Using electrode potentials to make predictions
Feasibility
With electrode potentials it is possible to predict whether or not a specific redox reaction can take place
To do this you need to compare the E0 for the two half equations
Can a solution of Cu2+ ions oxidise Pb(s) to Pb2+(aq)?
Write out both half equations and look up the E0 values
Cu2+(aq) + 2e- ⇌ Cu(s) E0 = +0.34V (forwards, reduced)
Pb2+(aq) + 2e- ⇌ Pb(s) E0 = -0.11V (backwards, oxidised)
Decide which will be reduction and oxidation in the usual way
Yes, a solution of Cu2+ ions oxidise Pb(s)
Work out the E0 of the cell
0.34 - - 0.11 = +0.45V
Limitations to predictions
Limitation 1:
Predictions only tell us if a reaction is feasibly, they do not refer to the rate of reaction
A reaction may be feasible but too slow to measure
In other words it may have a very high activation energy
Limitation 2:
Our predictions are based on E0 values, these require standard conditions to be met
If the reaction is done under non-standard conditions then the reaction may not occur as predicted
Non-standard conditions
Under non-standard conditions the electrode potential (E) for the half equation will not be the same as the standard electrode potential (E0)
Half equations are written as equilibria
If the non-standard conditions would push the equilibrium to the right, then the E value will be less positive than E0.
E.g. Fe3+(aq) + e- ⇌ Fe2+(aq) E0 = +0.77V
If the concentration of Fe3+(aq) is higher than 1.00M then the electrode potential (E) will be more positive than E0.
If the E0 values are within 0.3V of each then under non-standard conditions the reaction may not go as predicted
The E values will differ from E0 and it may well be that the different reaction is now more positive
Consider the following equations
MnO2 + 4H+ + 2e- ⇌ 2H2O + Mn2+ E0 = +1.23V
Cl2 + 2e- ⇌ 2Cl- E0 = +1.36V
Using the E0 you would expect the first reaction to go backwards and the second forwards
This reaction is usually done with concentrated HCl
The first reaction will be pushed to the right as conc. of H+ is higher than 1.00M
MnO2 + 4H+ + 2e- ⇌ 2H2O + Mn2+ E0 = +1.33V
The second reaction will be pushed to the left as conc. of Cl- is higher than 1.00M
Cl2 + 2e- ⇌ 2Cl- E0 = +1.26V
Reactions go opposite way