Untitled - Kelly Walsh High School

276 CHEMISTRY FOR THE UTTERLY CONFUSED
Don’t Forget!
If you reverse the reaction, you must reverse the sign of the cell potential.
Water, the other potential reactant, appears in the following half-reactions:
2 H2O(l) 2 e l H2(g) 2 OH (aq) E 0.83 V
4 H (aq) O2(g) 4 e l 2 H2O(l) E 1.23 V
We need to reverse the second of these reactions to place the water on the reactant
side.
2 H 2O(l) l 4 H (aq) O 2(g) 4 e E 1.23 V
The reactions that may occur at the cathode are:
K (aq) 1e l K(s) E 2.93 V
2 H2O(l) 2 e l H2(g) 2 OH (aq) E 0.83 V
The reactions that may occur at the anode are:
2 F (aq) l F2(g) 2 e E 2.87 V
2 H2O(l) l 4 H (aq) O2(g) 4 e E 1.23 V
We must narrow the options. There will be only one cathode reaction and only
one anode reaction. How do we pick the correct half-reactions? If one of the
half-reactions were spontaneous (positive), we would pick it for that electrode.
(If more than were spontaneous, we would pick the largest positive value.) All
four half-reactions in this case are nonspontaneous (negative). This is typical
for electrolysis, because you are using electrical energy to force a nonspontaneous
process to take place.
We will begin with the cathode. The reaction that will occur will be the one
requiring the smaller amount of energy. This will be the less negative (higher)
value. We can eliminate all other reduction half-reactions. Therefore, the cathode
half-reaction must be:
2 H 2O(l) 2 e l H 2(g) 2 OH (aq) E 0.83 V