Please help me solve this problem:
What is the value of the equilibrium constant (K) for the reaction below 25^o C?
3I2(s) + 2Fe(s) --> 2Fe^3+(aq) + 6I^-
3 answers
What do you have for data? My crystal ball is hazy today.
I know that I'm suppose to use the formula Ecell = 0.0592V/n*logK and solve for K, but before I do that, I have to find the oxidation and reduction from the chemical equation that was given for me. But I just do not know how to approach it. And i'm pretty stuck.
That helps but not enough.
Ecell = Eocell + (0.0592/n)*logK and you can findEocell from the reactants and products. Furthermore, Ecell = Eocell if you are at standard conditions.
Look up the potentials for the following:
I2 + 2e ==> 2I^- Eo = ??(as a reduction)
Fe ==> Fe^3+ + 3e Eo = ??(as an oxidation)
Add the Eo redn to Eo oxdn to find Eo for the cell (at standard conditions)
Ecell = Eocell + (0.0592/n)*logK and you can findEocell from the reactants and products. Furthermore, Ecell = Eocell if you are at standard conditions.
Look up the potentials for the following:
I2 + 2e ==> 2I^- Eo = ??(as a reduction)
Fe ==> Fe^3+ + 3e Eo = ??(as an oxidation)
Add the Eo redn to Eo oxdn to find Eo for the cell (at standard conditions)
2 answers