[c0d3]

i would like to know what happen to a redox reaction when it is performed in acidic and what happen in alkaline condition. EG. H2O2+FeCl3--->
what will be the product in low pH and what will be the product in high pH

[kimyacı]

 As far as I know, redox redox reactions are equilibrium reactions and Le Châtelier’s Principle valid in these reactions. The equilibrium position of a chemical
reaction changes when the conditions under  which the reaction takes place change.
These conditions include adding or removing  reactants or products, changing the temperature and changing the pressure,changing the pH...A  change in any one of these or several of  these simultaneously shifts the equilibrium. A change in conditions can favor the forward reaction or reverse reaction. In the former case, more reactants form into products, while the latter causes more products to
change to reactants.

[c0d3]

Thanks for your reply. However, i am aware of Le Châtelier’s Principle and i understand equilibrium will shift to which side under other factor. Only pH factor that i can't predict how a reaction will proceed. That is because i can't sure H+ or OH- will react with which of my reactants

[Borek]

First of all - are you sure there will be a redox reaction?

[c0d3]

First of all - are you sure there will be a redox reaction?

that is just an example.maybe a bad one. however, my main point is to ask how pH going to shift Le-chatelier priciple.

[BluRay]

i would like to know what happen to a redox reaction when it is performed in acidic and what happen in alkaline condition. EG. H2O2+FeCl3--->
what will be the product in low pH and what will be the product in high pH

3H₂O₂ + 2FeCl₃ --> 3Cl2 + 2Fe(OH)₃
So, for this reaction pH comes into the scene only because of the equilibrium of dissociation of Fe(OH)₃ (so, very low, if pH is > ~ 6).

For a reaction of the kind: MnO₄^- + Fe²⁺ + H⁺ --> Fe³⁺ + Mn²⁺ + H2O (not balanced) pH is more directly involved because of the presence of H⁺ among the reagents, so a low pH favours the reaction. In other cases it's OH^- to be present among the reagents, so an high pH favours the reaction (you can also see it computing the values of the potentials with the Nernst equation, at various pH).

[BluRay]

In other cases it's OH^- to be present among the reagents,

for example:
Cl₂ + 2OH^- <--> ClO^- + Cl^- +H₂O

In other cases pH doesn't affect the reaction, for example:
Cu²⁺ + Zn --> Cu + Zn²⁺

Anyway, almost all redox reactions involve the presence of H⁺ or OH^- or other species which are acids/basic, so you have to be careful in the evaluation.

[c0d3]

what if i have more than 1 step in a reaction eg. fenton process. Reactants involve H+ in first step and OH- in second step. How pH involve in this case


p.s. ignore the example if you feel it is a very bad one

[BluRay]

what if i have more than 1 step in a reaction eg. fenton process. Reactants involve H+ in first step and OH- in second step. How pH involve in this case


p.s. ignore the example if you feel it is a very bad one

You write the overall reaction.
Example, you have the two reactions:

1. Cl₂ + 2OH^-  -->  Cl^- + ClO^- + H₂O

2. 3ClO^- + S^2- + 2H⁺  -->  3Cl^- + SO₂ + H₂O

To write the overall reacrion you have to multiply the first by 3 and then sum the two:

3Cl₂ + 6OH^-  -->  3Cl^- + 3ClO^- + 3H₂O          +
3ClO^- + S^2- + 2H⁺  -->  3Cl^- + SO₂ + H₂O     =
----------------------------------------
3Cl₂ + S^2- + 4OH^-  --> 6Cl^- + SO₂ + 2H₂O

Note that 6OH^- +  2H⁺ --> 4OH^- + 2H₂O and that I have cancealed equal amounts of compounds at the left and at the right side of the final equation.

So, in this specific case, the overall reaction has OH^- among the reagents and so is favoured by an high pH.

[c0d3]

What if my H+ and OH- cancel out completely? Is it we can treat it like a reaction that occur with the slightest effect of pH? The pH can be considered does not affect that particular reaction. Is it correct to think that way?

[Borek]

What if my H+ and OH- cancel out completely?

You mean when they are not present in the final reaction equation? They don't matter.

[c0d3]

I see. Thanks for all the information provided.

[Borek]

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