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Chapter 18: Problem 14

Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

Short Answer

Expert verified
The sum of the first two reduction potentials, \(0.771V\) and \(-0.447V\), does not equal the third potential \(-0.036V\) because it doesn't take into account the reaction quotients or concentrations. To obtain the correct potential, the Nernst equation should be used: \[E_3 = E_1 + E_2 - \frac{RT}{3F} \ln{\frac{1}{[Fe^3+]}}\] This equation includes the effects of ion concentrations on the reaction potentials, allowing accurate calculation of the third potential using the first two potentials.

Step by step solution

01

Look up the Reduction Potentials

First, we need to look up the reduction potentials of the ions: 1. Fe³⁺ to Fe²⁺: \(E_1^\circ = 0.771V\) 2. Fe²⁺ to Fe: \(E_2^\circ = -0.447V\) 3. Fe³⁺ to Fe: \(E_3^\circ = -0.036V\)
02

Attempt to Combine Reduction Potentials

Add the reduction potentials from step 1: \[E_1^\circ + E_2^\circ \ne E_3^\circ\] \[0.771V + (-0.447V) \ne -0.036V\] \[0.324V \ne -0.036V\] This shows that adding the first two potentials doesn't equal the third potential.
03

Analyze the Nernst Equation

To understand why merely adding the potentials doesn't give the correct result, we can look at the Nernst equation: \[E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln{Q}\] The Nernst equation describes the relationship between the cell potential and the ion concentrations. The mismatch between the added potentials and the true potential indicates that the reaction quotient or concentrations are at play here.
04

Consider the Reaction Quotients

We need to consider the reaction quotients when adding the reactions. Here are the half-reactions: 1. Fe³⁺ + e⁻ → Fe²⁺ 2. Fe²⁺ + 2e⁻ → Fe And the full reaction: Fe³⁺ + 3e⁻ → Fe To find the correct reaction quotient for the full reaction, we have to multiply the reaction quotients for the half-reactions: For half-reaction 1: \[Q_1 = \frac{[Fe^2+]}{[Fe^3+]}\] For half-reaction 2: \[Q_2 = \frac{1}{[Fe^2+]}\] For the full reaction: \[Q = Q_1 * Q_2 = \frac{1}{[Fe^3+]}\]
05

Use the Concentration Term to Adjust Potentials

Now that we found the reaction quotient for the full reaction, we can use the Nernst equation to properly combine the half-cell potentials. \[E_{cell} = E_{cell}^\circ - \frac{RT}{nF} \ln{Q}\] For the full reaction, the Nernst equation should look like this: \[E_3 = E_1 + E_2 - \frac{RT}{3F} \ln{\frac{1}{[Fe^3+]}}\] Given that we know \(E_1\), \(E_2\), and \(E_3\), it is indeed possible to use the first two potentials to calculate the third potential. It is important to take into account the concentration term, as merely adding the standard reduction potentials does not always give an accurate result.

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Most popular questions from this chapter

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode immersed in a solution with \(\left[\mathrm{Al}^{3+}\right]=1.0 M .\) Sodium hydroxide is added to the aluminum compartment, causing \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) to precipitate. After precipitation of Al(OH) \(_{3}\) has ceased, the concentration of \(\mathrm{OH}^{-}\) is \(1.0 \times 10^{-4} M\) and the measured cell potential is 1.82 \(\mathrm{V}\) . Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Al}(\mathrm{OH})_{3}\). $$\mathrm{Al}(\mathrm{OH})_{3}(s) \rightleftharpoons \mathrm{Al}^{3+}(a q)+3 \mathrm{OH}^{-}(a q) \quad K_{\mathrm{sp}}=?$$

Given the following two standard reduction potentials, $$\begin{array}{ll}{\mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M}} & {\mathscr{E}^{\circ}=-0.10 \mathrm{V}} \\ {\mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M}} & {\mathscr{E}^{\circ}=-0.50 \mathrm{V}}\end{array}$$ solve for the standard reduction potential of the half-reaction $$\mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+}$$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

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

What mass of each of the following substances can be produced in 1.0 h with a current of 15 A? a. \(\mathrm{Co}\) from aqueous \(\mathrm{Co}^{2+}\) b. \(\mathrm{Hf}\) from aqueous \(\mathrm{Hf}^{4+}\) c. \(\mathrm{I}_{2}\) from aqueous \(\mathrm{KI}\) d. \(\mathrm{Cr}\) from molten \(\mathrm{CrO}_{3}\)

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)$$ The two half-cell reactions are $$\mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-}$$ $$\mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-}$$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3}\) . Oxide ions can move through this solid at high temperatures (about \(800^{\circ} \mathrm{C} ) . \Delta G\) for the overall reaction at $800^{\circ} \mathrm{C}\( under certain concentration conditions is \)-380 \mathrm{kJ}$ . Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.
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