Question

Calculate the equilibrium constant for the oxidation of I-(aq) to IO3-(aq) by O2 at 25⁰C in an aqueous acid solution, according to the following chemical equation
2 I-(aq) + 3 O2(g) D 2 IO3-(aq)
Data:
E0 (V)
2 IO3-(aq) + 12 H+(aq) + 12 e- → 2 I-(aq) + 6 H2O(l) 1.09
O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) 1.23

Answer 1

(1) 2 IO3-(aq) + 12 H+(aq) + 12 e- → 2 I-(aq) + 6 H2O(l) 1.09
(2) O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) 1.23
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(2) multiply by 3
(2) 3O2(g) + 12 H+(aq) + 12 e- → 6 H2O(l) 1.23
-----------------------------------------------------------
Reverse (1)
(1) 2 I-(aq) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq) + 12 e- 1.09
-------------------------------------------------------------
(1) 2 I-(aq) + 6 H2O(l) → 2 IO3-(aq) + 12 H+(aq) + 12 e- 1.09
(2) 3O2(g) + 12 H+(aq) + 12 e- → 6 H2O(l) 1.23
cancel stuff out:
2 I-(aq) + 3O2(g) →2 IO3-(aq) E = cathode-anode = 1.23-1.09 = 0.24 V

then theres 2 ways to find K:
(1)

delta G= -nFE = -(12)(96,485.33)(0.24) = - 277877.75

delta G= -RTlnK = - 277877.75 = -(8.314)(298.15)lnK

K = 4.84*10^48

(2) Nernst equation:
at 25 degrees, its simplified to: Ecell=E(knot)-(0.0592/n)(logK)

at equilibrium Ecell=0, so E(knot)=(0.0592/n)(logK)

0.24 = (0.0592/12)logK
K= 4.45*10^48

there is a slight difference in the K values because of the simplification of the nernst equation.

Answer 2

intense, thank you!

Answer 3

What is delta G by the way?

Answer 4

haha i know, electrochemistry looks intense, but it isn't really.

\[\Delta G \] is gibbs free energy

Answer 5

oooo we never talked about gibbs free energy... :o

Answer 6

is it used in any other places other than for those two equation you used, -nFE and -RTlnK?

Answer 7

yep, it's used in physics, and thermodynamics

Answer 8

a big one is \[\Delta G = \Delta H - T \Delta S\]

Answer 9

o.O

Could you tell me this Gibbs Free energy really represents... ?

Answer 10

some energy thatès not directly equal to entropy or enthalpy?

Answer 11

it's defined as the free energy in the system, it can be seen as the stability of a molecule but its much more than that. I mean this is some deep stuff that I'm not comfortable with enough to explain it, but if you watch this vid, you might understand it more http://www.youtube.com/watch?v=DPjMPeU5OeM

Answer 12

thank you!

Answer 13

no prob dude