[Morphic flip]

How do you tell if Cr₂O₇^2- would oxidise:
I^- to I₂ at a pH of 6
H₂O₂ at a pH of 4

Thank`s for any help.

[Borek]

Use Nernst equation to calculate oxidation potentials for half reactions, then compare them.

[Morphic flip]

Thank`s, will try that.

[Morphic flip]

Do you mean like:
E= - (0.0592) log 10^-6= 0.36
E= - (0.0592) log 10^-4= 0.24
so the only thing that would change would be the log of the pH?
IE: one is pH6, the other is pH4
If I compare them to each other, one is lower, how would that dictate whether they would be oxidised or not?
Thank`s.

[Borek]

No. You have to write full half-reactions and use Nernst equation with reaction quotients - these will be pH dependent (at least for reactions where H⁺ takes place) and Nernst equation will give you an answer.

[Morphic flip]

I^- - e^- -->I₂ 

e= -  (0.0592) log {(I^-)/(I₂)}

I can`t see where the H⁺ will go

[Piov1984]

I^- - e^- -->I₂ 

e= -  (0.0592) log {(I^-)/(I₂)}

I can`t see where the H⁺ will go

this is a non balanced equation and you are not supposed to use I2 that is on his elemental state on the Nerst equation...
I think the first reaction you are talking about is:

Cr₂O₇^2- + 6I^- +14H⁺ = 2Cr³⁺ + 3I₂ + 7H₂O

[Borek]

I^- - e^- -->I₂ 

e= -  (0.0592) log {(I^-)/(I₂)}

I can`t see where the H⁺ will go

I havene never said ALL three reactions are pH dependent. I have specifically wrote that there will be pH depdndence only for the reactions where H⁺ are used or produced.

[Borek]

I think the first reaction you are talking about is:

Cr₂O₇^2- + 6I^- +14H⁺ = 2Cr³⁺ + 3I₂ + 7H₂O

Complete reaction equation is of no use when the idea is to calculate potentials of half reactions to compare them.

[Morphic flip]

Can you give me one of the examples I originally posted?
Or be more specific, I`m missing the point/confused now.

[Piov1984]

Complete reaction equation is of no use when the idea is to calculate potentials of half reactions to compare them.

That's true indeed, i wrote the full reaction only because Morphic said: "
I can`t see where the H+ will go" so i showed him where they "will go".

Anyway i can't understand well your question Morphic, do you want to know why H₂O₂ is oxidized to pH 4 while I^- is reduced at pH 6?
if the point is that i can say that the Reduction Standard Potential for I^- to I₂ is around +0,536 v while for H₂O₂ becoming O₂ is around +1,776v.
Concerning Cr₂O₇^2- we are talking about the half reaction:
Cr₂O₇^2- + 14H⁺ + 6e^- = 2Cr³⁺ + 7H₂O
the potential is around 1,33v

So if my valuation and Standard Potentials are correct, at standard condictions, the oxidation of I^- is spontaneous while the oxidation of H₂O₂ isn't (obviously if we are talking about the same processes). Anyway we should consider that O₂ is volatile - this bring to right the equilibrium - and the fact is that also the second complete reaction involves protons as we can see(if reaction is this!):
Cr₂O₇^2- + 8H⁺ + 3H₂O₂ = 2Cr³⁺ + 3O₂ + 7H₂O
for these facts i suppose the lower pH could be due to the most H⁺ needed to bring to right the equilibrium.. but this is only a supposition, any ideas?

[Morphic flip]

Thank`s for help so far guys.
I need to know if Cr₂O₇^2- will oxidise:

I ^- to I₂at a pH of 6
H₂O₂ at a pH of 4

I need to know how to work these out for future reference. What formula, like an example using what I`ve given, so I can alter the example for what
may ever crop up.

[Borek]

Concerning Cr₂O₇^2- we are talking about the half reaction:
Cr₂O₇^2- + 14H⁺ + 6e^- = 2Cr³⁺ + 7H₂O
the potential is around 1,33v

Not exactly. Potential is given by Nernst equation:

E = E₀ + 0.0591/6 log([Cr₂O₇^2- ][H⁺ ]¹⁴/[Cr³⁺ ]²)

or

E = 1.33 + 0.0591/6 log([Cr₂O₇^2- ]/[Cr³⁺ ]²) - 14*0.0591/6 pH

So if my valuation and Standard Potentials are correct, at standard condictions

Question is not about standard conditions - pH is not 0.

Morphic: calculate potential as shown above (do you understand where did it get from?) then compare it with potential of iodine system and that of hydrogen peroxide (it will be pH dependent - write H₂O₂ oxidation half reaction!)

[Albert]

for these facts i suppose the lower pH could be due to the most H⁺ needed to bring to right the equilibrium.. but this is only a supposition, any ideas?

Yes, it could be. Looking at the three half-reactions, those involving H₂O₂ and Cr₂O₇^2- require acidic conditions.

[Morphic flip]

I can see where the 0.0592 (nerst) and the 6 log (pH6) comes from.
Where did the 1.33 come from?