MECalculating E°cell for IO3-(aq) + Fe2+(aq) --> Fe3+(aq) + I2(s)

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Discussion Overview

The discussion revolves around calculating the standard cell potential, E°cell, for the reaction involving iodate ions (IO3-) and iron ions (Fe2+). Participants are exploring the half-reactions and the standard reduction potentials associated with them, focusing on how to derive the E°cell value for a specific unbalanced redox reaction.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the half-reaction for Fe3+ to Fe2+ and notes its standard reduction potential as +0.77 V.
  • Another participant attempts to balance the half-reaction for IO3- to I2 and mentions finding a standard potential of +1.19 V from an external website, questioning its source.
  • There is a request for clarification on how the E°cell value was derived, as it does not appear in standard tables, including a reference to a Wikipedia page on standard electrode potentials.
  • Participants discuss the balancing of the half-reaction, indicating that H2O and H+ ions are involved in the process, but do not provide a detailed explanation due to lack of resources.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the source of the E°cell value for the IO3- to I2 reaction, indicating a lack of consensus on how to derive or confirm this value from standard tables.

Contextual Notes

The discussion highlights limitations in available resources, as participants reference external websites and express confusion over the absence of certain reactions in standard tables. There are also unresolved steps in balancing the half-reaction and deriving the E°cell value.

salman213
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1. Using tabulated standard reduction potentials from your text, calculate the standard cell potential, E°cell (always positive for a galvanic cell), based on the following (unbalanced) reaction:

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


3.

well you see i made the half reactions

Fe3+ (aq) ---> Fe2+(aq) EØ was easily found +0.77 V

IO3-(aq) ---> I2(s)


balanced the 2nd one to

12 H + 2IO3 + 5e- ---> I2 + 6H2O E CELL = ??

I CANT FIND THIS REACTION'S ECELL IN MY TEXT BOOK BUT SOMEHOW I FOUND IT ON THIS WEBSITE TO BE EØ = +1.19V, HOW DID THEY GET THIS VALUE! PLEASE HELP
 
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any idea anyone??
 
ok, first let's clear up any confusion

is it ...

IO_{3}^{-}(aq)\rightarrow I_{2}(s)
 
all they simply did was, balance out the equation with (first) H2O on the product side, then H+ on the reactant side

i do not have my Chemistry book with me atm, so i cannot give you an in-depth explanation.
 

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