20.6 Effect of Concentration on Cell EMF

We have seen that voltaic cells are based on spontaneous chemical reactions. We have also seen how to calculate the emf of a cell when the reactants and products are under standard conditions. However, as a voltaic cell is discharged, the reactants of the reaction are consumed and the products are generated, so the concentrations of these substances change. The emf progressively drops until E = 0, at which point we say the cell is "dead." At that point the concentrations of the reactants and products cease to change; they are at equilibrium. In this section we will examine how the cell emf depends on the concentrations of the reactants and products of the cell reaction. The emf generated under nonstandard conditions can be calculated by using an equation first derived by Walther Nernst (1864-1941), a German chemist who established many of the theoretical foundations of electrochemistry.

The Nernst Equation

The dependence of the cell emf on concentration can be obtained from the dependence of the free-energy change on concentration. Recall that the free-energy change, G, is related to the standard free-energy change, :

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[20.13]

The quantity Q is the reaction quotient, which has the form of the equilibrium constant expression except that the concentrations are those that exist in the reaction mixture at a given moment.

Substituting G = -nFE (Equation 20.11) into Equation 20.13 gives

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Solving this equation for E gives the Nernst equation:

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[20.14]

This equation is customarily expressed in terms of common (base 10) logarithms:

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[20.15]

At T = 298 K the quantity 2.303 RT/F equals 0.0592 V, so the equation simplifies to:

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[20.16]

We can use this equation to find the emf produced by a cell under nonstandard conditions or to determine the concentration of a reactant or product by measuring the emf of the cell.

As an example of how Equation 20.16 might be used, consider the following reaction, which we have discussed earlier:

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In this case n = 2 (two electrons are transferred from Zn to Cu2+), and the standard emf is +1.10 V. Thus, at 298 K the Nernst equation gives

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[20.17]

Recall that pure solids are excluded from the expression for Q. From Equation 20.17 it is evident that the emf increases as [Cu2+] increases and as [Zn2+] decreases. For example, when [Cu2+] is 5.0 M and [Zn2+] is 0.050 M, we have

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Notice that we have increased the concentration of the reactant (Cu2+) and decreased the concentration of the product (Zn2+) relative to standard conditions. These concentration changes increase the emf of the cell (E = +1.16 V) relative to standard conditions ( = +1.10 V). We could have anticipated this result by applying Le Châtelier's principle.

In general, if the concentrations of reactants increase relative to those of products, the cell reaction becomes more spontaneous and the emf increases. Conversely, if the concentrations of products increase relative to reactants, the emf decreases. As voltaic cell operates, reactants are converted into products, which increases the value of Q and causes the emf to decrease.

SAMPLE EXERCISE 20.11

Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 when [Cr2O72–] = 2.0 M, [H+] = 1.0 M, [I] = 1.0 M, and [Cr3+] = 1.0 10–5 M:

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SOLUTION The standard emf for this reaction was calculated in Sample Exercise 20.6: = 0.79 V. As you will see if you refer back to that exercise, n = 6. The reaction quotient, Q, is

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Using Equation 20.16, we have

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This result is qualitatively what we expect: Because the concentration of Cr2O72– (a reactant) is above 1 M and the concentration of Cr3+ (a product) is below 1 M, the emf is greater than E°.

PRACTICE EXERCISE

Calculate the emf generated by the cell described in the practice exercise accompanying Sample Exercise 20.6 when [Al3+] = 4.0 10–3 M and [I] = 0.010 M. Answer: E = +2.36 V

SAMPLE EXERCISE 20.12

If the voltage of a Zn—H+ cell (like that in Figure 20.11) is 0.45 V at 25°C when [Zn2+] = 1.0 M and PH2 = 1.0 atm, what is the concentration of H+?

SOLUTION The cell reaction is

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The standard emf is = +0.76 V. It is apparent that n = 2 for this reaction, and we are given that [Zn2+] = 1.0 M and PH2 = 1.0 atm (recall that gas pressures must be in atm for calculations involving Q). We now apply Equation 20.16:

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From the properties of logarithms (Appendix A.2 of the textbook):

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We can therefore write

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Solving for log [H+] gives

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This example shows how a voltaic cell whose cell reaction involves H+ can be used to measure [H+] or pH. A pH meter is a specially designed voltaic cell with a voltmeter calibrated to read pH directly.

PRACTICE EXERCISE

What is the pH of the solution in the cathode compartment of the cell pictured in Figure 20.11 when PH2 = 1.0 atm, [Zn2+] in the anode compartment is 0.10 M, and cell emf is 0.542 V? Answer: pH = 4.19

Concentration Cells

In each of the voltaic cells that we have looked at thus far, the reactive species at the anode has been different from the one at the cathode. However, the fact that cell emf depends on concentration means that a voltaic cell with a nonzero emf can be constructed using the same species in both the anode and cathode compartments as long as the concentrations are different. A cell based solely on the emf generated because of a difference in a concentration is called a concentration cell.

