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Chapter 18: Problem 87

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 \mathrm{M}), \mathrm{Cr}^{3+}(2.0 \mathrm{M})\right|\left|\mathrm{Co}^{2+}(0.20 \mathrm{M})\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$ \begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow & \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K &=2.79 \times 10^{7} \end{aligned} $$ 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 the given galvanic cell is approximately 0.848V, and the Gibbs free energy change (\(\Delta G\)) is approximately -1.63 × 10^5 J/mol.

Step by step solution

01

Identify the half-cell reactions

Before we proceed, let's identify the half-cell reactions occurring at the anode and the cathode in the given galvanic cell. Observe that this cell has the following structure: $$\mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 \mathrm{M}), \mathrm{Cr}^{3+}(2.0 \mathrm{M})\right|\left|\mathrm{Co}^{2+}(0.20 \mathrm{M})\right| \mathrm{Co}$$ At the anode (left-side), we have Cr2+ being oxidized to Cr3+: \[ \mathrm{Cr}^{2+}(aq) \longrightarrow \mathrm{Cr}^{3+}(aq) + e^- \] At the cathode (right-side), we have Co2+ being reduced to Co: \[ \mathrm{Co}^{2+}(aq) + 2e^- \longrightarrow \mathrm{Co}(s) \] Now that we have our half-cell reactions, let us proceed.
02

Using the Nernst Equation

The Nernst Equation is given by: \[ E_{cell} = E^\circ_{cell} - \frac{0.0592}{n} \log Q \] To use this equation, we need to find the values for \(E^\circ_{cell}\), n, and Q.
03

Calculate the standard cell potential, \(E^\circ_{cell}\)

We can find the standard cell potential by using the relationship between the equilibrium constant (K) and standard cell potential: \(E^\circ_{cell} =\frac{0.0592}{n} \log K\) We are given the equilibrium constant, K = 2.79 × 10^7, and the stoichiometry of the electrons exchanged in the overall reaction is 2, so n = 2. Now we can calculate the standard cell potential: \(E^\circ_{cell} =\frac{0.0592 \times \log (2.79 \times 10^7)}{2} = 0.902V\)
04

Calculate the reaction quotient (Q)

The reaction quotient, Q, is given by: \(Q = \frac{[\mathrm{Cr}^{3+}]^{2}}{[\mathrm{Cr}^{2+}][\mathrm{Co}^{2+}]}\) Using the given concentrations for Cr2+, Cr3+, and Co2+, we can calculate Q: \(Q = \frac{(2.0)^{2}}{(0.3)(0.2)} =\frac {4}{0.06} = 66.67\)
05

Calculate the cell potential, E_{cell}

Now we have all the values required to use the Nernst Equation. Let's substitute these values into the equation: \(E_{cell} = 0.902 - \frac{0.0592}{2} \log 66.67 = 0.902 - 0.0296 \times 1.823 = 0.902 - 0.054 = 0.848V\)
06

Calculate the Gibbs free energy change, ∆G

Now we will use the relationship between the cell potential and Gibbs free energy change: \(\Delta G = -nF E_{cell}\) Where n is the stoichiometry of the electrons exchanged (n = 2), F is the Faraday constant (F = 96,485 C/mol), and \(E_{cell}\) is the cell potential (0.848 V): \(\Delta G = -2 \times 96485 \times 0.848 = -163140.104 J/mol ≈ -1.63 \times 10^5 J/mol\) Thus, we have found the cell potential (\(\mathscr{E}\)) and Gibbs free energy change (\(\Delta G\)) for the given galvanic cell to be approximately 0.848V and -1.63 × 10^5 J/mol, respectively.

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Most popular questions from this chapter

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? a. molten KF b. molten \(\mathrm{CuCl}_{2}\) c. molten \(\mathrm{MgI}_{2}\)

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 \(\mathscr{E}_{\text {cell }}^{\circ}\) 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 \(\left[\mathrm{Zn}^{2+}\right]=0.10 M\) and \(\left[\mathrm{Fe}^{2+}\right]=1.0 \times 10^{-5} M\)

The following standard reduction potentials have been det mined for the aqueous chemistry of indium: $$ \begin{array}{cl} \mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V} \end{array} $$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\operatorname{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q) ?\)

Consider the galvanic cell based on the following halfreactions: $$ \begin{array}{ll} \mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au} & \mathscr{E}^{\circ}=1.50 \mathrm{~V} \\ \mathrm{Tl}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Tl} & \mathscr{E}^{\circ}=-0.34 \mathrm{~V} \end{array} $$

One of the few industrial-scale processes that produce organic compounds electrochemically is used by the Monsanto Company to produce 1,4 -dicyanobutane. The reduction reaction is $$ 2 \mathrm{CH}_{2}=\mathrm{CHCN}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN} $$ The \(\mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) is then chemically reduced using hydrogen gas to \(\mathrm{H}_{2} \mathrm{~N}-\left(\mathrm{CH}_{2}\right)_{6}-\mathrm{NH}_{2}\), which is used in the production of nylon. What current must be used to produce 150. \(\mathrm{kg} \mathrm{NC}-\left(\mathrm{CH}_{2}\right)_{4}-\mathrm{CN}\) per hour?
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