10.4 The nernst equation (Page 2/6)

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Now that the connection has been made between the free energy and cell potentials, nonstandard concentrations follow. Recall that

Δ G = Δ G ° + R T ln Q

where Q is the reaction quotient (see the chapter on equilibrium fundamentals). Converting to cell potentials:

− n F E_{cell} = − n F E_{cell}^{°} + R T ln Q or E_{cell} = E_{cell}^{°} - \frac{R T}{n F} ln Q

This is the Nernst equation . At standard temperature (298.15 K), it is possible to write the above equations as

E_{cell} = E_{cell}^{°} - \frac{0.0 257 V}{n} ln Q or E_{cell} = E_{cell}^{°} - \frac{0.0592 V}{n} log Q

If the temperature is not 273.15 K, it is necessary to recalculate the value of the constant. With the Nernst equation, it is possible to calculate the cell potential at nonstandard conditions. This adjustment is necessary because potentials determined under different conditions will have different values.

Cell potentials at nonstandard conditions

Consider the following reaction at room temperature:

Co (s) + {Fe}^{2+} (a q, 1.94 M) ⟶ {Co}^{2+} (a q, 0.15 M) + Fe (s)

Is the process spontaneous?

Solution

There are two ways to solve the problem. If the thermodynamic information in Appendix G were available, you could calculate the free energy change. If the free energy change is negative, the process is spontaneous. The other approach, which we will use, requires information like that given in Appendix L . Using those data, the cell potential can be determined. If the cell potential is positive, the process is spontaneous. Collecting information from Appendix L and the problem,

\begin{array}{l} Anode (oxidation): Co (s) ⟶ {Co}^{2+} (a q) + {2e}^{−} E_{{Co}^{2+} /Co}^{°} = −0.28 V \\ Cathode (reduction): {Fe}^{2+} (a q) + {2e}^{−} ⟶ Fe (s) E_{{Fe}^{2+} /Fe}^{°} = −0.447 V \\ E_{cell}^{°} = E_{cathode}^{°} - E_{anode}^{°} = −0.447 V - (−0.28 V) = −0.17 V \end{array}

The process is not spontaneous under standard conditions. Using the Nernst equation and the concentrations stated in the problem and n = 2,

Q = \frac{{[Co}^{2+}]}{{[Fe}^{2+}]} = \frac{0.15 M}{1.94 M} = 0.077

E_{cell} = E_{cell}^{°} - \frac{0.0592 V}{n} log Q

E_{cell} = −0.1 7 V - \frac{0.0592 V}{2} log 0.077

E_{cell} = −0.17 V + 0.033 V = −0.014 V

The process is (still) nonspontaneous.

Check your learning

What is the cell potential for the following reaction at room temperature?

Al (s) │ {Al}^{3+} (a q, 0.15 M) ║ {Cu}^{2+} (a q, 0.025 M) │ Cu (s)

What are the values of n and Q for the overall reaction? Is the reaction spontaneous under these conditions?

Answer:

n = 6; Q = 1440; E _cell = +1.97 V, spontaneous.

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Finally, we will take a brief look at a special type of cell called a concentration cell . In a concentration cell, the electrodes are the same material and the half-cells differ only in concentration. Since one or both compartments is not standard, the cell potentials will be unequal; therefore, there will be a potential difference, which can be determined with the aid of the Nernst equation.

Concentration cells

What is the cell potential of the concentration cell described by

Zn (s) │ {Zn}^{2+} (a q, 0.10 M) ║ {Zn}^{2+} (a q, 0.50 M) │ Zn (s)

Solution

From the information given:

\begin{array}{l} \underline{\begin{array}{l} Anode: Zn (s) ⟶ {Zn}^{2+} (a q, 0.10 M) + {2e}^{−} E_{anode}^{°} = −0.7618 V \\ Cathode: {Zn}^{2+} (a q, 0.50 M) + {2e}^{−} ⟶ Zn (s) E_{cathode}^{°} = −0.7618 V \end{array}} \\ Overall: {Zn}^{2+} (a q, 0.50 M) ⟶ {Zn}^{2+} (a q, 0.10 M) E_{cell}^{°} = 0.000 V \end{array}

The standard cell potential is zero because the anode and cathode involve the same reaction; only the concentration of Zn ²⁺ changes. Substituting into the Nernst equation,

E_{cell} = 0.000 V - \frac{0.0592 V}{2} log \frac{0.10}{0.50} = +0.021 V

and the process is spontaneous at these conditions.

