# 20.1: Electrode Potentials and their Measurement

Learning Objectives

• To distinguish between galvanic and electrolytic cells.

In any electrochemical process, electrons flow from one chemical substance to another, driven by an oxidation–reduction (redox) reaction. A redox reaction occurs when electrons are transferred from a substance that is oxidized to one that is being reduced. The reductant is the substance that loses electrons and is oxidized in the process; the oxidant is the species that gains electrons and is reduced in the process. The associated potential energy is determined by the potential difference between the valence electrons in atoms of different elements.

Because it is impossible to have a reduction without an oxidation and vice versa, a redox reaction can be described as two half-reactions, one representing the oxidation process and one the reduction process. For the reaction of zinc with bromine, the overall chemical reaction is as follows:

$Zn_{(s)} + Br_{2(aq)} \rightarrow Zn^{2+}_{(aq)} + 2Br^−_{(aq)} \label{19.1}$

The half-reactions are as follows:

reduction half-reaction:

$Br_{2(aq)} + 2e^− \rightarrow 2Br^−_{(aq)} \label{19.2}$

oxidation half-reaction:

$Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^− \label{19.3}$

Each half-reaction is written to show what is actually occurring in the system; Zn is the reductant in this reaction (it loses electrons), and Br2 is the oxidant (it gains electrons). Adding the two half-reactions gives the overall chemical reaction (Equation $$\ref{19.1}$$). A redox reaction is balanced when the number of electrons lost by the reductant equals the number of electrons gained by the oxidant. Like any balanced chemical equation, the overall process is electrically neutral; that is, the net charge is the same on both sides of the equation.

Note

In any redox reaction, the number of electrons lost by the reductant equals the number of electrons gained by the oxidant.

In most of our discussions of chemical reactions, we have assumed that the reactants are in intimate physical contact with one another. Acid–base reactions, for example, are usually carried out with the acid and the base dispersed in a single phase, such as a liquid solution. With redox reactions, however, it is possible to physically separate the oxidation and reduction half-reactions in space, as long as there is a complete circuit, including an external electrical connection, such as a wire, between the two half-reactions. As the reaction progresses, the electrons flow from the reductant to the oxidant over this electrical connection, producing an electric current that can be used to do work. An apparatus that is used to generate electricity from a spontaneous redox reaction or, conversely, that uses electricity to drive a nonspontaneous redox reaction is called an electrochemical cell.

There are two types of electrochemical cells: galvanic cells and electrolytic cells. Galvanic cells are named for the Italian physicist and physician Luigi Galvani (1737–1798), who observed that dissected frog leg muscles twitched when a small electric shock was applied, demonstrating the electrical nature of nerve impulses. A galvanic (voltaic) cell uses the energy released during a spontaneous redox reaction (ΔG < 0) to generate electricity. This type of electrochemical cell is often called a voltaic cell after its inventor, the Italian physicist Alessandro Volta (1745–1827). In contrast, an electrolytic cell consumes electrical energy from an external source, using it to cause a nonspontaneous redox reaction to occur (ΔG > 0). Both types contain two electrodes, which are solid metals connected to an external circuit that provides an electrical connection between the two parts of the system (Figure $$\PageIndex{1}$$). The oxidation half-reaction occurs at one electrode (the anode), and the reduction half-reaction occurs at the other (the cathode). When the circuit is closed, electrons flow from the anode to the cathode. The electrodes are also connected by an electrolyte, an ionic substance or solution that allows ions to transfer between the electrode compartments, thereby maintaining the system’s electrical neutrality. In this section, we focus on reactions that occur in galvanic cells. Figure $$\PageIndex{1}$$: Electrochemical Cells. A galvanic cell (left) transforms the energy released by a spontaneous redox reaction into electrical energy that can be used to perform work. The oxidative and reductive half-reactions usually occur in separate compartments that are connected by an external electrical circuit; in addition, a second connection that allows ions to flow between the compartments (shown here as a vertical dashed line to represent a porous barrier) is necessary to maintain electrical neutrality. The potential difference between the electrodes (voltage) causes electrons to flow from the reductant to the oxidant through the external circuit, generating an electric current. In an electrolytic cell (right), an external source of electrical energy is used to generate a potential difference between the electrodes that forces electrons to flow, driving a nonspontaneous redox reaction; only a single compartment is employed in most applications. In both kinds of electrochemical cells, the anode is the electrode at which the oxidation half-reaction occurs, and the cathode is the electrode at which the reduction half-reaction occurs.

