Electrochemistry: Electrochemical Cells

E17 Objective

! Construct electrochemical cells based on two half-reactions that are physically separated so that electrons are transferred externally between the two half-cells.

! Examine the relation between cell voltages and the concentrations of the components of the electrochemical cell.

! Construct a non-spontaneous reaction using electrochemical half-cells measure the cell voltage when coupled and calculate the value of an equilibrium constant, Ksp.

Discussion In our everyday lives, we use electrochemical cells or their products without thinking about them. When you turn the ignition key of your car, the starter motor is powered by the current from a lead–acid battery (see text, Chapter 20). Some cars (Tesla) run solely on stored electrical energy. The chromium plate on the trim of some cars was deposited electrochemically. The aluminum in the engine parts was produced in an electrolytic cell. Electrochemical reactions are involved in all of these processes, which either use electrical energy to produce chemical substances or vice versa. Some electrochemical cells Oxidation-reduction reactions take place in electrochemical cells. For example, when zinc is oxidized by copper(II) ion, the zinc atom loses two electrons and the copper(II) ion gains two electrons. We can express this electron transfer as two separate half-reactions: Oxidation: Zn(s) → Zn2+ + 2 e- Reduction: Cu2+ + 2 e- → Cu(s) The sum of the two half-reactions gives the overall (net) chemical reaction Zn(s) + Cu2+ → Cu(s) + Zn2+ As discussed in Experiment 16-Redox Reactions, the overall (net) reaction does not contain any electrons because all of the electrons lost by the zinc are gained by copper(II) ion. An electrochemical cell is simply a device used to physically separate an electrochemical reaction into two component half-reactions in such a way that the electrons are transferred through an external circuit rather than directly between reactants in the same solution. If the chemical reaction proceeds spontaneously, creating a current flow of electrons in the external circuit, we have what is referred to as a Voltaic or Galvanic cell. It’s interesting that we can construct electrochemical cells based on chemical reactions that we do not ordinarily regard as oxidation–reduction reactions but precipitation. Consider the electrochemical (precipitation) reaction: Ag+(aq) + Cl-(aq) → AgCl(s) can constructed from an electrochemical cell based on the following half-reactions, (reduction): Ag+(aq) + e- → Ag(s) (oxidation): Ag(s) + Cl-(aq) → AgCl(s) + e-

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In the electrochemical cell, the net reaction is the combination of Ag+ with Cl- to form AgCl, which we ordinarily think of as a precipitation reaction, not an oxidation–reduction reaction. Or consider the cell based on these two half-reactions: (reduction): Cu2+(1 M) + 2 e- → Cu(s) (oxidation): Cu(s) → Cu2+(0.1 M) + 2 e- (net cell reaction): Cu2+(1 M) → Cu2+(0.1 M) This is called a concentration electrochemical cell, in which there appears to be a net transfer of Cu2+ ions from the more concentrated to the more dilute solution. This appears not to be a chemical reaction at all! In a general an electrochemical cell can be based on any reaction or process that can be separated into two half-reactions involving the transfer of electrons to or from an external circuit. An apparatus for carrying out the reaction of Zn with Cu2+ in an electrochemical cell is shown in Figure-1. In the cell, oxidation takes place at the zinc electrode (the anode),1 liberating electrons to the external circuit. Reduction takes place at the copper electrode (the cathode),2 consuming electrons

Figure-1. The Daniell cell, a simple electrochemical cell that transforms the energy liberated by a chemical reaction into electrical energy. coming from the external circuit. By isolating each half reaction in its own compartment, we have arranged things so that electron transfer must take place through the external circuit made of metallic wire. It is not possible for the electrons to travel long distances through the solution because they are 1 This statement may be regarded as a definition of anode: The anode is the electrode at which oxidation takes place. 2 This statement may be regarded as a definition of cathode: The cathode is the electrode at which reduction takes place.

