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Chapter 18: Problem 16

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

Short Answer

Expert verified
The statement is false. Concentration cells work because the actual cell potential, described by the Nernst equation (\(E_{cell} = E掳_{cell} - \frac{RT}{nF} \ln{Q}\)), is dependent on the concentration of the reacting species. The standard reduction potentials, determined under standard conditions, serve as reference values for the redox tendency of the species involved but do not directly depend on concentration.

Step by step solution

01

Understanding Concentration Cells

A concentration cell is a type of electrochemical cell, where both the half-cells (anode and cathode) have the same chemical species but at different concentrations. The cell generates its voltage because of the difference in concentrations of the reacting species. The overall cell reaction involves the transfer of ions from one half-cell to the other, resulting in a measurable voltage across the electrodes.
02

Understanding Standard Reduction Potentials

Standard reduction potential, denoted by E掳, is a measure of the tendency of a chemical species to acquire electrons and get reduced. These potentials are determined under standard conditions, which include a fixed ion concentration of 1 M, a temperature of 298.15 K (25掳C), and a pressure of 1 atm. By convention, E掳 values are measured relative to the standard hydrogen electrode (SHE), which is assigned a value of zero volts. A positive E掳 value suggests a species has a greater tendency to get reduced, while a negative value indicates a lesser tendency.
03

Relationship between Concentration and Standard Reduction Potentials

To understand the impact of concentration on the cell potential, we apply the Nernst equation, which relates the actual cell potential (E_cell) to the standard cell potential (E掳_cell) and the concentrations of the reacting species. The Nernst equation is given by: \[E_{cell} = E掳_{cell} - \frac{RT}{nF} \ln{Q}\] Here, R is the gas constant, T is the absolute temperature, n is the number of electrons transferred, F is the Faraday constant, and Q is the reaction quotient of the cell. The equation illustrates that the actual cell potential depends on the concentrations of the reacting species and varies with changes in concentration.
04

Analyzing the Statement

The given statement claims that concentration cells work because standard reduction potentials are dependent on the concentration. This statement is not entirely accurate. Concentration cells work because the actual cell potential, described by the Nernst equation, is dependent on the concentration. The standard reduction potential is determined under standard conditions (1 M concentration) and serves as a reference point. However, the actual cell potential is influenced by the differences in concentration between the half-cells, which drives the cell voltage.
05

Conclusion

The given statement is false. Concentration cells work because the actual cell potential, described by the Nernst equation, is dependent on the concentration of the reacting species. In contrast, the standard reduction potentials are determined under standard conditions and serve as reference values for the redox tendency of the species involved.

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Most popular questions from this chapter

The free energy change for a reaction, \(\Delta G,\) is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{E}\), for the reaction. This is surprising because \(\mathscr{E}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

In the electrolysis of an aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{ll} {\text{}} & \quad{ \mathscr{E}^{\circ} } \\ \hline {\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}} & {2.01 \mathrm{V}} \\ {\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}} & {1.23 \mathrm{V}} \\ {2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}} & {-0.83 \mathrm{V}} \\\ {\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}} & {-2.71 \mathrm{V}}\end{array}$$

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0\(M \mathrm{M}^{2+} .\) Solution \(\mathrm{B}\) in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment 0.0100 mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and 0.0100 mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$\mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s)$$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be 0.44 \(\mathrm{V}\) at \(25^{\circ} \mathrm{C} .\) Assume that the process $$\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}$$ has a standard reduction potential of \(-0.31 \mathrm{V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C} .\)

The overall reaction and equilibrium constant value for a hydrogen-oxygen fuel cell at 298 \(\mathrm{K}\) is $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad K=1.28 \times 10^{83}$$ a. Calculate \(8^{\circ}\) and \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) for the fuel cell reaction. b. Predict the signs of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the fuel cell reaction. c. As temperature increases, does the maximum amount of work obtained from the fuel cell reaction increase, decrease, or remain the same? Explain.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium $$2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$$ Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C} :\) $$3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.957 \mathrm{V}$$ $$\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \mathscr{E}^{\circ}=0.775 \mathrm{V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and NO \(_{2}\) mixture with only 0.20\(\% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and 1.00 atm? Assume that no other gases are present and that the change in acid concentration can be neglected.
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