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Chapter 18: Problem 134

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\\ \hline{M^{4+}+4 e^{-} \longrightarrow M} & {0.66} \\ {N^{3+}+3 e^{-} \longrightarrow N} & {0.39}\end{array}$$ What is the value of \(\Delta G^{\circ}\) and \(K\) for this cell?

Short Answer

Expert verified
For the given galvanic cell with half-reactions \(M^{4+} + 4e^- \rightarrow M\) and \(N^{3+} + 3e^- \rightarrow N\), the overall balanced redox reaction is \(3M^{4+} + 12N^{3+} \rightarrow 3M + 4N\). With standard electrode potentials of 0.66V and 0.39V, the standard cell potential is \(0.27V\). This yields a standard Gibbs free energy change, \(\Delta G^{\circ}\), of \(-312000 J/mol\) and an equilibrium constant, K, of \(3.4 \times 10^{42}\).

Step by step solution

01

1. Determine the overall balanced redox reaction

To find the overall balanced redox reaction, we need to combine the given half-reactions while ensuring that the same number of electrons are gained and lost, hence we need to multiply the first half-reaction by 3 and the second half-reaction by 4 to balance the electrons: \(3(M^{4+} + 4e^- \rightarrow M) \\ 4(N^{3+} + 3e^- \rightarrow N) \) Now, we can sum up the half-reactions to obtain the full redox reaction: \(3M^{4+} + 12e^- + 12 N^{3+} \rightarrow 3M + 12e^- + 4N \) The electrons cancel out, and the balanced redox reaction is \(3M^{4+} + 12N^{3+} \rightarrow 3M + 4N \)
02

2. Calculate the standard cell potential

We will use the standard electrode potentials given in the problem (\(\color{blue}{\mathscr{E}_1^{\circ} = 0.66V }\) and \(\color{blue}{\mathscr{E}_2^{\circ} = 0.39V}\)) to calculate the standard cell potential: \[\mathscr{E}_{cell}^{\circ} = \mathscr{E}_{cathode}^{\circ} - \mathscr{E}_{anode}^{\circ}\] Since a spontaneous redox reaction has a positive cell potential (\(\mathscr{E}_{cell}^{\circ} > 0\)), we assign the half-cell with a higher standard electrode potential as the cathode and the other as the anode. \(\mathscr{E}_{cell}^{\circ} = 0.66V - 0.39V = \color{blue}{0.27V}\)
03

3. Calculate the Gibbs free energy change

We can calculate the standard Gibbs free energy change (\(\Delta G^{\circ}\)) using the standard cell potential (\(\mathscr{E}_{cell}^{\circ}\)) and the number of electrons transferred (n) in the following equation: \[\Delta G^{\circ} = -n F \mathscr{E}_{cell}^{\circ}\] where F is the Faraday constant, which is approximately 96485 C/mol of electrons. In this case, the number of electrons transferred is 12 (from the balanced redox reaction). Therefore, \(\Delta G^{\circ} = -(12)(96485C/mol)(0.27V)\) \(\Delta G^{\circ} = \color{blue}{-312000 J/mol}\)
04

4. Calculate the equilibrium constant (K)

We can find the equilibrium constant (K) using the standard Gibbs free energy change (\(\Delta G^{\circ}\)) and the equation: \[K = e^{-\frac{\Delta G^{\circ}}{RT}}\] where R is the universal gas constant, approximately 8.314 J/(mol K), and T is the temperature in Kelvin. Assuming a temperature of 298 K, \(K = e^{-\frac{-312000 J/mol}{(8.314 J/(mol K))(298 K)}}\) \(K = \color{blue}{3.4 \times 10^{42}}\) The equilibrium constant, K, for the cell is approximately 3.4 x 10^42, and the standard Gibbs free energy change is -312000 J/mol.

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Most popular questions from this chapter

An electrochemical cell is set up using the following unbalanced reaction: $$\mathrm{M}^{a+}(a q)+\mathrm{N}(s) \longrightarrow \mathrm{N}^{2+}(a q)+\mathrm{M}(s)$$ The standard reduction potentials are: $$\mathrm{M}^{a+}+a \mathrm{e}^{-} \longrightarrow \mathrm{M} \quad \mathscr{E}^{\circ}=0.400 \mathrm{V}$$ $$\mathrm{N}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{N} \quad \mathscr{E}^{\circ}=0.240 \mathrm{V}$$ The cell contains 0.10\(M \mathrm{N}^{2+}\) and produces a voltage of 0.180 \(\mathrm{V}\) . If the concentration of \(\mathrm{M}^{a+}\) is such that the value of the reaction quotient \(Q\) is \(9.32 \times 10^{-3},\) calculate \(\left[\mathrm{M}^{a+}\right] .\) Calculate \(w_{\text { max }}\) for this electrochemical cell.

A solution at \(25^{\circ} \mathrm{C}\) contains 1.0\(M \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-4} M\) \(\mathrm{Ag}^{+}\). Which metal will plate out first as the voltage is gradually increased when this solution is electrolyzed? (Hint: Use the Nernst equation to calculate \(\mathscr{E}\) for each half-reaction.)

The black silver sulfide discoloration of silverware can be removed by heating the silver article in a sodium carbonate solution in an aluminum pan. The reaction is $$3 \mathrm{Ag}_{2} \mathrm{S}(s)+2 \mathrm{Al}(s) \rightleftharpoons 6 \mathrm{Ag}(s)+3 \mathrm{S}^{2-}(a q)+2 \mathrm{Al}^{3+}(a q)$$ a. Using data in Appendix \(4,\) calculate \(\Delta G^{\circ}, K,\) and \(\mathscr{E}^{\circ}\) for the above reaction at $25^{\circ} \mathrm{C} .\left[\text { For } \mathrm{Al}^{3+}(a q), \Delta G_{\mathrm{f}}^{\circ}=-480 . \mathrm{kJ} / \mathrm{mol.}\right]$ b. Calculate the value of the standard reduction potential for the following half-reaction: $$2 \mathrm{e}^{-}+\mathrm{Ag}_{2} \mathrm{S}(s) \longrightarrow 2 \mathrm{Ag}(s)+\mathrm{S}^{2-}(a q)$$

Balance the following oxidation–reduction reactions that occur in basic solution using the half-reaction method. a. $\mathrm{PO}_{3}^{3-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{PO}_{4}^{3-}(a q)+\mathrm{MnO}_{2}(s)$ b. $\operatorname{Mg}(s)+\mathrm{OCl}^{-}(a q) \rightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Cl}^{-}(a q)$ c. $\mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+(a q) \rightarrow$ $$\mathrm{HCO}_{3}(a q)+\mathrm{Ag}(s)+\mathrm{NH}_{3}(a q)$$

A solution at \(25^{\circ} \mathrm{C}\) contains $1.0 M \mathrm{Cd}^{2+}, 1.0 M \mathrm{Ag}^{+}, 1.0 \mathrm{M}\( \)\mathrm{Au}^{3+},\( and 1.0 \)\mathrm{M} \mathrm{Ni}^{2+}$ in the cathode compartment of an electrolytic cell. Predict the order in which the metals will plate out as the voltage is gradually increased.
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