Calculating 'n' in Nernst Equation for Fe3+ and I- Reaction

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Homework Statement



How do I find out the 'n' of Nernst Equation for this reaction?

2Fe3+ + 3I- -------> 2Fe2+ + I3-
 
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writing the half equations

2Fe+3 + 2e- --------> 2Fe+2

3I- --------> I3- + 2e-

so there is total exchange of 2 electrons ,so n = 2
 
How did you find that 2 e are exchanged in the second reaction?
I tried balancing the equation by ion-electron method. What is the oxidation state of I in I3-? I guess it is -1/3 for each atom and -1 overall. I have no idea how you got 2 e :confused:
 
i too did it using ion-electron method
Step 1
balance the no of atoms -"so we put a 3 on left I-"

Step 2
balance the charge - "there's an excess of charge(2 units) on right side so we add 2 e-there"
 
I think there is a problem in your step 2.
We add electrons in ion-electron method to balance oxidation state. To balance the charge, we add H+ or OH-
 
Is the answer wrong??
you use H+ and OH- when it is given that the reaction is carried out in acidic or basic medium.Nothing is given so we have to consider it being carried out in neutral medium.Apply the ion-electron balancing for neutral medium method.Waise there's no need of thinking about the medium.This reaction is a simple inter-molecular redox and so it is balanced before you reach that step of adding H+ or OH-.And why are you even bothering about knowing the oxidation state of I in I3- ;that is required if you are applying the oxidation number method of balancing.
 
So do you mean that we add H+ or OH- to balance charge only if the medium is acidic/basic and we add electrons when it is neutral?
 
I always balanced the oxidation sate in ion-electron method by adding charges. Then added H+ or OH- to balance charges on both sides. It worked well always.

But the correct method is different. Thanks.