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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Reduction Potentials
When studying electrochemistry, understanding standard reduction potentials is crucial. These potentials measure the tendency of a chemical species to gain electrons and thus be reduced. They are typically measured in volts (V). Standard reduction potentials are determined under standard conditions, which include 25°C, 1 atmosphere for gases, and 1 mol/L concentrations for solutions.
Consider a reduction half-reaction like the one for iron, where ions are reduced to solid iron:
  • Fe²⁺ + 2e⁻ → Fe(s)
Each half-reaction is assigned a standard reduction potential, derived from experimental measurements. In our example, the standard reduction potential for Fe²⁺ to Fe is E° = -0.44 V.
These values allow us to compare the electron affinity of different ions. More positive values indicate a greater tendency to gain electrons. On the other hand, a more negative potential means the ion is less inclined to be reduced. This information enables us to predict reaction spontaneity when paired with the concept of Gibbs Free Energy.
Half-Reactions
Half-reactions break down the steps of redox (reduction-oxidation) reactions, crucial for understanding how electrons are transferred between substances. Unlike complete reactions, half-reactions separately depict the oxidation or reduction process.
In a half-reaction, you show either the loss or gain of electrons. This clarity helps in balancing reactions and calculating potentials or energies. Let’s examine the oxidation half-reaction for iron:
  • Fe(s) → Fe²⁺ + 2e⁻
For this process, two electrons are lost as iron oxidizes from a solid form to ions. In the reduction half-reaction:
  • Fe²⁺ + 2e⁻ → Fe(s)
Here, the reverse occurs as electrons are gained. It's important to note that the number of electrons gained in the reduction half-reaction must match those lost in oxidation to maintain charge balance. By using half-reactions, chemists can derive equations describing complex chemical processes in simpler terms.
Faraday's Constant
Faraday's Constant is a key concept in electrochemistry, representing the electric charge per mole of electrons. It is one of the factors used to calculate Gibbs Free Energy changes in redox reactions.
The constant is approximately equal to 96485 C/mol. It connects the charge flow in electrochemical reactions to the amount of substance undergoing change, facilitating the calculation of energy changes in mole-based terms. In essence, it quantifies the total electric charge carried by one mole of electrons.
Applying Faraday's Constant in Gibbs Free Energy calculations, we use the formula:
  • ΔG° = -nFℰ°
Here, ΔG° represents the change in Gibbs Free Energy, n is the number of moles of electrons transferred, and ℰ° is the electromotive force or cell potential.
This formula helps to quantify how much free energy is available or required for a reaction under standard conditions, thus predicting the reaction's spontaneity in electrochemical systems.

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Most popular questions from this chapter

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of \(2.50\) A passing for \(50.0\) min. If \(2.618 \mathrm{~g}\) Ru is produced at the cathode, what is the charge on the ruthenium ions in solution?

When magnesium metal is added to a beaker of \(\mathrm{HCl}(a q)\), a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when \(\mathrm{Mg}\) is added directly to the beaker of \(\mathrm{HCl}\) ? How can you harness this reaction to do useful work?

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Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$ \operatorname{AgI}(s)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s)+\mathrm{I}^{-}(a q) $$ (Hint: Reference the \(K_{\text {sp }}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+} .\) )

Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is \(\begin{aligned} \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow & \mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g) \end{aligned}\) Balance this equation.

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