## Galvanic (Voltaic) Cells

To illustrate the basic principles of a galvanic cell, let’s consider the reaction of metallic zinc with cupric ion (Cu2+) to give copper metal and Zn2+ ion. The balanced chemical equation is as follows:

$Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)} \label{19.4}$

We can cause this reaction to occur by inserting a zinc rod into an aqueous solution of copper(II) sulfate. As the reaction proceeds, the zinc rod dissolves, and a mass of metallic copper forms. These changes occur spontaneously, but all the energy released is in the form of heat rather than in a form that can be used to do work.

This same reaction can be carried out using the galvanic cell illustrated in part (a) in Figure $$\PageIndex{3}$$. To assemble the cell, a copper strip is inserted into a beaker that contains a 1 M solution of Cu2+ ions, and a zinc strip is inserted into a different beaker that contains a 1 M solution of Zn2+ ions. The two metal strips, which serve as electrodes, are connected by a wire, and the compartments are connected by a salt bridge, a U-shaped tube inserted into both solutions that contains a concentrated liquid or gelled electrolyte. The ions in the salt bridge are selected so that they do not interfere with the electrochemical reaction by being oxidized or reduced themselves or by forming a precipitate or complex; commonly used cations and anions are Na+ or K+ and NO3 or SO42−, respectively. (The ions in the salt bridge do not have to be the same as those in the redox couple in either compartment.) When the circuit is closed, a spontaneous reaction occurs: zinc metal is oxidized to Zn2+ ions at the zinc electrode (the anode), and Cu2+ ions are reduced to Cu metal at the copper electrode (the cathode). As the reaction progresses, the zinc strip dissolves, and the concentration of Zn2+ ions in the Zn2+ solution increases; simultaneously, the copper strip gains mass, and the concentration of Cu2+ ions in the Cu2+ solution decreases (part (b) in Figure $$\PageIndex{3}$$). Thus we have carried out the same reaction as we did using a single beaker, but this time the oxidative and reductive half-reactions are physically separated from each other. The electrons that are released at the anode flow through the wire, producing an electric current. Galvanic cells therefore transform chemical energy into electrical energy that can then be used to do work. Figure $$\PageIndex{2}$$: The Reaction of Metallic Zinc with Aqueous Copper(II) Ions in a Galvanic Cell. (a) A galvanic cell can be constructed by inserting a copper strip into a beaker that contains an aqueous 1 M solution of Cu2+ ions and a zinc strip into a different beaker that contains an aqueous 1 M solution of Zn2+ ions. The two metal strips are connected by a wire that allows electricity to flow, and the beakers are connected by a salt bridge. When the switch is closed to complete the circuit, the zinc electrode (the anode) is spontaneously oxidized to Zn2+ ions in the left compartment, while Cu2+ ions are simultaneously reduced to copper metal at the copper electrode (the cathode). (b) As the reaction progresses, the Zn anode loses mass as it dissolves to give Zn2+(aq) ions, while the Cu cathode gains mass as Cu2+(aq) ions are reduced to copper metal that is deposited on the cathode.

The electrolyte in the salt bridge serves two purposes: it completes the circuit by carrying electrical charge and maintains electrical neutrality in both solutions by allowing ions to migrate between them. The identity of the salt in a salt bridge is unimportant, as long as the component ions do not react or undergo a redox reaction under the operating conditions of the cell. Without such a connection, the total positive charge in the Zn2+ solution would increase as the zinc metal dissolves, and the total positive charge in the Cu2+ solution would decrease. The salt bridge allows charges to be neutralized by a flow of anions into the Zn2+ solution and a flow of cations into the Cu2+ solution. In the absence of a salt bridge or some other similar connection, the reaction would rapidly cease because electrical neutrality could not be maintained.

