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 due to the difference in concentrations of the same species in each half-cell, which generates a voltage. Although the actual cell potential (E) is influenced by the concentration of reactants and products as described by the Nernst equation, the standard reduction potential (E°) remains constant and is not dependent on concentration.

Step by step solution

01

1. Understand concentration cells

A concentration cell is a type of electrochemical cell where both the cathode and anode are made of the same material, while the electrolyte solutions have different concentrations. The voltage generated by a concentration cell is due to the difference in concentrations of the same species in each half-cell. The overall cell potential of a concentration cell is a result of the concentration gradient of the half cells, which drives the movement of the ions and creates a charge flow.
02

2. Explore standard reduction potentials

The standard reduction potential (E°) is a measurement of the inherent voltage (or tendency) of a half-cell reaction to occur under standard conditions (i.e., temperature, pressure, and concentration). The value of the standard reduction potential is determined with respect to the hydrogen electrode, which has a potential of 0 V by definition.
03

3. Investigate the dependency of standard reduction potentials on concentration

Standard reduction potentials (E°) are defined under standard conditions, and they are constants for a given half-cell reaction. However, the actual cell potential (E) may vary depending on the concentrations of the reacting species. This relationship is governed by the Nernst equation: \[E = E° - \frac{RT}{nF} \ln Q\] Where: - E is the actual cell potential - E° is the standard cell potential - R is the gas constant - T is the absolute temperature - n is the number of moles of electrons exchanged in the redox reaction - F is the Faraday constant - Q is the reaction quotient, which represents the ratios of the concentrations of reactants and products From the Nernst equation, we can see that the actual cell potential (E) is affected by the concentration of reactants and products. However, the standard reduction potential (E°) remains constant.
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4. Assess the statement and provide an explanation

The statement "Concentration cells work because standard reduction potentials are dependent on concentration" is false. Although the actual cell potential (E) is influenced by the concentration of reactants and products, the standard reduction potential (E°) remains constant. In concentration cells, the voltage generated is due to the concentration gradient of the half-cells and not because of a change in the standard reduction potentials.

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Most popular questions from this chapter

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at \(25^{\circ} \mathrm{C}\) is \(0.195 \mathrm{V}\). What is \(\left[\mathrm{Cu}^{2+}\right] ?\) (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

A galvanic cell is based on the following half-reactions: $$ \begin{array}{ll} \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ}=-0.440 \mathrm{V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ}=0.000 \mathrm{V} \end{array} $$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} \mathrm{M}\) and the hydrogen compartment contains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00\) atm, and a weak acid, HA, at an initial concentration of \(1.00 \mathrm{M}\). If the observed cell potential is \(0.333 \mathrm{V}\) at \(25^{\circ} \mathrm{C},\) calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

Consider the following half-reactions: \(\begin{aligned} \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} & & & \mathscr{E}^{\circ}=1.188 \mathrm{V} \\ \mathrm{PtCl}_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow & \mathrm{Pt}+4 \mathrm{Cl}^{-} & & \mathscr{E}^{\circ}=0.755 \mathrm{V} \\ \mathrm{NO}_{3}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} & \longrightarrow \mathrm{NO}+2 \mathrm{H}_{2} \mathrm{O} & & \mathscr{E}^{\circ}=0.96 \mathrm{V} \end{aligned}\) Explain why platinum metal will dissolve in aqua regia (a mixture of hydrochloric and nitric acids) but not in either concentrated nitric or concentrated hydrochloric acid individually.

The electrolysis of \(\mathrm{BiO}^{+}\) produces pure bismuth. How long would it take to produce \(10.0 \mathrm{g}\) Bi by the electrolysis of a \(\mathrm{BiO}^{+}\) solution using a current of \(25.0 \mathrm{A} ?\)

Consider the standard galvanic cell based on the following half-reactions: $$\begin{array}{c} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{array}$$ The electrodes in this cell are \(\mathrm{Ag}(s)\) and \(\mathrm{Cu}(s) .\) Does the cell potential increase, decrease, or remain the same when the following changes occur to the standard cell? a. \(\mathrm{CuSO}_{4}(s)\) is added to the copper half-cell compartment (assume no volume change). b. \(\mathrm{NH}_{3}(a q)\) is added to the copper half-cell compartment. [Hint: \(\left.\mathrm{Cu}^{2+} \text { reacts with } \mathrm{NH}_{3} \text { to form } \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) .\right]\) c. \(\mathrm{NaCl}(s)\) is added to the silver half-cell compartment. [Hint: \(\left.\mathrm{Ag}^{+} \text {reacts with } \mathrm{Cl}^{-} \text {to form } \mathrm{AgCl}(s) .\right]\) d. Water is added to both half-cell compartments until the volume of solution is doubled. e. The silver electrode is replaced with a platinum electrode. $$ \mathrm{Pt}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt} \quad \mathscr{E}^{\circ}=1.19 \mathrm{V} $$

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