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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 (ΔG) and cell potential (E) is given by ΔG = -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 ΔG is simply k times the original ΔG.

Step by step solution

01

Recall the relationship between ΔG and cell potential E

The relationship between Gibbs Free Energy (ΔG) and cell potential (E) is given by the following formula: \[ΔG = -nFE\] where: - ΔG 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 ΔG

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 (ΔG) 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 ΔG for the new reaction is given by: \[ΔG' = -k \cdot n \cdot F \cdot E\] Since k is constant, and E remains the same (as it does not depend on the coefficients in the balanced redox equation), the new ΔG' is just k times the original ΔG: \[ΔG' = k \cdot ΔG\]
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 ΔG and E, given by ΔG = -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 ΔG.

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Most popular questions from this chapter

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 \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} \mathrm{M}\)

The compound with the formula TII \(_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{T}^{3+}+2 \mathrm{e}^{-} \longrightarrow & \mathrm{Tl}^{+} & & \mathscr{E}^{\circ}=1.25 \mathrm{V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} & \longrightarrow 3 \mathrm{I}^{-} & & \mathscr{E}^{\circ}=0.55 \mathrm{V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. a. What is the potential of the cell at \(25^{\circ} \mathrm{C}\) if the copper electrode is placed in a solution in which \(\left[\mathrm{Cu}^{2+}\right]=\) \(2.5 \times 10^{-4} \mathrm{M} ?\) b. The copper electrode is placed in a solution of unknown \(\left[\mathrm{Cu}^{2+}\right] .\) The measured potential at \(25^{\circ} \mathrm{C}\) is \(0.195 \mathrm{V}\). What is \(\left[\mathrm{Cu}^{2+}\right] ?\) (Assume \(\mathrm{Cu}^{2+}\) is reduced.)

The measurement of \(\mathrm{pH}\) using a glass electrode obeys the Nernst equation. The typical response of a pH meter at \(25.00^{\circ} \mathrm{C}\) is given by the equation $$ \mathscr{E}_{\text {meas }}=\mathscr{E}_{\text {ref }}+0.05916 \mathrm{pH} $$ where \(\mathscr{E}_{\text {ref }}\) contains the potential of the reference electrode and all other potentials that arise in the cell that are not related to the hydrogen ion concentration. Assume that \(\mathscr{E}_{\mathrm{ref}}=0.250 \mathrm{V}\) and that \(\mathscr{E}_{\text {meas }}=0.480 \mathrm{V}\) a. What is the uncertainty in the values of \(\mathrm{pH}\) and \(\left[\mathrm{H}^{+}\right]\) if the uncertainty in the measured potential is \(\pm 1 \mathrm{mV}\) \((\pm 0.001 \mathrm{V}) ?\) b. To what precision must the potential be measured for the uncertainty in \(\mathrm{pH}\) to be \(\pm 0.02 \mathrm{pH}\) unit?

You have a concentration cell in which the cathode has a silver electrode with 0.10 \(M\) Ag \(^{+}\). The anode also has a silver electrode with \(\mathrm{Ag}^{+}(a q), 0.050 \space \mathrm{M}\space \mathrm{S}_{2} \mathrm{O}_{3}^{2-},\) and \(1.0 \times 10^{-3} \mathrm{M}\) \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-} .\) You read the voltage to be 0.76 \(\mathrm{V}\) a. Calculate the concentration of \(\mathrm{Ag}^{+}\) at the anode. b. Determine the value of the equilibrium constant for the formation of \(\mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}\) $$\mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \quad K=?$$
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