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Chapter 18: Problem 51

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 \(18.1\).

Short Answer

Expert verified
The standard cell potential (\(\mathscr{C}^{\circ}\)) for the given half-reaction is calculated to be \(-0.828\ \mathrm{V}\) using the given values of \(\Delta G_{\mathrm{f}}^{\circ}\). To confirm our calculation, compare this value with the one given in Table 18.1 for the same half-reaction.

Step by step solution

01

Calculate the \(\Delta G^{\circ}\) for the reaction

We will use the following relation for calculating the Gibbs free energy change of the reaction: \[\Delta G^{\circ} = \sum \Delta G_{f}^{\circ} (\text{products}) - \sum \Delta G_{f}^{\circ}(\text{reactants})\] For the given reaction: \[\Delta G^{\circ} = [\Delta G_{\mathrm{f}}^{\circ}(H_2) + 2\Delta G_{\mathrm{f}}^{\circ}(OH^{-})] - [2\Delta G_{\mathrm{f}}^{\circ}(H_2O) + 2\Delta G_{\mathrm{f}}^{\circ}(e^{-})]\] Using the given values of \(\Delta G_{f}^{\circ}\): \[\Delta G^{\circ} = [(0.0) + 2(-157)] - [2(-237) + 2(0.0)] = -314 + 474 = 160 \mathrm{kJ/mol}\]
02

Calculate the \(\mathscr{C}^{\circ}\) for the reaction using the Nernst equation

The relationship between the Gibbs free energy change and standard cell potential is given by the Nernst equation: \[\Delta G^{\circ} = -nFE^{\circ}\] Where \(n\) represents the number of electrons transferred, \(F\) is the Faraday's constant (96,485 C/mol), and \(E^{\circ}\) is the standard cell potential. For the given half-reaction, \(n=2\) (two electrons are transferred): \[160000 = -2(96485)(E^{\circ})\] Now, we need to solve for \(E^{\circ}\): \[E^{\circ} = -\frac{160000}{2(96485)} = -0.828 \mathrm{V}\]
03

Compare the calculated value with that in Table 18.1

We've calculated the value \(\mathscr{C}^{\circ} = -0.828\). Now, we need to compare it with the value given in Table 18.1. You can consult the table in your textbook and look for the corresponding standard cell potential for the same half-reaction. If the value closely matches the calculated value of \(-0.828 \mathrm{V}\), it confirms the correctness of our calculation.

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Consider the following half-reactions: $$\begin{array}{cc}\mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{array}$$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 \mathrm{M}\) in chloride ion and \(0.020 \mathrm{M}\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

Consider only the species (at standard conditions) $$\mathrm{Na}^{+}, \mathrm{Cl}^{-}, \mathrm{Ag}^{+}, \mathrm{Ag}, \mathrm{Zn}^{2+}, \mathrm{Zn}, \mathrm{Pb}$$ in answering the following questions. Give reasons for your answers. (Use data from Table 18.1.) a. Which is the strongest oxidizing agent? b. Which is the strongest reducing agent? c. Which species can be oxidized by \(\mathrm{SO}_{4}^{2-}(a q)\) in acid? d. Which species can be reduced by \(\mathrm{Al}(s)\) ?
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