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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

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$ \mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q) $$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+} .\) )

Combine the equations $$ \Delta G^{\circ}=-n F \mathscr{E}^{\circ} \quad \text { and } \quad \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} $$ to derive an expression for \(\mathscr{E}^{\circ}\) as a function of temperature. Describe how one can graphically determine \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) from measurements of \(\mathscr{E}^{\circ}\) at different temperatures, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. What property would you look for in designing a reference half-cell that would produce a potential relatively stable with respect to temperature?

Balance the following oxidation-reduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)\) d. \(\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}^{-}(a q)\) e. \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)\)

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-} $$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ}\) $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}(l) &=-237 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g) &=0.0 \\ \mathrm{OH}^{-}(a q) &=-157 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{e}^{-} &=0.0 \end{aligned} $$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 17-1.

Sketch the galvanic cells based on the following overall reactions. Show the direction of electron flow, and identify the cathode and anode. Give the overall balanced equation. Assume that all concentrations are \(1.0 M\) and that all partial pressures are 1.0 atm. a. \(\mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q)\) b. \(\mathrm{Cu}^{2+}(a q)+\mathrm{Mg}(s) \rightleftharpoons \mathrm{Mg}^{2+}(a q)+\mathrm{Cu}(s)\)
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