We start by predicting the feasibility of the reaction. Remember that the prediction of reaction feasibility tells us nothing about the rate of reaction. You might like to review the section on predicting reaction feasibility before you progress.
Using the standard electrode potential data we find:
|
Pt(s)½2I-(aq), I2(aq)::Fe3+(aq), Fe2+(aq)½Pt(s) |
+ 0.23V |
As this has a positive Ecell value, the reaction, as it is written from left to right in the cell diagram, is feasible. This is the reaction:
|
2Fe3+(aq) + 2I-(aq) ⇒ 2Fe2+(aq) + I2(aq) |
|
We should really use the actual Ecell value, as the one obtained from standard electrode potential data only applies under standard conditions. The reaction that took place in the test tube was not under standard conditions. The solutions were not 1.0 M and the temperature was not 298 K. Using the actual data obtained in experiment, we still obtain a positive value (+ 0.05V) showing that this reaction is feasible under the conditions used. You should see from the photograph that the iron half-cell is the positive one. |
|