Consider the cell shown in Figure 20.16. One compartment consists of a strip of nickel metal immersed in a 1.00 M solution of Ni2+(aq). The other compartment also has a Ni(s) electrode, but immersed in a 1.00 10–3 M solution of Ni2+(aq). The two compartments are connected by a salt bridge and by an external wire with a voltmeter. The half-cell reactions are the reverse of one another:

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Figure 20.16 A concentration cell based on the Ni2+—Ni cell reaction.

It is apparent that the standard emf for this cell is zero: = (cathode) - (anode) = (-0.28 V) - (-0.28 V) = 0 V. But this cell operates under nonstandard conditions because of the difference in concentration of Ni2+(aq) in the two compartments. In fact, the driving force for the cell is provided by the difference in concentration: Operation of the cell tends to equalize the concentrations of Ni2+ in both compartments. Thus, the compartment with the more dilute solution will be the anode compartment; oxidation of Ni(s) at the anode increases the concentration of Ni2+(aq) in the more dilute solution. Conversely, the compartment with the more concentrated solution will be the cathode compartment, for the reduction of Ni2+(aq) at the cathode decreases the concentration of Ni2+(aq) in that compartment. The overall cell reaction is therefore:

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We can calculate the emf of a concentration cell by using the Nernst equation. For this particular cell we see that n = 2. We also see that the expression for the reaction quotient for the overall reaction is Q = [Ni2+]dilute / [Ni2+]concentrated. We can calculate the emf at 298 K by using Equation 20.16:

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This concentration cell generates an emf of nearly 0.09 V even though = 0 for the cell. The difference in concentration provides the driving force for the cell. As the cell operates, the concentration of the dilute solution in the anode compartment increases and that of the concentrated solution in the cathode compartment decreases. When the concentrations in the two compartments become the same, the value of Q = 1 and E = 0.

The idea of generating a potential by a difference in concentration is the basis for the operation of pH meters (Figure 16.6). It is also a critical aspect in the regulation of the heartbeat in mammals.

SAMPLE EXERCISE 20.13

A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has PH2 = 1.00 atm and an unknown concentration of H+(aq). Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, PH2 = 1.00 atm). At 298 K, the cell voltage is measured to be 0.211 V and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. What is [H+] for the solution in electrode 1? What is its pH?

SOLUTION The fact that current flows from electrode 1 to electrode 2 tells us that electrode 1 is the anode of the cell and electrode 2 is the cathode. The electrode reactions are therefore as follows, with the concentration of H+(aq) in electrode 1 represented with the unknown x:

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Because is the same for both half-cells, = 0 V; this is a concentration cell. We recognize that n = 2 for the cell and that the reaction quotient for the overall reaction is given by

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We use Equation 20.16 to solve for the value of x:

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Thus, in electrode 1 [H+] = 2.8 10–4 M. The concentration of H+ in electrode 1 is lower than that in electrode 2, which is why electrode 1 is the anode of the cell: The oxidation of H2 to H+(aq) increases [H+] in electrode 1. We essentially calculated the pH of the solution while we were solving the problem: pH = -log(2.8 10–4) = 3.56

PRACTICE EXERCISE

A concentration cell is constructed with two Zn(s)—Zn2+(aq) half-cells. The first cell has [Zn2+] = 1.35 M and the second cell has [Zn2+] = 3.75 10–4 M. (a) Which half-cell is the anode of the cell? (b) What is the emf of the cell? Answers: (a) the second cell; (b) 0.105 V

Cell EMF and Chemical Equilibrium

The Nernst equation helps us understand why the emf of a voltaic cell drops as it is discharged: As reactants are converted to products, the value of Q increases, so the value of E decreases. The cell emf eventually reaches E = 0. Because G = -nFE (Equation 20.11), it follows that G = 0 when E = 0. Recall that a system is at equilibrium when G = 0. Thus, when E = 0, the cell reaction has reached equilibrium; no net reaction is occurring in the voltaic cell.

At equilibrium the reaction quotient equals the equilibrium constant: Q = K at equilibrium. Substituting E = 0 and Q = K into the Nernst equation (Equation 20.14) gives

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At 298 K this equation simplifies to

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which can be rearranged to give

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[20.18]

This useful equation tells us that the equilibrium constant for a redox reaction can be obtained from the value of the standard emf for the reaction.

SAMPLE EXERCISE 20.14

Using the standard reduction potentials listed in Appendix E of the textbook, calculate the equilibrium constant for the oxidation of Fe2+ by O2 in acidic solution, according to the following reaction:

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SOLUTION We observe that O2 is reduced and Fe2+ is oxidized in the reaction. We can determine in the usual way:

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Thus, = (1.23 V) - (0.77 V) = 0.46 V, and n = 4. Using Equation 20.18, we have

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The large magnitude of K indicates that Fe2+ ions are unstable in acidic solutions in the presence of O2 (unless a suitable reducing agent is present).

PRACTICE EXERCISE

Using standard reduction potentials (Appendix E), calculate the equilibrium constant at 25°C for the reaction

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Answer: K = 1.2 10–10