Electrochemical Cell – Part II:
Concentration Cell
Introduction
A concentration cell is an electrochemical cell in which the two half-reactions are
identical but involve solutions at different concentrations. Take a lead concentration cell
for example. The half-reactions for this cell are:
Pb(s) → Pb2+(C1) + 2ePb2+(C2) +2e- → Pb(s)
Eº= +0.13 V
Eº= -0.13 V
Where C1 and C2 are the concentrations of Pb2+ ions around each electrode. The overall
cell reaction is:
Pb2+(C2) → Pb2+(C1)
If C1=C2, the electromotive force (emf) of this cell will be exactly zero no matter what
values C1 and C2 actually have. An emf will develop, however, when the concentrations
of Pb2+ ions around the electrodes are different (C1≠C2). We can also say that the overall
cell reaction will be spontaneous in the direction in which it is written when C2>C1.
What makes this concentration cell work? It works because the process:
Pb2+(C2) → Pb2+(C1)
is spontaneous whether it occurs inside or outside of the electrochemical cell. The
following example should help you understand the spontaneity of this process when it
occurs outside of a voltaic cell. Suppose we have two solutions of Pb2+ ions separated by
a porous barrier. The concentrations of Pb2+ ions in these solutions will again be
represented by C1 and C2. If C2>C1, the natural tendency is for ions to flow through the
barrier in such a way the C2 decreases and C1 increases. Ions will continue to flow from
one side of the barrier to the other until the concentrations of the solutions become equal.
This process, which will occur without any outside intervention, is clearly spontaneous.
As a result, it must also be spontaneous in the voltaic cell.
What emf will be delivered by a concentration cell? The emf can be calculated from the
Nernst equation. At 25ºC, this equation is:
Ecell = Eºcell –(0.0592/n)logQ
where n is the number of electrons transferred in the overall cell reaction and A is the
reaction quotient. For the lead concentration cell we have:
Eºcell = 0.13 – 0.13 = 0.00 V
Q = C1/C2
If C1 = 0.010 M and C2 = 1.0 M, the emf will be:
Ecell = Eºcell –(0.0592/2)log[0.010]/[1.0]
Ecell = 0.059 V
because n = 2 for the lead concentration cell. If, however, C1 = 0.00010 M and C2 = 1.0
M again, the emf will increase, so that its value is:
Ecell = –(0.0592/2)log[0.0010]/[1.0]
Ecell = 0.118 V
These examples show that the emf of a concentration cell increases as the disparity
between the concentrations of the two solutions increases.
Table of Standard Electrode (Reduction) Potentials in Aqueous Solution at 25ºC.
(for chemicals used in this experiment)
Cathode (Reduction) Half-Reaction
Ag+(aq) + e- ↔ Ag(s)
Standard Potential, Eº (Volts)
0.80
Electrochemical Cell – Part II:
Concentration Cell
Procedure
**NOTE: Use instructions provide by the instructor – DO NOT USE INSTRUCTIONS
IN CHEMLAB! To remove the instructions on the screen, and free-up more working
area, perform the following operation: click on the OPTIONS tab; then click on LAB
ONLY. The instructions “disappear” and all of the area is now lab space.**
1. Select an Electrochemical Cell from the Equipment menu.
1. Choose Silver for the Left Electrode and for the Right Electrode.
2. Select the Half-Cell Assembly and add 4 ml of 0.1 M Silver nitrate to the test tube
from the Chemical menu.
3. Obtain a 100 ml beaker and add 40 ml of 1.0 M Silver nitrate from the Chemical
menu.
4. Place the Half-Cell Assembly within the beaker by selecting and combining them.
5. Read the voltage from the meter. Record the value and note the positive electrode
(i.e. Left or Right) in Observations.
6. Empty the 100 ml beaker and add 40 ml of 0.1 M Silver nitrate. (First you must
Remove the Half-Cell Assembly from it.)
7. Repeat steps (4) and (5)
8. Pour the contents of the 100 ml beaker into a 600 ml beaker and fill with water to
the 400 ml level.
9. Repeat steps (4) and (5)
Electrochemical Cell
Part II: Concentration Cell
Name:__________
Sect:____________
Observations
Data
Note: All observations taken at STP (25oC and 100 kPa)
Cell Concentrations
Measured
Voltage
Ag / 1.0 M Ag(NO3) // 0.1 M Ag(NO3) / Ag
V
Ag / 0.1 M Ag(NO3) // 0.1 M Ag(NO3) / Ag
V
Ag / 0.01 M Ag(NO3) // 0.1 M Ag(NO3) / Ag
V
Questions
1. Proved a definition for Concentration Cell
2. What is the overall cell reaction of a silver concentration cell?
3. Why is this reaction spontaneous when the half-cell concentrations are different?
4. Identify the anode and the cathode when the half-cell concentrations are different.
Use care in your answer because both electrodes are identical metals.
5. Use the Nernst equation to calculate the expected emf from the silver concentration
cell for your three systems. Show calculations on a separate sheet of paper. Fill in
table:
Cell Concentrations
Calculated
Voltage
Ag / 1.0 M Ag(NO3) // 0.1 M Ag(NO3) / Ag
V
Ag / 0.1 M Ag(NO3) // 0.1 M Ag(NO3) / Ag
V
Ag / 0.01 M Ag(NO3) // 0.1 M Ag(NO3) / Ag
V
6. Compare the calculated emf with the observed emf. Provide a tentative explanation
of any major disagreement.
Discussion