A voltmeter can be used to measure the difference in electrical potential between the two compartments. Opening the switch that connects the wires to the anode and the cathode prevents a current from flowing, so no chemical reaction occurs. With the switch closed, however, the external circuit is closed, and an electric current can flow from the anode to the cathode. The potential (Ecell) of the cell, measured in volts, is the difference in electrical potential between the two half-reactions and is related to the energy needed to move a charged particle in an electric field. In the cell we have described, the voltmeter indicates a potential of 1.10 V (part (a) in Figure $$\PageIndex{3}$$). Because electrons from the oxidation half-reaction are released at the anode, the anode in a galvanic cell is negatively charged. The cathode, which attracts electrons, is positively charged.

Note

A galvanic (voltaic) cell converts the energy released by a spontaneous chemical reaction to electrical energy. An electrolytic cell consumes electrical energy from an external source to drive a nonspontaneous chemical reaction.

Not all electrodes undergo a chemical transformation during a redox reaction. The electrode can be made from an inert, highly conducting metal such as platinum to prevent it from reacting during a redox process, where it does not appear in the overall electrochemical reaction. This phenomenon is illustrated in Example $$\PageIndex{1}$$.

Example $$\PageIndex{1}$$

A chemist has constructed a galvanic cell consisting of two beakers. One beaker contains a strip of tin immersed in aqueous sulfuric acid, and the other contains a platinum electrode immersed in aqueous nitric acid. The two solutions are connected by a salt bridge, and the electrodes are connected by a wire. Current begins to flow, and bubbles of a gas appear at the platinum electrode. The spontaneous redox reaction that occurs is described by the following balanced chemical equation:

$3Sn_{(s)} + 2NO^-_{3(aq)} + 8H^+_{(aq)} \rightarrow 3Sn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}$

For this galvanic cell,

1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is the positive electrode and which is the negative electrode.

Given: galvanic cell and redox reaction

Asked for: half-reactions, identity of anode and cathode, and electrode assignment as positive or negative

Strategy:

1. Identify the oxidation half-reaction and the reduction half-reaction. Then identify the anode and cathode from the half-reaction that occurs at each electrode.
2. From the direction of electron flow, assign each electrode as either positive or negative.

Solution:

1. A In the reduction half-reaction, nitrate is reduced to nitric oxide. (The nitric oxide would then react with oxygen in the air to form NO2, with its characteristic red-brown color.) In the oxidation half-reaction, metallic tin is oxidized. The half-reactions corresponding to the actual reactions that occur in the system are as follows:

reduction: NO3(aq) + 4H+(aq) + 3e → NO(g) + 2H2O(l)

oxidation: Sn(s) → Sn2+(aq) + 2e

Thus nitrate is reduced to NO, while the tin electrode is oxidized to Sn2+.

1. Because the reduction reaction occurs at the Pt electrode, it is the cathode. Conversely, the oxidation reaction occurs at the tin electrode, so it is the anode.
2. B Electrons flow from the tin electrode through the wire to the platinum electrode, where they transfer to nitrate. The electric circuit is completed by the salt bridge, which permits the diffusion of cations toward the cathode and anions toward the anode. Because electrons flow from the tin electrode, it must be electrically negative. In contrast, electrons flow toward the Pt electrode, so that electrode must be electrically positive.

Exercise $$\PageIndex{1}$$

Consider a simple galvanic cell consisting of two beakers connected by a salt bridge. One beaker contains a solution of MnO4 in dilute sulfuric acid and has a Pt electrode. The other beaker contains a solution of Sn2+ in dilute sulfuric acid, also with a Pt electrode. When the two electrodes are connected by a wire, current flows and a spontaneous reaction occurs that is described by the following balanced chemical equation:

$2MnO^−_{4(aq)} + 5Sn^{2+}_{(aq)} + 16H^+_{(aq)} \rightarrow 2Mn^{2+}_{(aq)} + 5Sn^{4+}_{(aq)} + 8H_2O_{(l)}$

For this galvanic cell,

1. write the half-reaction that occurs at each electrode.
2. indicate which electrode is the cathode and which is the anode.
3. indicate which electrode is positive and which is negative.

1. MnO4(aq) + 8H+(aq) + 5e → Mn2+(aq) + 4H2O(l); Sn2+(aq) → Sn4+(aq) + 2e
2. The Pt electrode in the permanganate solution is the cathode; the one in the tin solution is the anode.
3. The cathode (electrode in beaker that contains the permanganate solution) is positive, and the anode (electrode in beaker that contains the tin solution) is negative.

