Suppose we place a solution of NaF in an electrolytic cell.
The possible reactions at the cathode are
Recall that the more positive (or less negative) the value of , the more favorable the reduction is.
Thus, the reduction of H2O to H2 is far more favorable than the reduction of Na+ to Na. Hydrogen gas is produced at the cathode.
The possible reactions at the anode are the oxidation of F– or of H2O:
Because oxidation is the reverse of reduction, the more negative (or less positive) the value of , the more favorable the oxidation is.
Thus, the values tell us that it is far easier to oxidize H2O than to oxidize F–. In fact, it is easier still to oxidize the OH–(aq) that is produced at the cathode:
Regardless of whether H2O or OH– is oxidized, O2(g) is produced at the anode in preference to F2(g).
We see that the electrolysis of NaF(aq) leads only to the reduction and oxidation of H2O. The NaF(aq)serves only as an electrolyte to allow the electricity to be conducted through the electrolytic cell.
Calculation of the minimum emf required for the above electrolysis-–
The overall reaction is as follows:
This process, is simply the reverse of the reaction in the H2-O2 fuel cell.
Notice that = (cathode) – (anode) is negative, reminding us that the process is not spontaneous, but must be driven by an outside source of energy.