Electrochemical Cells and Potentials

Learning Goals

You will learn how electrochemical cells produce electrical energy and how to estimate the standard potential of a cell.


We could construct an electrochemical cell to either produce electrical energy to do work or to use electrical energy as a driving force for an electrochemical reaction. A galvanic cell is an electrochemical cell that produces electrical energy. Electrons flow from regions of high electrical potential to regions of low electrical potential. The electromotive force that drives the electrons is due to the difference in electrical potential between electrodes (given the symbol E). The electrical work is given by the product of the charge and the potential energy difference: W = C * E. We use the symbol C for the charge, which is expressed in coulombs. The units of E are volts or joules/coulomb so the product of C and E has the units of joules.

We want to maximize W so we want to maximize C and E. E is an extensive property, which means that it is a function of the concentration of the reactants and products. We want to specify standard conditions and we will be consistent and use the same standard conditions that we did for enthalpy and the Gibbs free energy. We will then use the symbol Eo if the reactants and products are pure solids, or if they are in solution at a concentration of 1.0 M, or if they are gases at 1.0 bar. Eo is positive for product-favored reactions and defined in terms of our reaction Gibbs free energy as:

Gorxn = -nFE, where F is the Faraday constant (0.6485309 x 104 J/vmol)

For example, for the reaction:

Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s), Eo = 1.10 v, so Gorxn

would be negative in accord with our previous requirement for a reaction that is spontaneous in the product direction. If the reaction is reversed the absolute value of Eo and Gorxn remains the same, but the sign is reversed.

We know that for a redox reaction, the net reaction is the sum of the half-reactions such as:
Anode, oxidation Zn(s) Zn2+(aq) + 2e-
Cathode, reduction Cu2+(aq) + 2e- Cu(s)
Net Zn(s) + Cu2+(aq) Cu(s) + Zn2+(aq)

If we knew the potentials of each half-cell, then we could add them to predict the cell potential. However we can't measure the potential of a half-cell by itself for what is measured is the potential difference between two cellsórather like trying to predict the maximum speed that a new car design could obtain without considering the motor. But resourceful scientists don't give up easily. If we can't measure the half-cell potential alone, can we develop a standard half-cell that could be hooked up to any other half-cell and then give us a measure of the needed half-cell potential in reference to the standard? The answer is yes. An ingenious standard half-cell was developed that can serve as both a reducing agent or as an oxidizing agent to any other half-cell. The standard half-cell is the standard hydrogen electrode (SHE):

2 H3O+(aq, 1M) + 2e- H2(g, 1bar) + 2 H2O(liq)

As written, H2 is the reducing agent (it is oxidized) and H3O+ is the oxidizing agent (it is reduced). The potential of this half-cell is assigned the value of 0.00 so that whatever voltage is measured when this cell is hooked up to another cell is assigned as the half-cell potential of the other half-cell. If the SHE is attached to another half-cell so that H2 is the oxidizing agent and H3O+ is the reducing agent, then the above reaction is reversed, but the assigned potential is still 0.00.

We can now attach a half-cell to the SHE and then determine the potential of the half-cell. We can use a cell diagram to indicate two half-cells connected by a salt bridge that shows the oxidized and reduced components of each half-cell:

Zn2+(aq, 1M)/Zn(s) | | H2(g, 1bar)/H3O+(aq, 1M)

A voltmeter is used to measure the potential of this cell. The voltmeter is constructed so as to only read a value if its positive terminal is attached to the positive electrode of the cell and its negative terminal is attached to the negative electrode of the cell. Thus the voltmeter tells us the voltage (always positive) and the source of electrons in the cell. Figure 21.9 illustrates this situation. Since Zn is a better reducing agent than H2 gas, Zn is the electrode. The measured voltage is 0.76 v and this value is then assigned as the Zn2+/Zn half-cell standard potential Eo.

Similarly we can determine that the half-cell potential, Eo, for the Cu/Cu2+ half-cell is 0.76 volts.

Review Question

Calculate the Eo for the cell:

Zn(s) + Ag+(aq, 1M) Zn2+(aq, 1M) + Ag(s)

On-Line Activity

Here's a good site for review of this material: review


Web Author: Dr. Leon L. Combs
Copyright ©2000 by Dr. Leon L. Combs - ALL RIGHTS RESERVED