redox half equations

ClO- +2H+ +2e --> Cl- +H₂O
ClO- + 2H₂O --> ClO₃- +4H+ +4e

To find out if it is spontaneous I would need to use the Eº values to calculate the Eº reaction, but I can't remember if I do anything to the signs of the Eº to calculate.
I recall that Eº is to measure the tendency of being reduced. So for the reduction of ClO- to Cl- the value is +0.89V so the sign doesn't change since in the question it is being reduced. The second equation, however, has an  Eº of +0.5V, if it is being reduced, and in this question it gets oxidised. Do I make it the Eº -0.5V since it is being oxidised rather than reduced? So then would Eº reaction = Eº reduction - Eº oxidation =+0.89--0.5=1.39V?

PS:Trying to revise for chemistry after 1 year, hence might be a stupid question

[mjc123]

First, note that it says "in basic conditions", so your half-equations should involve OH^- rather than H⁺, e.g.
ClO^- + H₂O + 2e^-  Cl^- + 2OH^-
Secondly, you are confusing two different ways of working out E°_reaction. You can
EITHER write the two half-reactions as reductions (with the given E° values) and subtract one from the other (as the overall reaction equation is one half-reaction minus the other), so that
E°_rxn = E°₁(redn) - E°₂(redn)
OR write one half-reaction as a reduction and the other as an oxidation (with E°(ox) = -E°(redn)) and add them together. Then
E°_rxn = E°₁(redn) + E°₂(ox) = E°₁(redn) - E°₂(redn)
Either way, you would get in this case E° = 0.89 - 0.5 = 0.39V

[Zed1m]

First, note that it says "in basic conditions", so your half-equations should involve OH^- rather than H⁺, e.g.
ClO^- + H₂O + 2e^-  Cl^- + 2OH^-
Secondly, you are confusing two different ways of working out E°_reaction. You can
EITHER write the two half-reactions as reductions (with the given E° values) and subtract one from the other (as the overall reaction equation is one half-reaction minus the other), so that
E°_rxn = E°₁(redn) - E°₂(redn)
OR write one half-reaction as a reduction and the other as an oxidation (with E°(ox) = -E°(redn)) and add them together. Then
E°_rxn = E°₁(redn) + E°₂(ox) = E°₁(redn) - E°₂(redn)
Either way, you would get in this case E° = 0.89 - 0.5 = 0.39V

Thank you, I get it now!