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Chapter 18: Problem 83

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$\mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co}$$ The overall reaction and equilibrium constant value are $$2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow{2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s)} \quad K=2.79 \times 10^{7}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

Short Answer

Expert verified
The cell potential, \(\mathscr{E}\), for this galvanic cell is found using the Nernst Equation: \(\mathscr{E} = \mathscr{E}^{\circ} - \cfrac{RT}{nF} \ln K\). Balancing the redox reaction, the number of electrons transferred, \(n\), is 4. Plugging in the given values and constants, we find \(\mathscr{E} \approx 0.721\, \mathrm{V}\). To calculate the change in Gibbs free energy, \(\Delta G\), use the relationship \(\Delta G = -nF\mathscr{E}\), resulting in \(\Delta G \approx -278\, \mathrm{kJ\, mol^{-1}}\).

Step by step solution

01

Determine the standard potential for the cell reaction

Use the Nernst Equation: \(\mathscr{E} = \mathscr{E}^{\circ} - \cfrac{RT}{nF} \ln K\), where \(\mathscr{E}\) is the cell potential, \(\mathscr{E}^{\circ}\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant, and \(K\) is the equilibrium constant. We are given \(K\), and we can find \(n\) by balancing the redox reaction. Then, we can solve for \(\mathscr{E}^{\circ}\). Step 2: Balance the redox reaction and find the number of electrons transferred (n)
02

Balance the redox reaction

The overall cell reaction is given as \(2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow{2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s)}\). Here, two electrons are transferred from each \(Cr^{2+}\) ion (total of 4 electrons) to the \(Co^{2+}\) ion. So, \(n = 4\). Step 3: Calculate the cell potential, \(\mathscr{E}\)
03

Calculate the cell potential using the Nernst Equation

Use the Nernst Equation: \(\mathscr{E} = \mathscr{E}^{\circ} - \cfrac{RT}{nF} \ln K\). We now have the values for all variables except \(\mathscr{E}^{\circ}\): \(T = 25^{\circ} \mathrm{C} + 273.15 = 298.15\, \mathrm{K}\) (convert Celsius to Kelvin), \(n = 4\), \(K = 2.79 \times 10^{7}\), \(R = 8.314\, \mathrm{J\, mol^{-1}K^{-1}}\), and \(F = 96485\, \mathrm{C\, mol^{-1}}\). Plug these values into the Nernst Equation and solve for \(\mathscr{E}\). Step 4: Calculate \(\Delta G\)
04

Calculate \(\Delta G\) using the relationship between \(\mathscr{E}\) and \(\Delta G\)

Use the relationship: \(\Delta G = -nF\mathscr{E}\), where \(\Delta G\) is the change in Gibbs free energy, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant, and \(\mathscr{E}\) is the cell potential. Plug in the values for \(n\), \(F\), and \(\mathscr{E}\) calculated in previous steps, and solve for \(\Delta G\).

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Most popular questions from this chapter

In making a specific galvanic cell, explain how one decides on the electrodes and the solutions to use in the cell.

Zirconium is one of the few metals that retains its structural integrity upon exposure to radiation. For this reason, the fuel rods in most nuclear reactors are made of zirconium. Answer the following questions about the redox properties of zirconium based on the half-reaction \(\mathrm{ZrO}_{2} \cdot \mathrm{H}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Zr}+4 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-2.36 \mathrm{~V}\) a. Is zirconium metal capable of reducing water to form hydrogen gas at standard conditions? b. Write a balanced equation for the reduction of water by zirconium metal. c. Calculate \(\mathscr{8}^{\circ}, \Delta G^{\circ}\), and \(K\) for the reduction of water by zirconium metal. d. The reduction of water by zirconium occurred during the accident at Three Mile Island, Pennsylvania, in \(1979 .\) The hydrogen produced was successfully vented and no chemical explosion occurred. If \(1.00 \times 10^{3} \mathrm{~kg} \mathrm{Zr}\) reacts, what mass of \(\mathrm{H}_{2}\) is produced? What volume of \(\mathrm{H}_{2}\) at \(1.0 \mathrm{~atm}\) and \(1000 .{ }^{\circ} \mathrm{C}\) is produced? e. At Chernobyl, USSR, in 1986 , hydrogen was produced by the reaction of superheated steam with the graphite reactor core: $$\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ A chemical explosion involving the hydrogen gas did occur at Chernobyl. In light of this fact, do you think it was a correct decision to vent the hydrogen and other radioactive gases into the atmosphere at Three Mile Island? Explain.

Sketch a galvanic cell, and explain how it works. Look at Figs. \(18.1\) and \(18.2 .\) Explain what is occurring in each container and why the cell in Fig. \(18.2\) "works" but the one in Fig. \(18.1\) does not.

Consider the galvanic cell based on the following half-reactions: $$\begin{array}{ll}\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array}$$ a. Determine the overall cell reaction and calculate \(\mathscr{E}_{\mathrm{ccll}}^{\circ}\) b. Calculate \(\Delta G^{\circ}\) and \(K\) for the cell reaction at \(25^{\circ} \mathrm{C}\). c. Calculate \(\mathscr{E}_{\text {cell }}\) at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{Au}^{3+}\right]=1.0 \times 10^{-2} M\) and \(\left[\mathrm{Tl}^{+}\right]=1.0 \times 10^{-4} \mathrm{M}\)

Estimate \(\mathscr{C}^{\circ}\) 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 \(\mathscr{b}^{\circ}\) given in Table \(18.1\).
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