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

[salman213]

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

 

[salman213]

any idea anyone??

 

[rocomath]

ok, first let's clear up any confusion

is it ...

$$IO_{3}^{-}(aq)\rightarrow I_{2}(s)$$

 

[rocomath]

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.

 

[salman213]

yes i know how they balanced it but i don't understand how they got that Ecell value. In a table for example here

http://en.wikipedia.org/wiki/Table_of_standard_electrode_potentials

that reaction

12 H + 2IO3 + 5e- ---> I2 + 6H2O

is not there so how did they get the ecell value for it?