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Chapter 18: Problem 23

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, \(\mathscr{b}\), for the reaction. This is surprising because \(\mathscr{B}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

Short Answer

Expert verified
An extensive property, like free energy change (\(\Delta G\)), depends on the amount of matter in a system, while an intensive property, like cell potential (\(\mathscr{E}\)), does not. Although they seem unrelated, the two can be connected using the equation \(\Delta G = -nFE\), where \(n\) is the number of electrons transferred in the reaction, and \(F\) is Faraday's constant. The product of \(n\) and \(F\) represents the total charge transferred, an extensive property. By substituting the values of \(n\), \(F\), and \(\mathscr{E}\) into the equation, we can calculate the extensive property \(\Delta G\) from the intensive property \(\mathscr{E}\), demonstrating that under certain conditions, extensive and intensive properties can be interconnected.

Step by step solution

01

Definition of Extensive Property

Extensive properties are the properties that depend on the amount of matter (mass or volume) in a system. These properties are additive, which means that the value of the property of the entire system is the sum of the values of the property for its individual components. Examples of extensive properties include mass, volume, and enthalpy.
02

Definition of Intensive Property

Intensive properties, on the other hand, do not depend on the amount of matter in the system. They are independent of the mass or volume and are characteristic of the substance itself. Intensive properties can help in identifying a substance as they remain the same regardless of the quantity present. Examples of intensive properties include temperature, pressure, and cell potential (\(\mathscr{E}\)).
03

Connecting Extensive and Intensive Properties

Though it might seem counterintuitive, extensive and intensive properties can be connected under certain conditions. In the case of calculating the free energy change (\(\Delta G\)) from the cell potential (\(\mathscr{E}\)), we use the following equation: \[\Delta G = -nFE\] In this equation, \(n\) represents the number of electrons transferred in the reaction, \(F\) is the Faraday's constant (which measures the charge of one mole of electrons), and \(\mathscr{E}\) is the cell potential. The product of \(n\) and \(F\) gives the total charge transferred during the reaction, which is an extensive property that depends on the amount of matter involved in the reaction. The cell potential, \(\mathscr{E}\), is an intensive property, as it doesn't depend on the amount of matter.
04

Calculating \(\Delta G\) from \(\mathscr{E}\)

To calculate the free energy change (\(\Delta G\)) from the cell potential (\(\mathscr{E}\)), we first need to determine the number of electrons (\(n\)) involved in the reaction and then multiply this value by the Faraday constant (\(F\)). After that, we can use the equation mentioned above: \[\Delta G = -nFE\] By plugging in the values of \(n\), \(F\), and \(\mathscr{E}\), we can calculate the extensive property \(\Delta G\) from the intensive property \(\mathscr{E}\). This shows that under certain conditions, extensive and intensive properties can be connected and used to determine each other.

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Most popular questions from this chapter

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M ?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is \(1.62 \mathrm{~V} .\) Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

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Nerve impulses are electrical "signals" that pass through neurons in the body. The electrical potential is created by the differences in the concentration of \(\mathrm{Na}^{+}\) and \(\mathrm{K}^{+}\) ions across the nerve cell membrane. We can think about this potential as being caused by a concentration gradient, similar to what we see in a concentration cell (keep in mind that this is a very simple explanation of how nerves work; there is much more involved in the true biologic process). A typical nerve cell has a resting potential of about \(-70 \mathrm{mV}\). Let's assume that this resting potential is due only to the \(\mathrm{K}^{+}\) ion concentration difference. In nerve cells, the \(\mathrm{K}^{+}\) concentration inside the cell is larger than the \(\mathrm{K}^{+}\) concentration outside the cell. Calculate the \(\mathrm{K}^{+}\) ion concentration ratio necessary to produce a resting potential of \(-70 . \mathrm{mV}\). $$\frac{\left[\mathrm{K}^{+}\right]_{\text {inside }}}{\left[\mathrm{K}^{+}\right]_{\text {outside }}}=?$$

What reactions take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. \(1.0 M \mathrm{NiBr}_{2}\) solution b. \(1.0 \mathrm{M} \mathrm{AlF}_{3}\) solution c. \(1.0 \mathrm{M} \mathrm{MnI}_{2}\) solution

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?
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