23.2. Cell Potential http://www.ck12.org
When using these values, keep the following points in mind:
1.E° values apply to the reaction as written in the left-to-right direction. If the reaction is reversed, the sign
changes, but the value stays the same.- The more positiveE° is, the greater the tendency for the material to be to reduced.
- Under standard-state conditions, any species on the left side of a half-cell reaction will spontaneously react
with any species on the right side of a half-cell reaction that is ranked below it. - Multiplying an entire half-reaction by a constant doesnotchange the associated reduction potential. This is
different than the rules for manipulating∆Hrxnvalues.
Calculating Standard Cell Potentials
In order to function, any electrochemical cell must consist of two half-cells.Table23.2 can be used to determine
the reactions that will occur and the standard cell potential for any combination of two half-cells without actually
constructing the cell. The half-cell with the higher reduction potential according to the table will undergo reduction,
while the half-cell with the lower reduction potential will undergo oxidation. If those specifications are followed,
the overall cell potential will be a positive value. The cell potential must be positive in order for the redox reaction
in the cell to be spontaneous. If a negative cell potential were calculated, the reaction would not be spontaneous.
However, that reaction would be spontaneous in the reverse direction.
Example 23.1
Calculate the standard cell potential of a voltaic cell that uses the Ag|Ag+and Sn|Sn^2 +half-cell reactions. Write the
balanced equation for the overall cell reaction that occurs. Identify the anode and the cathode.
Answer:
The silver half-cell has a reduction potential of +0.80 V, whereas the tin half-cell has a reduction potential of -0.14
V. Reduction will happen in the silver half-cell, because its standard reduction potential is more positive. The tin
half-cell will undergo oxidation.
Oxidation (anode): Sn(s)→Sn^2 +(aq)+2e−
Reduction (cathode): Ag+(aq)+e−→Ag(s)Before adding the two reactions together, the number of electrons lost in the oxidation must equal the number of
electrons gained in the reduction. The silver half-cell reaction must be multiplied by two. Then, adding in the tin
half-reaction, we get the full redox process:
Sn(s)+2Ag+(aq)→Sn^2 +(aq)+2Ag(s)The overall cell potential can be calculated as follows:
E◦cell=E◦red−E◦oxid
E◦cell= + 0 .80 V−(− 0 .14 V) = + 0 .94 VThe standard cell potential is positive, so the reaction is spontaneous as written. Tin is oxidized at the anode, while
Ag+is reduced at the cathode. Note that the voltage for the silver ion reduction is not doubled even though the
reduction half-reaction had to be doubled to balance the overall redox equation.