Calculating ξ° value for a galvanic cell

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SUMMARY

The discussion centers on calculating the standard cell potential (ξ°) for the galvanic cell reaction IO3(aq) + Fe2+(aq) <--> Fe3+(aq) + I2(aq). Participants identified the half-reactions involved, with the reduction of Fe2+ to Fe3+ having a standard reduction potential of 0.77V. A key point of contention was the appropriate standard reduction potential for the iodine half-reaction, with confusion surrounding whether to use I2(s) + 2e- --> 2I- or another form. The conversation highlights the importance of stoichiometry in determining the correct reaction pathway.

PREREQUISITES
  • Understanding of galvanic cells and electrochemical reactions
  • Familiarity with standard reduction potentials
  • Knowledge of half-reaction notation and stoichiometry
  • Basic principles of oxidation and reduction in electrochemistry
NEXT STEPS
  • Research standard reduction potentials for iodine species, specifically I2 and I-
  • Study the Nernst equation and its application in calculating cell potentials
  • Explore stoichiometric calculations in redox reactions
  • Investigate the implications of reaction conditions on galvanic cell performance
USEFUL FOR

Chemistry students, electrochemists, and researchers involved in redox reaction analysis and galvanic cell design will benefit from this discussion.

Phyzwizz
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Calculate the ξ° value for the galvanic cell:
IO3(aq) + Fe2+(aq) <--> Fe3+(aq) + I2(aq)

So I found the half reactions to be:

5Fe2+(aq) --> 5Fe3+(aq) + 5e- Reducing agent 0.77V
6H+(aq) + 5Fe2+(aq) + 5e- --> I2(aq) + 3H2O Oxidizing agent

The problem I have is that I can't find the standard reduction potential for the cation's half reaction. Do I just use the potential for I2(s) + 2e- --> 2I- or this something special that I'm missing?

Thanks!
 
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Phyzwizz said:
Calculate the ξ° value

ξ°? and not E0?

IO3(aq) + Fe2+(aq) <--> Fe3+(aq) + I2(aq)

IO3(aq)? and not IO3-?

It can be difficult to not reduced iodine further to I- in the presence of Fe2+, but perhaps with right stoichiometry reaction will stop at I2.

5Fe2+(aq) --> 5Fe3+(aq) + 5e- Reducing agent 0.77V

Why 5?

6H+(aq) + 5Fe2+(aq) + 5e- --> I2(aq) + 3H2O Oxidizing agent

This is completely off - please check if that's what you meant.

The problem I have is that I can't find the standard reduction potential for the cation's half reaction.

Not surprising - reaction you wrote doesn't make sense.
 

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