Galvanic Cells

Galvanic cells harness the electrical energy available from the electron transfer in a redox reaction to perform useful electrical work. The key to gathering the electron flow is to separate the oxidation and reduction half-reactions, connecting them by a wire, so that the electrons must flow through that wire. That electron flow, called a current, can be sent through a circuit which could be part of any number of electrical devices such as radios, televisions, watches, etc.

The figure below shows two typical setups for galvanic cells. The left hand cell diagram shows and oxidation and a reduction half-reaction joined by both a wire and a porous disk, while the right hand cell diagram shows the same cell substituting a salt bridge for the porous disk.

The salt bridge or porous disk is necessary to maintain the charge neutrality of each half-cell by allowing the flow of ions with minimal mixing of the half-cell solutions. As electrons are transferred from the oxidation half-cell to the reduction half-cell, a negative charge builds in the reduction half-cell and a positive charge in the oxidation half-cell. That charge buildup would serve to oppose the current from anode to cathode-- effectively stopping the electron flow--if the cell lacked a path for ions to flow between the two solutions.

The above figure points out that the electrode in the oxidation half-cell is called the anode and the electrode in the reduction half-cell is called the cathode. A good mnemonic to help remember that is "The Red Cat ate An Ox" meaning reduction takes place at the cathode and oxidation takes place at the anode.

The anode, as the source of the negatively charged electrons is usually marked with a minus sign (-) and the cathode is marked with a plus sign (+). Physicists define the direction of current flow as the flow of positive charge based on an 18th century understanding of electricity. As we now know, negatively charged electrons flow in a wire. Therefore, chemists indicate the direction of electron flow on cell diagrams and not the direction of current. To make that point clear, the direction of electron flow is indicated on with a arrow and the symbol for an electron, e^- .

Instead of drawing a cell diagram such as or chemists have devised a shorthand way of completely describing a cell called line notation. This notation scheme places the constituents of the cathode on the right and the anode components on the left. The phases of all reactive species are listed and their concentrations or pressures are given if those species are not in their standard states (i.e. 1 atm. for gasses and 1M for solutions). All phase interfaces are noted with a single line ( | ) and multiple species in a single phase are separated by commas. For example, a half-cell containing 1M solutions of CuO and HCl and a Pt electrode for the reduction of Cu²⁺ would be written as:

Pt (s) | Cu²⁺ (aq), H⁺ (aq)

Note that the spectator ions, oxide and chloride, have been left out of the notation and that the anode is written to the far left.

The salt bridge or porous disk is shown in the notation as a double line ( || ). Therefore, a cell that undergoes the oxidation of magnesium by Al³⁺ could have the following cell notation if the anode is magnesium and the cathode is aluminum:

Mg (s) | Mg²⁺ (aq) || Al³⁺ (aq) | Al (s)

One can measure the cell potential, E_cell, in volts, of any galvanic cell with the aid of a potentiometer. However, it is impossible to directly measure the potential of each individual half-cell. Chemists, however, have devised a method to measure the ability of a chemical species to reduce another by compiling tables of standard reduction potentials, E^o (the ^o indicates that the reaction is at standard state). Arbitrarily assigning a value of exactly zero to the potential of the standard hydrogen electrode allows us to measure the E^o of any half- reaction. That measurement is made by constructing a galvanic cell between the SHE and the unknown half-cell at standard state conditions. For example, when the following cell is constructed (see Heading for a review of the line notation), an E^o_cell of 0.34 V is observed (note the setup of the SHE as the anode because Cu²⁺ has a greater reduction potential than H⁺):

Pt (s) | H₂ (g) | H⁺ (aq) || Cu²⁺ (aq) | Cu (s)

Because the SHE has a potential of exactly zero volts, as defined above, the reaction:

has a value of 0.34 V for its E^o (recall that E^o_cell = E^o_SHE + E^o). Fortunately, every important reduction potential has been measured and tabulated. Useful lists of reduction potentials are available in most introductory chemistry texts, including yours. In this SparkNote, all potentials will be given to you if you need them.

Those tables of standard reduction potentials list all half-reactions as reductions. Half- reactions with the largest reduction potential are placed at the top of the list and the smallest (most negative) reduction potentials are at the bottom. Those species on the left-hand side of the equations at the top of the list are the most easily reduced (like F₂, or H₂O₂) and those at the bottom are the least readily reduced (like Li⁺).

Take a look at the list of standard reduction potentials in your chemistry text. An intuitive trend should be obvious when looking at the data--electronegative species (those with the greatest attraction for electrons) are easily reduced, i.e. given an electron. The most electronegative atom, F, has the largest reduction potential whereas one of the least electronegative atoms, Li, has the smallest reduction potential.

By compiling a list of standard reduction potentials of all possible reductions, one can, at least in theory, calculate the cell potential, E^o_cell, of any arbitrary redox reaction. By knowing the sign of E^o_cell, we can predict whether a reaction is spontaneous at standard conditions. If E^o_cell is positive, then the reaction is spontaneous. Conversely, if E^o_cell is negative, then the reaction is non-spontaneous as written but spontaneous in the reverse direction (see Thermodynamics, Electrical Work and Cell Potential for an explanation of why that is so).

To compute the cell potential of a reaction at standard conditions, E^o_cell, you do not need to balance the equation of your redox reaction. However, as we will learn in Thermodynamics of Electrochemistry, if the reaction is not conducted at standard state, then it is essential to balance the redox reaction to compute its cell potential. For now, let's assume that all reactions are conducted at standard conditions unless otherwise specified.

When asked to compute the cell potential for a reaction, you will need to be able to separate the overall reaction into its oxidation and reduction half-reactions as in .

Once those half reactions are separated, then find the reduction potential for the reaction written as a reduction. As you can see in , one reaction is written as an oxidation. For that reaction, you need to calculate its oxidation potential. To calculate an oxidation potential, simply reverse the sign of the E^o for the corresponding reduction reaction (just the oxidation written in the opposite direction). Simply add the reduction potential of the reduction and the oxidation potential of the oxidation to calculate the E^o_cell. It is important to note here that E^o's are intrinsic properties of reactions and, therefore, do not depend on the stoichiometry of the reaction. That means that you DO NOT multiply the E^o of a reaction by the coefficient used to balance the overall redox reaction. A proof of that point is provided in Thermodynamics of Electrochemistry#. Multiplying the value of E^o for a half-reaction is the number one mistake made in calculating E^o_cell. Please, don't let that happen to you! Simply read off the values of E^o for the oxidation and reduction half-reactions and add those two values together, as in . /PARAGRAPH