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Chapter 17: 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
In conclusion, cell potentials are not multiplied by the coefficients in the balanced redox equation because they are intensive properties, meaning they do not depend on the amounts of substances involved in the reaction. Additionally, the relationship between Gibbs Free Energy (螖G) and cell potential (E) is given by 螖G = -nFE, which does not involve the stoichiometric coefficients. When the entire redox equation is multiplied by a factor k, the cell potential E remains unchanged, and the new 螖G is simply k times the original 螖G.

Step by step solution

01

Recall the relationship between 螖G and cell potential E

The relationship between Gibbs Free Energy (螖G) and cell potential (E) is given by the following formula: \[螖G = -nFE\] where: - 螖G is the Gibbs Free Energy (in joules or J), - n is the number of moles of electrons transferred in the redox reaction, - F is the Faraday's constant (96,485 C/mol), and - E is the cell potential (in volts or V).
02

Understand the meaning of cell potential E

The cell potential E is the difference in potential energy between the two half-reactions of a redox reaction. It determines the capacity of the redox reaction to do work. E is an intensive property, which means it does not depend on the amount or extent of the substances involved in the reaction. It is measured in volts (V) and is independent of the coefficients in the balanced redox equation.
03

Show how coefficients in the balanced redox equation affect 螖G

Now, let's analyze a generic balanced redox equation, given by: \(aA + bB \rightarrow cC + dD\) According to the stoichiometric coefficients (a, b, c, and d), the reaction can be written in terms of two half-reactions: Reduction: \(aA + ne^{-} \rightarrow bB\) Oxidation: \(cC \rightarrow dD + ne^{-}\) Now, let's calculate the Gibbs Free Energy change (螖G) for the entire redox reaction. If we multiply the balanced redox equation by any factor k, the reaction would become: \(kaA + kbB \rightarrow kcC + kdD\) Notice that when we multiply the entire redox equation by k, it affects the number of moles of electrons transferred in the reaction as well. The new n would be kn. The 螖G for the new reaction is given by: \[螖G' = -k \cdot n \cdot F \cdot E\] Since k is constant, and E remains the same (as it does not depend on the coefficients in the balanced redox equation), the new 螖G' is just k times the original 螖G: \[螖G' = k \cdot 螖G\]
04

Conclude why cell potentials are not multiplied by coefficients in the balanced redox equation

As shown above, the cell potential E remains the same for the redox reaction despite the coefficients being multiplied by k. This is because E is an intensive property that is independent of the amounts of substances involved in the reaction. Also, the relationship between 螖G and E, given by 螖G = -nFE, is not affected by the stoichiometric coefficients in the balanced redox equation. Thus, it is clear that cell potentials are not multiplied by the coefficients in the balanced redox equation due to their independence from the amounts of substances and their direct relationship with 螖G.

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Most popular questions from this chapter

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 \(\mathscr{C}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} \quad \quad \mathscr{E}^{\circ}=0.68 \mathrm{V}\) b. \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \mathscr{E}^{\circ}=-1.18 \mathrm{V}\) \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036 \mathrm{V}\)

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} & \mathscr{E}^{\circ}=-0.76 \mathrm{V} \\ \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.44 \mathrm{V} \end{array} $$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\text {cell. }}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{C}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Zn}^{2+}\right]=0.10 \mathrm{M}\) and \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M}\)

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces \(1.00 \mathrm{kg}\) water at \(25^{\circ} \mathrm{C} ?\) Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

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