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much too reactive, reacting rapidly with water to reduce it (e- + H2O → ½ H2 + OH-). The two half-cells are connected by a salt bridge. The current flow in solution through the salt bridge (K2SO4) consists of positive ions moving in one direction and negative ions moving in the opposite direction. This ionic current flow in solution is a direct result of electron transfer that takes place at the surface of the electrodes. As current flows in the cell, there is a tendency for excess positive charge (in the form of Zn2+ ions) to accumulate in solution around the zinc anode as zinc atoms are oxidized. Likewise, excess negative charge (in the form of SO4

2- ions) accumulates around the copper cathode as Cu2+ ions are removed from solution by reduction to copper metal. These excess charges create an electric field that causes the ions to migrate through the salt bridge, positive ions (cations) migrating toward the cathode and negative ions (anions) migrating toward the anode. This migration of ions between the two compartments of the cell constitutes the cell current in the solution. To summarize, ions are the charge carriers in solution, and electrons are the charge carriers in the external circuit.3 It is important to realize that the net chemical result of the operation of the cell shown in Figure-1 is exactly the same as the net stoichiometric reaction: One Cu2+ ion is reduced for each zinc atom that is oxidized. But there are some important nonchemical differences between the direct reaction and the reaction carried out in the cell. By using the cell, we are able to convert a most of the chemical energy of the reaction directly into electrical energy that can be used to do useful work (like running an electric motor). The direct chemical redox reaction wastes all of the chemical energy as heat—the random thermal motion of the metallic atoms and the ions in solution. There is another practical difference in the two reactions. In the direct reaction, copper is plated out on the zinc metal as the reaction proceeds, so that after a period of time the zinc atoms get coated with a layer of copper until the reaction slows down and practically stops. In the cell, the reaction can proceed until the cell reaches equilibrium (where the cell voltage is zero). For this reaction, where equilibrium lies far to the right, current would flow either until the Zn anode is practically consumed or until practically all of the Cu2+ ions are plated out on the copper cathode. Line Notation for a Cell We can always construct, at least in principle, a half-cell that corresponds to a particular redox half-reaction. For example, the redox half-reactions that take place in the Daniell cell are Cu2+ + 2 e- → Cu(s) Zn(s) → Zn2+ + 2 e- The sum of these two half-reactions gives the net chemical reaction Zn(s) + Cu2+ → Zn2+ + Cu(s) We will represent such a cell by a line diagram Zn | ZnSO4 || CuSO4 | Cu in which a vertical bar (|) represents a phase boundary and the dashed vertical bar (| or !) represents the boundary between two miscible ionic solutions (liquid junction). A double dashed vertical bar (||) will be used to represent a double liquid junction through an intermediate ionic solution called a salt bridge. Salt bridges are used to minimize the liquid junction potential and to prevent mixing of the components of two half-cells. 3 In the wires, electrons are the charge carriers, and the flow of current in a wire consists entirely of a flow of negative charge. Since current is conventionally defined as a flow of positive charge, the conventional current flow is opposite to the electron flow. The movement of an electron in one direction in the wire is equivalent to the movement of a hypothetical positive charge in the opposite direction.

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Cell Voltage4 The volt (V) is the unit of electrical potential, or driving force. The product of the voltage times the charge of the electron, e, is a measure of the work done when this unit electric charge is transferred from one substance to another. The voltage of a cell—sometimes called its electromotive force (emf) or potential—is thus a quantitative value expressing the tendency of the chemical reaction occurring in the cell to take place. The magnitude of this voltage depends on the relative strengths of the oxidizing and reducing agents. If the oxidizing agent has an affinity for electrons that is stronger than the tendency of the reducing agent to hold electrons, the electrical potential, or voltage, is correspondingly large. Standard Electrode Potentials although we cannot measure a single half-cell potential, we can construct a scale of half-cell potentials by choosing a single reference half-cell and measuring the potential of all other half-cells with respect to it. The reference half-cell chosen is based on the half-reaction 2 H+ + 2 e- " H2(g) This redox couple is arbitrarily assigned a potential of zero, so that the total cell voltage is ascribed to the other couple. For example, in a cell composed of the Zn2+/Zn half-reaction and the H+/H2 half-reaction, where all species are at unit activity,5 the potential of the cell is found to be -0.763 V, with the zinc electrode being more negative than the hydrogen electrode. This value, -0.763 V, is called the standard electrode potential for the Zn2+/Zn couple (see Figure 2). All standard electrode potentials are the values of the voltage obtained when all substances in solution are present at unit activity (approximately 1 M), all gases are at unit fugacity (approximately 1 atm pressure), and the temperature is at a fixed, convenient value, usually 25 ˚C. Figure 2 On the potential scale, the H+/H2 half-reaction is arbitrarily assigned the value zero. The zinc electrode has a voltage of -0.76 V measured against this standard hydrogen electrode (SHE). This value is assigned as the standard electrode potential of the Zn2+/Zn couple. This procedure is analogous to measuring elevation from sea level (rather than from the center of the earth), with sea level being assigned zero in the scale of elevation. The saturated calomel electrode (SCE) is more often used as a practical reference electrode.