## Constructing Cell Diagrams

Because it is somewhat cumbersome to describe any given galvanic cell in words, a more convenient notation has been developed. In this line notation, called a cell diagram, the identity of the electrodes and the chemical contents of the compartments are indicated by their chemical formulas, with the anode written on the far left and the cathode on the far right. Phase boundaries are shown by single vertical lines, and the salt bridge, which has two phase boundaries, by a double vertical line. Thus the cell diagram for the Zn/Cu cell shown in part (a) in Figure $$\PageIndex{3}$$ is written as follows: Figure $$\PageIndex{3}$$.

A cell diagram includes solution concentrations when they are provided.

Galvanic cells can have arrangements other than the examples we have seen so far. For example, the voltage produced by a redox reaction can be measured more accurately using two electrodes immersed in a single beaker containing an electrolyte that completes the circuit. This arrangement reduces errors caused by resistance to the flow of charge at a boundary, called the junction potential. One example of this type of galvanic cell is as follows:

$Pt_{(s)}\, | \, H_{2(g)} | HCl_{(aq)}\,|\, AgCl_{(s)} \,Ag_{(s)} \label{19.5}$

This cell diagram does not include a double vertical line representing a salt bridge because there is no salt bridge providing a junction between two dissimilar solutions. Moreover, solution concentrations have not been specified, so they are not included in the cell diagram. The half-reactions and the overall reaction for this cell are as follows:

cathode reaction:

$AgCl_{(s)} + e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)} \label{19.6}$

anode reaction:

$\dfrac{1}{2}\,\mathrm{H_{2(g)}}\rightarrow\mathrm{H^+_{(aq)}}+\mathrm{e^-} \label{19.7}$

overall:

$\mathrm{AgCl_{(s)}}+\dfrac{1}{2}\mathrm{H_{2(g)}}\rightarrow\mathrm{Ag_{(s)}}+\mathrm{Cl^-_{(aq)}}+\mathrm{H^+_{(aq)}} \label{19.8}$

A single-compartment galvanic cell will initially exhibit the same voltage as a galvanic cell constructed using separate compartments, but it will discharge rapidly because of the direct reaction of the reactant at the anode with the oxidized member of the cathodic redox couple. Consequently, cells of this type are not particularly useful for producing electricity.

Example $$\PageIndex{2}$$

Draw a cell diagram for the galvanic cell described in Example 1. The balanced chemical reaction is as follows:

$3Sn_{(s)} + 2NO^−_{3(aq)} + 8H^+_{(aq)} \rightarrow 3Sn^{2+}_{(aq)} + 2NO_{(g)} + 4H_2O_{(l)}$

Given: galvanic cell and redox reaction

Strategy:

Using the symbols described, write the cell diagram beginning with the oxidation half-reaction on the left.

Solution

The anode is the tin strip, and the cathode is the Pt electrode. Beginning on the left with the anode, we indicate the phase boundary between the electrode and the tin solution by a vertical bar. The anode compartment is thus Sn(s)∣Sn2+(aq). We could include H2SO4(aq) with the contents of the anode compartment, but the sulfate ion (as HSO4) does not participate in the overall reaction, so it does not need to be specifically indicated. The cathode compartment contains aqueous nitric acid, which does participate in the overall reaction, together with the product of the reaction (NO) and the Pt electrode. These are written as HNO3(aq)∣NO(g)∣Pt(s), with single vertical bars indicating the phase boundaries. Combining the two compartments and using a double vertical bar to indicate the salt bridge,

$Sn_{(s)}\,|\,Sn^{2+}_{(aq)}\,||\,HNO_{3(aq)}\,|\,NO_{(g)}\,|\,Pt_{(s)}$

The solution concentrations were not specified, so they are not included in this cell diagram.

Exercise $$\PageIndex{2}$$

Draw the cell diagram for the following reaction, assuming the concentration of Ag+ and Mg2+ are each 1 M:

$Mg_{(s)} + 2Ag^+_{(aq)} \rightarrow Mg^{2+}_{(aq)} + 2Ag_{(s)}$

$Mg(s) \,|\,Mg^{2+}(aq, 1 M \,||\,Ag^+(aq, 1 M)\,|\,Ag(s)$