Check your answer: In a concentration cell, the standard cell potential will always be zero. To get a positive cell potential (spontaneous process) the reaction quotient Q must be<1. Q <1 in this case, so the process is spontaneous.

Check your learning

What value of Q for the previous concentration cell would result in a voltage of 0.10 V? If the concentration of zinc ion at the cathode was 0.50 M , what was the concentration at the anode?

Answer:

Q = 0.00042; [Zn ²⁺ ] _cat = 2.1 $\times$ 10 ⁻⁴ M .

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Key concepts and summary

Electrical work ( w _ele ) is the negative of the product of the total charge ( Q ) and the cell potential ( E _cell ). The total charge can be calculated as the number of moles of electrons ( n ) times the Faraday constant ( F = 96,485 C/mol e ⁻ ). Electrical work is the maximum work that the system can produce and so is equal to the change in free energy. Thus, anything that can be done with or to a free energy change can also be done to or with a cell potential. The Nernst equation relates the cell potential at nonstandard conditions to the logarithm of the reaction quotient. Concentration cells exploit this relationship and produce a positive cell potential using half-cells that differ only in the concentration of their solutes.

Key equations

$E_{cell}^{°} = \frac{R T}{n F} ln K$
$E_{cell}^{°} = \frac{0.02 57 V}{n} ln K = \frac{0.0 592 V}{n} log K (at 298.15 K)$
$E_{cell} = E_{cell}^{°} - \frac{R T}{n F} ln Q (Nernst equation)$
$E_{cell} = E_{cell}^{°} - \frac{0.02 57 V}{n} ln Q = E_{cell}^{°} - \frac{0.05 92 V}{n} log Q (at 298.15 K)$
Δ G = − nFE _cell
$Δ G^{°} = − n F E_{cell}^{°}$
$w_{ele} = w_{max} = − n F E_{cell}$

Chemistry end of chapter exercises

For the standard cell potentials given here, determine the Δ G ° for the cell in kJ.

(a) 0.000 V, n = 2

(b) +0.434 V, n = 2

(a) 0 kJ/mol; (b) −83.7 kJ/mol; (c) +235.3 kJ/mol

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For the Δ G ° values given here, determine the standard cell potential for the cell.

(a) 12 kJ/mol, n = 3

(b) −45 kJ/mol, n = 1

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Determine the standard cell potential and the cell potential under the stated conditions for the electrochemical reactions described here. State whether each is spontaneous or nonspontaneous under each set of conditions at 298.15 K.

(a) $Hg (l) + S^{2−} (a q, 0.10 M) + 2 {Ag}^{+} (a q, 0.25 M) ⟶ 2 Ag (s) + HgS (s)$

(b) The galvanic cell made from a half-cell consisting of an aluminum electrode in 0.015 M aluminum nitrate solution and a half-cell consisting of a nickel electrode in 0.25 M nickel(II) nitrate solution.

(c) The cell made of a half-cell in which 1.0 M aqueous bromine is oxidized to 0.11 M bromide ion and a half-cell in which aluminum ion at 0.023 M is reduced to aluminum metal. Assume the standard reduction potential for Br ₂ ( l ) is the same as that of Br ₂ ( aq ).

(a) standard cell potential: 1.50 V, spontaneous; cell potential under stated conditions: 1.43 V, spontaneous; (b) standard cell potential: 1.405 V, spontaneous; cell potential under stated conditions: 1.423 V, spontaneous; (c) standard cell potential: −2.749 V, nonspontaneous; cell potential under stated conditions: −2.757 V, nonspontaneous

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Determine Δ G and Δ G ° for each of the reactions in the previous problem.

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Use the data in Appendix L to determine the equilibrium constant for the following reactions. Assume 298.15 K if no temperature is given.

(a) $AgCl (s) ⇌ {Ag}^{+} (a q) + {Cl}^{−} (a q)$

(b) $CdS (s) ⇌ {Cd}^{2+} (a q) + S^{2−} (a q) at 377 K$

(d) $H_{2} O (l) ⇌ H^{+} (a q) + {OH}^{−} (a q) at 25 °C$

(a) 1.7 $\times$ 10 ⁻¹⁰ ; (b) 2.6 $\times$ 10 ⁻²¹ ; (c) 8.9 $\times$ 10 ¹⁹ ; (d) 1.0 $\times$ 10 ⁻¹⁴

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Source: OpenStax, Ut austin - principles of chemistry. OpenStax CNX. Mar 31, 2016 Download for free at http://legacy.cnx.org/content/col11830/1.13

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