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R-5
2. Method 2: Using the Half-cell Reactions
The second method of calculation uses equation (7). Since experimentally we set up two distinct half-
cells, we will apply the Nernst equation to the individual half-cells, and will work with half-cell reaction
quotients rather than cell reaction quotients. Therefore, half-cell potentials must be determined for
both the reduction and oxidation reactions. The respective Nernst equations are:
E
red
= E
red
-
0.
0592
n
red
log Q
red
(11)
and
E
ox
= E
ox
-
0.
0592
n
ox
log Q
ox
(12)
The subscript “red” refers to reduction and “ox” refers to oxidation. We will then be able to use
equation (7) to calculate E
cell
.
NOTE:
E
cell
= E
red
+ E
ox
. = (E
red
-
0.
0592
n
red
log Q
red
) + (E
ox
-
0.
0592
n
ox
log Q
ox
)
(13)
E
cell
= E
red
+ E
ox
-
(
0.
0592
n
red
log Q
red
)
-
(
0.
0592
n
ox
log Q
ox
)
E
cell
= E
cell
-
(
0.
0592
n
red
log Q
red
)
-
(
0.
0592
n
ox
log Q
ox
)
C. Reference and Test Solutions
An alternate use of the Nernst equation is to use the measured cell potentials and equation (7) or (8) to
determine unknown concentrations of a metal ion. Let us assume that you want to determine the
concentration of the metal ion in a test solution. First, you must determine which half-cell is the anode,
and which is the cathode. The way to do this is discussed in the next section under Experimental
Method. In this experiment, dipping a metal electrode into a 1.00 M solution of its own metal cations
will form the reference half-cell.
If using equation (7), then you will calculate the reference half-cell potential:
