Calculate the reduction potential for the following reaction.

Question

Answer 1

You must first find a table of standard (measured) electrode potentials. This can be found in any chemistry textbook. You then need to split the reaction into two half-reactions, or half-cellos. One will be an oxidation and one will be a reduction. Relying on the definitions of oxidation (loss of electrons, the charge becomes more positive) and reduction (the gain of electrons, the charge becomes more negative), you can set up these reactions. So Fe--->Fe⁺² + 2e^- and I₂ + 2e^----> 2I^-. In a reduction, put the electrons on the reactant (the left side) and in an oxidation put the electrons on the product side. Now write one half-cell under the other and cancel electrons: Fe---> Fe²⁺ + 2e^-

I₂ + 2e^- ---> 2I-

Now look up the reduction potentials and use the following: E_cell= E₀(cathode (the reduction reaction)-E₀ (anode, the oxidation reaction. From the tables, 0.54-(-0.45)=0.99V

Answer 2

Standard reduction potentials:

Fe²⁺ + 2e- ==> Fe Eº = -0.41 V (anode - oxidation)

I₂ + 2e- ==> 2I^- Eº = 0.54 V (cathode - reduction)

Eºcell = cathode - anode = 0.54 - (-0.41)

Eºcell = 0.95 V