Untitled - Kelly Walsh High School

Electrochemistry 273
do is reverse the overall reaction and change the sign on the E cell to positive.
The other method involves using the Nernst equation with the individual halfreactions,
then combining them depending on whether or not it is a galvanic
cell. The only disadvantage to the second method is that you must use the
Nernst equation twice. Either method should lead you to the correct answer.
Calculate the potential of the following half-cell:
Cr 2O 7 2 (aq) 14 H (aq) 6 e l 2 Cr 3 (aq) 7 H2O(l) E 1.33 V
containing 0.10 M K 2Cr 2O 7, 0.20 M Cr 3 (aq), and 1.0 10 4 M H (aq).
Solution:
E E 0.0592
n log a
E 1.33 V 0.0592
6
18-5 Electrolytic Cells
ignoring H 2O
Electrolytic cells use electricity from an external source to produce a desired
redox reaction. Electroplating and the recharging of an automobile battery are
examples of electrolytic cells.
In the operation of both galvanic and electrolytic cells, there is a reaction occurring
on the surface of each electrode. For example, the following reaction takes
place at the cathode of a cell:
Cu 2 (aq) 2 e l Cu(s)
The rules of stoichiometry also apply in this case. In electrochemical cells, we
must consider not only the stoichiometry related to chemical formulas, but also
the stoichiometry related to electric currents. The half-reaction under consideration
not only involves 1 mol of each of the copper species, but also 2 mol of
electrons. We can construct a mole ratio that includes moles of electrons or we
could construct a mole ratio using faradays. A faraday (F) is a mole of electrons.
Thus, we could use either of the following ratios for the copper half-reaction:
1 mol Cu 2
2 mol
[Cr 3 ] 2
[Cr 2 O 7 2 ][H ] 14 b
[0.20] 2
log a
[0.10] [1 104 ] 14b 0.78 V
1 mol Cu2
2F