Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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 "Concentration cells work because standard reduction potentials are dependent on concentration" is false. Concentration cells work due to the concentration gradient of ions in the solution, and while the Nernst equation allows us to account for changes in concentration when determining the actual cell potential, the standard reduction potentials themselves do not change with concentration.

Step by step solution

01

Understand Concentration Cells

: A concentration cell is a type of electrochemical cell in which both electrodes are made of the same material, and the only difference between the two half-cells is the concentration of the ion in the solution. The spontaneous redox reaction in this cell generates an electric current due to the concentration gradient of the ions.
02

Understand Standard Reduction Potentials

: Standard reduction potential is the tendency of a chemical species to gain electrons and undergo reduction. It is measured under standard conditions (25°C, 1 atm, and 1M concentration). The standard reduction potential represents the cell potential of a particular electrode when connected to a standard electrode (usually the hydrogen electrode) under standard conditions.
03

Understand the relationship between concentration and standard reduction potentials

: Standard reduction potentials, by definition, are measured under standard conditions, meaning that the concentration of the ions is taken to be 1M. However, in real-world electrochemical systems, concentrations are not always 1M. The Nernst equation allows us to determine the actual cell potential of the system under non-standard conditions by considering changes in concentration. The Nernst equation: \(E = E° - \frac{RT}{nF} \ln Q\) In this equation, E is the cell potential, E° is the standard cell potential, R is the ideal gas constant, T is the temperature, n is the number of moles of electrons transferred, F is the Faraday constant, and Q is the reaction quotient.
04

Evaluate the provided statement

: The given statement is that concentration cells work because standard reduction potentials are dependent on concentration. This statement is partially accurate, but not entirely true. Concentration cells work due to the concentration gradient of the ions present in the cell, which generates an electric current. The Nernst equation allows us to account for the changes in concentration in determining the actual cell potential. However, the standard reduction potentials themselves are measured under standard conditions and, by definition, do not change with concentration. Therefore, the statement is false. In conclusion, the statement "Concentration cells work because standard reduction potentials are dependent on concentration" is false. Concentration cells work due to the concentration gradient of ions in the solution, and while the Nernst equation allows us to account for changes in concentration when determining the actual cell potential, the standard reduction potentials themselves do not change with concentration.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the following half-reaction, \(\mathscr{E}^{\circ}=-2.07 \mathrm{~V}\) : $$\mathrm{AlF}_{6}^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Al}+6 \mathrm{~F}^{-}$$ Using data from Table \(18.1\), calculate the equilibrium constant at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{Al}^{3+}(a q)+6 \mathrm{~F}^{-}(a q) \rightleftharpoons \mathrm{AlF}_{6}^{3-}(a q) \quad K=?$$

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{I}^{-}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{NO}(g)\) c. \(\mathrm{Br}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{Mn}^{2+}(a q)\) d. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \rightarrow \mathrm{CH}_{2} \mathrm{O}(a q)+\mathrm{Cr}^{3+}(a q)\)

In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.

Electrolysis of an alkaline earth metal chloride using a current of \(5.00 \mathrm{~A}\) for \(748 \mathrm{~s}\) deposits \(0.471 \mathrm{~g}\) of metal at the cathode. What is the identity of the alkaline earth metal chloride?

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad 8^{\circ}=1.09 \mathrm{~V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\) \(\mathscr{b}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E ^ { \circ }}=1.60 \mathrm{~V}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free