4 The voltmeter used to measure the cell voltage must draw only a small current from the cell, in order not to load the cell and change the concentrations at the electrode surface. The cell voltage under load will be smaller than the open-circuit voltage. 5 The activity of an ion in solution is usually less than the molar concentration because of ionic interactions. It may be simply thought of as the “effective concentration.”

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The Standard Hydrogen Electrode (SHE) and Practical Reference Electrodes A practical hydrogen reference electrode is shown in Figure 3, but it is not possible to construct a standard hydrogen electrode (SHE) whose composition corresponds to the arbitrarily chosen reference state.6 This is because the choice of the standard state for an electrolyte is a solution with a concentration that is 1 m and in which the ionic activity coefficient is also equal to unity. The standard state thus corresponds to a hypothetical solution (one that cannot be made in the laboratory), because the activity coefficients of real solutions are usually less than 1 for 1 m solutions. In order to obtain a standard electrode potential (E˚), we extrapolate measured cell voltages to conditions that correspond to the standard reference state. This procedure is too cumbersome for everyday measurements, so in practice we measure cell potentials versus reference half-cells whose potentials have been very accurately determined with respect to the SHE. The saturated calomel electrode (SCE) is a popular reference electrode based on the half-reaction Hg2Cl2(s) + 2 e- → 2 Hg(l) + 2 Cl- (sat. KCl(aq)) Hg | Hg2Cl2(s) | sat. KCl(aq)|| E = 0.244 V versus SHE at 25 ˚C

The saturated KCl electrolyte reduces liquid-junction potentials to a small and reproducible value (a few millivolts or less). Half-cell potentials measured versus the saturated calomel electrode are easily converted to the standard hydrogen electrode scale by adding +0.244 V (see Figure 2). Calculating the Standard Potential of an Electrochemical Cell When two half-cells are combined to make an electrochemical cell, the standard E˚ of the cell will equal the difference of the standard half-cell potentials of the two redox couples. For example, the E˚ for the electrochemical cell formed from the half-cell couples Zn2+/Zn and Cl2/Cl- is +2.122 V. Note that this is just the difference between the two standard half-cell potentials on the scale of redox potentials shown in Figure 36-2. The net reaction that occurs when the cell operates spontaneously is Zn(s) + Cl2(g) # Zn2+ + 2 Cl- E˚cell= E˚(red) - E˚(ox) = 1.359 V - (-0.763) V

Figure 3 Hydrogen gas, adsorbed on the platinum electrode = 2.122 V and in contact with 1 M H+, forms the reference half-cell. When this half-cell is coupled with a Zn electrode in contact with 1 M Zn2+ to form the cell Pt|H2|H+||Zn2+| Zn, the meter indicates that the zinc electrode is more negative than the hydrogen electrode. The cell voltage will not be the same as the theoretical E˚(-0.76 V versus SHE for the Zn2+/Zn half-reaction) because of activity and junction potential effects.

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The Effect of Concentration The voltage of the cell can be effected by changes in the concentration or pressure of the reactants and products when they are not in their standard states. The quantitative relationship between the voltage of a cell and the concentrations of the reactants and products, was introduced in 1889 by Walther Nernst (1864-1941) and is known as the Nernst Equation, Ecell = E˚cell – RT/nF(ln Q) or Ecell = E˚cell – 2.303RT/nF(log Q) 2.303RT/F = 0.05916 V at 25˚C (298.15 K) where Ecell is the measured voltage of the cell, E˚cell is the standard cell potential as calculated from the standard electrode potentials of the half-reactions, R is the universal gas constant (8.3144 J/mol·K), T is the absolute temperature, n is the nuber of meles of electrons transferred per mole of net electrochemical reaction, F is the (Faraday constant) number of coulombs of charge per mole of electrons (96,485 C), and Q is the reaction quotient of the products over the reactants. As an example of the application of the Nernst equation consider the Zn2+/Zn and Cl2/Cl- couples under non-standard conditions, Zn(s) + Cl2(g, 4 atm) → Zn2+(0.01 M) + 2 Cl-(1 M). Ecell = E˚cell – 0.05916 V/n{log [Zn2+][Cl-]2/P(Cl2)} Ecell = 2.122 V - 0.05916 V/2{log (0.010)(1)2/(4.0)} = 2.122 V + 0.077 V = 2.199 V Calculating an equilibrium constant by measuring the voltage of an electrochemical cell Electrochemical cells can be used to determine the equilibrium constant of chemical reactions. As an example of this, let’s consider a cell we construct in this experiment, composed of the half-reactions CuCO3(s) + 2 e- → Cu(s) + CO3

