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Chapter 18: Problem 9

Explain why cell potentials are not multiplied by the coefficients in the balanced redox equation. (Use the relationship between \(\Delta G\) and cell potential to do this.)

Short Answer

Expert verified
Cell potentials are not multiplied by the coefficients in the balanced redox equation because reduction potentials are an intensive property, meaning they are independent of the extent of the reaction or the amount of the substance. As a result, doubling or tripling the reaction doesn't change the reduction potential. The cell potential is given by the difference between the reduction potentials of the half-reactions, and since coefficients don't affect this, they don't appear in the cell potential calculation. The number of moles of electrons transferred (n) impacts the calculation of Gibbs free energy (\(\Delta G\)) in the relationship \(\Delta G = -nFE_{cell}\), but not the cell potential itself.

Step by step solution

01

State the relationship between Gibbs free energy and cell potential

The relationship between Gibbs free energy (\(\Delta G\)) and cell potential (\(E_{cell}\)) is given by the equation: \[\Delta G = -nFE_{cell}\] where \(\Delta G\) is the Gibbs free energy, \(n\) is the number of moles of electrons transferred in the redox reaction, \(F\) is Faraday's constant (approximately 96485 C/mol), and \(E_{cell}\) is the cell potential.
02

Express cell potential as a difference of reduction potentials

Cell potential is equal to the difference between the standard reduction potentials of the half-reactions involved in the redox reaction: \[E_{cell} = E_{cathode} - E_{anode}\] Here, \(E_{cathode}\) is the reduction potential of the half-reaction occurring at the cathode, while \(E_{anode}\) is the reduction potential of the half-reaction occurring at the anode.
03

Explain why the balanced redox equation has no effect on the coefficient

In the balanced redox equation, each half-reaction is multiplied by a coefficient to ensure that the number of electrons transferred in each half-reaction is equal. However, when we look at the cell potential expression (difference between the reduction potentials), the coefficients do not appear. The reason is that the reduction potentials of half-reactions are an intensive property, meaning they are independent of the extent of the reaction or the amount of the substance. In other words, doubling or tripling the reaction (by multiplying the balanced half-reaction with a coefficient) doesn't change the reduction potential, so the coefficients will not appear when calculating the cell potential.
04

Conclude that the cell potential doesn't depend on the coefficients

Thus, we conclude that cell potentials are not multiplied by the coefficients in the balanced redox equation, since the coefficients don't affect the intensive property such as the reduction potentials. As a result, only the number of moles of electrons transferred (\(n\)) impacts the calculation of Gibbs free energy in the relationship between \(\Delta G\) and \(E_{cell}\), but not the cell potential itself.

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How long will it take to plate out each of the following with a current of \(100.0 \mathrm{~A}\) ? a. \(1.0 \mathrm{~kg} \mathrm{Al}\) from aqueous \(\mathrm{Al}^{3+}\) b. \(1.0 \mathrm{~g} \mathrm{Ni}\) from aqueous \(\mathrm{Ni}^{2+}\) c. \(5.0 \mathrm{~mol} \mathrm{Ag}\) from aqueous \(\mathrm{Ag}^{+}\)

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 \(-70 . \mathrm{mV}\). $$\frac{\left[\mathrm{K}^{+}\right]_{\text {inside }}}{\left[\mathrm{K}^{+}\right]_{\text {outside }}}=?$$
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