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Chapter 18: Problem 54

The equation \(\Delta G^{\circ}=-n F \mathscr{E}^{\circ}\) also can be applied to halfreactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and \(\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ}\right.\) for \(\left.\mathrm{e}^{-}=0 .\right)\)

Short Answer

Expert verified
The standard Gibbs free energy of formation (ΔGf°) for Fe²⁺(aq) and Fe³⁺(aq) can be calculated using the equation ΔG° = -nFℰ° and the standard reduction potentials. The balanced half-reactions are: \(Fe^{2+}(aq) + 2e^- \rightarrow Fe(s)\) with \(E°_{Fe^{2+}/Fe} = -0.44 V\) \(Fe^{3+}(aq) + 3e^- \rightarrow Fe(s)\) with \(E°_{Fe^{3+}/Fe} = -0.036 V\) Using these values, we find that the ΔGf° for Fe²⁺(aq) is approximately 84965.6 J/mol and the ΔGf° for Fe³⁺(aq) is approximately 10460.42 J/mol.

Step by step solution

01

Write the balanced half-reactions for Fe²⁺ and Fe³⁺

The balanced half-reactions for Fe²⁺ and Fe³⁺ are: \(Fe^{2+}(aq) + 2e^- \rightarrow Fe(s)\) \(Fe^{3+}(aq) + 3e^- \rightarrow Fe(s)\) Now we have the half-reactions, and we know the number of electrons involved (n) in each reaction.
02

Find the standard reduction potentials for the half-reactions

You can find the standard reduction potentials for these half-reactions in a standard reduction potential table: For Fe²⁺: \(E°_{Fe^{2+}/Fe} = -0.44 V\) For Fe³⁺: \(E°_{Fe^{3+}/Fe} = -0.036 V\)
03

Calculate the ΔGf° for Fe²⁺ and Fe³⁺

Use the equation ΔG° = -nFℰ° to find the ΔGf° for each species. Faraday's constant (F) is equal to 96485 C/mol. For Fe²⁺: ΔG° = -nFℰ° ΔG° = -(2 mol)(96485 C/mol)(-0.44 V) ΔG° = 84965.6 J/mol For Fe³⁺: ΔG° = -nFℰ° ΔG° = -(3 mol)(96485 C/mol)(-0.036 V) ΔG° = 10460.42 J/mol Thus, the ΔGf° for Fe²⁺(aq) is approximately 84965.6 J/mol and the ΔGf° for Fe³⁺(aq) is approximately 10460.42 J/mol.

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Most popular questions from this chapter

Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) c. \(\mathrm{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}{ }^{2-}(a q)\) d. \(\operatorname{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow}\) \(\operatorname{CrO}_{4}{ }^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Base }}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)

The Ostwald process for the commercial production of nitric acid involves the following three steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Which reactions in the Ostwald process are oxidationreduction reactions? b. Identify each oxidizing agent and reducing agent.

Calculate the \(\mathrm{pH}\) of the cathode compartment for the following reaction given \(\mathscr{E}_{\text {cell }}=3.01 \mathrm{~V}\) when \(\left[\mathrm{Cr}^{3+}\right]=0.15 \mathrm{M}\), \(\left[\mathrm{Al}^{3+}\right]=0.30 M\), and \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right]=0.55 M\) \(2 \mathrm{Al}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q) \longrightarrow\) \(2 \mathrm{Al}^{3+}(a q)+2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l)\)

Balance the following oxidation-reduction reactions that occur in basic solution. a. \(\mathrm{Cr}(s)+\mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}(\mathrm{OH})_{3}(s)\) b. \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{S}^{2-}(a q) \rightarrow \mathrm{MnS}(s)+\mathrm{S}(s)\) c. \(\mathrm{CN}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{CNO}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{~V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).
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