2- (1 M) _______Cu(s) → Cu2+ (1 M) + 2 e- CuCO3(s) → Cu2+ (1 M) + CO3

2- (1 M) (Net Cell Reaction) (1) The cell can be represented by the line notation Cu | Cu2+ || CO3

2- (1 M) | CuCO3(s) | Cu (2) The voltage of the cell is given by the Nernst equation: Ecell = E˚cell – 2.303RT/nF(log Q) (3) with Q = [Cu2+][ CO3

2-] = 1 (4) Because Q = 1 for the cell represented by Reaction (1), Equation (3) will reduce to Ecell = E˚. If the cell is allowed to operate spontaneously, Ecell will gradually decrease until it reaches zero. At this point, the chemical reaction has reached equilibrium and Q → Keq, the equilibrium constant for the chemical reaction. Therefore, at equilibrium we can rewrite Equation (3) as

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Ecell = 0 = E˚cell – 2.303RT/nF(log Keq) (5) rearranging we can write log Keq = nE˚/2.303RT/F = nE˚/0.05916 V (6) So if we know or measure E˚cell we can calculate the equilibrium constant. Experimental Procedure Special Supplies: Voltmeter; porous porcelain cup (3 in. tall, 1 in. and ½ in. in diameter) for half-cells; 150 mL Beaker for half-cell; strips of Cu, Zn, and Fe metal (or a large iron nail) for use as metal electrodes; 10 cm graphite rods for use as inert electrodes. Chemicals: 1.0 M copper(II) nitrate, Cu(NO3)2; 0.10 M copper(II) nitrate, Cu(NO3)2; 0.10 M zinc nitrate, Zn(NO3)2; 0.10 M iron(II) sulfate, FeSO4, prepared just at the beginning of the experiment); 0.10 M iron(III) chloride, FeCl3; bromine water, Br2 (saturated solution); 0.10 M potassium bromide, KBr; 0.05 M iodine, I2, dissolved in methanol; 0.10 M potassium iodide, KI; 6 M ammonia, NH3; 1 M sodium sulfide, Na2S; 1 M sodium carbonate, Na2CO3; 6 M HCl for cleaning. 1. Standard Cell Potentials In this part of the experiment you will measure the potentials of several cells composed of redox couples that you studied in Experiment 16.

Figure4.PorousCupsfor½-Cells Figure 5. Cell formed from nested porous cup/beaker. (a) M2+/M Half-Cells. For the experimental part of this section, prepare half-cells for Cu, Zn, and Fe by adding ~10 mL of 0.1 M Zn(NO3)2 to 150 mL beaker, ~10 mL 0.1 M Cu(NO3)2, and ~10 mL 0.1 M FeSO4 (freshly prepared), into separate porous cups (large and small) and placing electrodes of the appropriate metal in each of the three half-cells. (Clean the electrodes with sandpaper or steel wool. Clean an iron nail for the iron electrode, and alligator clips (as needed) with 6 M HCl, and rinse with deionized water.) (b) A Simple Daniell Cell. To make a simple simple Daniell cell, Zn | Zn(NO3)2||Cu(NO3)2 | Cu, place Cu2+/Cu half-cell-porous cup (nested) into the 150-mL beaker containing the Zn2+/Zn half-cell. Which of these two metals will give up its electrons more readily? To determine the answer, connect the metal electrodes by means of alligator clips and leads to the voltmeter. Read the meter carefully to the

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nearest 0.1 V, noting which electrode is the negative terminal (the electron source or anode). (As further proof of this polarity, recall the comparative behavior of Zn metal in Cu2+ solution, and of Cu metal in Zn2+ solution, from Experiment 16, part 1 of the Experimental Procedure.) Explain fully all aspects of the operation of the Daniell cell and complete a diagram of this cell in your Report form. Be sure you understand the following: (1) What constitutes the electric current in the wire? (2) What constitutes the electric current in the solution? (3) Why must there be actual contact of the two solutions? (4) What are the chemical reactions at each electrode? In the same manner, put the appropriate half-cell in the beaker (half-cell) or nest two porous cups to form the cells Fe | FeSO4||Cu(NO3)2 | Cu Zn | Zn(NO3)2||FeSO4 | Fe Measure and record the voltage of each cell, noting carefully which electrode is positive. Save the Zn2+/Zn and Cu2+/Cu half-cells for later use. (c) Nonmetal Half-Cells. When the oxidized and reduced form of a redox couple are water soluble, electrical contact is made by placing an inert electrode in the solution. It conducts electrons to and from the external circuit. We will use graphite rods but most often, a platinum or gold wire is used. (Graphite is somewhat porous, which presents a problem if the electrode will be transferred from one solution to another; it is hard to rinse off solution that has entered the pores.) Prepare in three small porous cups the following solutions, which contain equimolar amounts of the oxidized and reduced forms of the redox couple: 3 mL Br2 water (sat. soln.) + 3 mL 0.1 M KBr 3 mL 0.1 M FeCl3 + 3 mL 0.1 M FeSO4 3 mL 0.05 M I2 (in methanol) + 3 mL 0.1 M KI (Because I2 reacts almost quantitatively with an excess of I- to form I3

