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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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Most popular questions from this chapter

Consider a cell based on the following half-reactions: $$\begin{aligned}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{C}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Fe}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+} & \mathscr{C}^{\circ}=0.77 \mathrm{~V} \end{aligned}$$ a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 M\), the cell potential is observed to be \(0.31 \mathrm{~V}\). Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$\mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q)$$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathbf{I}_{3}^{-}(a q)\) c. \(\begin{aligned} \mathrm{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\mathrm{Acid}}{\longrightarrow} \\ & \mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}^{2-}(a q) \\ \text { d. } \mathrm{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow} \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \\ \text { e. } \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Base }}{\longrightarrow} \\\ \mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q) \end{aligned}\)

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic and an electrolytic cell.

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(\mathrm{g})\) 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 \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $$\begin{array}{r}\mathscr{E}^{\circ}=0.957 \mathrm{~V}\end{array}$$ \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) $${8}^{\circ}=0.775 \mathrm{~V}$$ a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a NO and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) ? Assume that no other gases are present and that the change in acid concentration can be neglected.

Aluminum is produced commercially by the electrolysis of \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in the presence of a molten salt. If a plant has a continuous capacity of \(1.00\) million A, what mass of aluminum can be produced in \(2.00 \mathrm{~h} ?\)
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