Untitled - Kelly Walsh High School

Electrochemistry 275
Don’t Forget!
Oxidation is an anode process, even in electrolysis.
The cathode reaction comes directly from a table of standard reduction potentials,
while the anode reaction is the reverse reaction from such a table.
However, how did we know that it was the fluorine reaction requiring reversal?
The fluorine half-reaction in the table is:
F 2 2 e l 2 F
If we examine the reactant, we find that the compound, KF, is an ionic compound
containing potassium ions and fluoride ions. For this reason, we could
replace the KF(l) in the original equation with K (l) F (l). These two ions,
either alone or in combination, are the only substances, other than electrons,
that can appear on the reactant side of the half-reactions. One of these ions,
the fluoride ion, appears in the fluorine half-reaction. Since KF, and therefore
F , is a reactant, we must reverse the fluorine half-reaction to place the fluoride
ion on the reactant side. The original KF has no F 2, so F 2 cannot be a
reactant.
What happens if we replace our reactant, KF(l), with KF(aq)? This apparently
minor change makes a big difference in the results. The potassium ions and the
fluoride ions are still present, so they are still under consideration, but we also
need to consider water. Water appears in many places in a table of reduction
potentials. We must examine every place it appears alone on a side or with one
of the ions we know to be present, K and F . The potassium and fluorine halfreactions,
with their reduction potentials are:
K (aq) 1 e l K(s) E 2.93 V
F2(g) 2 e l 2F (aq) E 2.87 V
We need to reverse the fluorine half-reaction to place the fluoride ion on the
reactant side:
2 F (aq) l F 2(g) 2 e E 2.87 V