-, the last mixture is essentially 0.025 M in I3

- and 0.025 M in I-.) Insert the graphite inert electrode into the first non-metal half-cell prepared above and nest the half-cell in to Zn | Zn2+ half-cell in the 150 mL beaker. Measure and record the voltages of the following cells, being sure to note which electrode is positive: Zn | Zn(NO3)2 || Br2, KBr | C Zn | Zn(NO3)2 || FeCl3, FeSO4 | C Zn | Zn(NO3)2 || KI3, KI | C Use the same graphite rod for each half-cell in turn, rinsing it in water between measurements. For all six cells for which you made measurements, compare the voltage of each cell (including the sign) with that calculated by combining the E˚s of the two half-cells that constitute the cell. In each case, the E˚cell is the algebraic difference of the E˚ values for the two half-cells. (Why is it necessary to subtract one E˚ from the other?) You will probably notice that the measured cell voltage is less than the calculated E˚cell. The discrepancies may be attributed to differences in the activities of the ions from the hypothetical state of unit activity(a) and possible electrical resistance at the physical and liquid junctions. However, note that for the seven half-cells studied, the measured

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values are still sufficient to establish a table of redox couples that is consistent with the order shown in the table of standard reduction potentials. 2. The Effect of Concentration Make a Cu2+/Cu half-cell by placing 50 mL of 0.1 M Cu(NO3)2 in a 150-mL beaker, along with a copper electrode. Nest a Zn2+/Zn half-cell {Zn(s) + 0.1 M Zn(NO3)2} made in a porous cup into the beaker and connect the electrodes to a voltmeter. Read and record the voltage; then add, stirring continuously, about 5 mL of 6 M NH3 to the Cu2+ solution, until the deep blue Cu(NH3)4

2+ complex ion is obtained, thus reducing the concentration of Cu2+ {Cu(H2O)62+}. Read and

record the voltage. Further reduce the concentration of Cu2+ by adding, while stirring, an excess (10 mL) of 1 M Na2S to the copper half-cell. Again, read and record the voltage. Interpret the voltage changes you observe in terms of the Nernst equation and Le Châtelier principle. 3. Calculating an Equilibrium Constant from a Cell Voltage Measurement Place 50 mL of 1.0 M Na2CO3 and a clean copper strip (electrode) in a 150-mL beaker. Add 5 drops of 1.0 M Cu(NO3)2 to form a precipitate of CuCO3 and stir the solution. In a porous cup add ~10 mL 1.0 M Cu(NO3)2 and insert a clean copper electrode. Nest the porous cup into the beaker containing the precipitate of CuCO3, connect the voltmeter to the copper (electrodes) strips, and record the voltage. Which electrode is the positive electrode? Because the concentrations of all the reactants and products in the cell are 1 M, the cell voltage will be equal to E˚cell, neglecting activity coefficients. From Equation (6) you can calculate Ksp, the equilibrium constant for the reaction CuCO3(s) → Cu2+(aq) + CO3

2-(aq) In using Equation (6) you must know the value of n and the magnitude and sign of E˚cell. What is the value of n that appears in the half-reactions that are summed to give Reaction (1)? The cell diagram is shown in line notation of the cell (2). By convention, the cell voltage, Ecell, is defined as the potential of the right hand electrode (the Cu strip dipping in 1 M CO3

2-) measured with respect to the left-hand electrode. If the right-hand electrode is the negative electrode, Ecell is negative. As defined by this convention, is the E˚cell you measured positive or negative?

Electrochemistry: Electrochemical Cells Report Form

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E17 Name: ____________________ Partner’s Name: __________________(if any)_Lab Section: MW/TTH/M-TH/F (circle)

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Compare your value of Ksp to the literature value of 2.5 x 10-10.