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

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Discussion Overview

The discussion revolves around determining the value of 'n' in the Nernst Equation for the redox reaction involving Fe3+ and I-. Participants explore the half-reactions, balancing methods, and the implications of reaction medium on the balancing process.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asks how to find 'n' for the given reaction.
  • Another participant proposes that the total exchange of 2 electrons leads to n = 2 based on their half-equations.
  • A participant questions the calculation of electron exchange in the second half-reaction and discusses the oxidation state of iodine in I3-.
  • Another participant describes their method of balancing the reaction using the ion-electron method, detailing steps for balancing atoms and charge.
  • One participant challenges the previous step regarding charge balancing, suggesting that H+ or OH- should be used only in acidic or basic media.
  • A participant argues that the reaction should be considered in a neutral medium and that the balancing should not require H+ or OH-.
  • Another participant provides a link to a resource that suggests H+ and OH- are used to balance atoms, while electrons are used to balance charge.
  • A participant reflects on their previous method of balancing and acknowledges the differences in approach.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate method for balancing the redox reaction, particularly regarding the use of H+ and OH- in relation to the reaction medium. There is no consensus on the correct approach to balancing the reaction.

Contextual Notes

Participants highlight the importance of the reaction medium in determining the balancing method, indicating that assumptions about the medium can affect the steps taken in the ion-electron method.

zorro
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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.
 

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