Question

Given the following half-reactions and associated standard reduction potentials: $$ \begin{array}{c}{\text { AuBr }_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{V}} \\ {\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

Short answer

(a) The largest positive emf is produced by the combination of the following half-cell reactions: Au(s)+4 Br⁻(aq) → AuBr₄⁻(aq)+3 e⁻ and IO⁻(aq)+H₂O(l)+2 e⁻ → I⁻(aq)+2 OH⁻(aq), with an emf of 1.35 V. (b) The smallest positive emf is produced by the combination of the following half-cell reactions: I⁻(aq)+2 OH⁻(aq) → IO⁻(aq)+H₂O(l)+2 e⁻ and Eu²⁺(aq) → Eu³⁺(aq)+ e⁻, with an emf of 0.92 V.

Step-by-step solution

Identifying reduction and oxidation for largest positive emf

For the largest positive emf, we want the half-reaction with the highest \(E_{red}\) to act as reduction and the one with the lowest \(E_{red}\) to act as oxidation. - Highest \(E_{red}\): IO⁻(aq)+H₂O(l)+2 e⁻ → I⁻(aq)+2 OH⁻(aq) with \(E_{red}^{\circ}=+0.49V\) - Lowest \(E_{red}\): AuBr₄⁻(aq)+3 e⁻ → Au(s)+4 Br⁻(aq) with \(E_{red}^{\circ}=-0.86V\) Now, we reverse the oxidation half-reaction because we need it to act as an oxidation.

Writing the largest positive emf reaction and calculating the emf

Largest positive emf reaction: Au(s)+4 Br⁻(aq) → AuBr₄⁻(aq)+3 e⁻ (Oxidation) IO⁻(aq)+H₂O(l)+2 e⁻ → I⁻(aq)+2 OH⁻(aq) (Reduction) Calculate the emf: \(E_{cell} = E_{red(R)} - E_{red(O)}\) \(E_{cell} = (+0.49) - (-0.86)\) \(E_{cell} = 1.35 V\)

Identifying reduction and oxidation for smallest positive emf

For the smallest positive emf, we want the half-reaction with the highest \(E_{red}\) to act as oxidation and the one with the second-lowest \(E_{red}\) to act as reduction. - Highest \(E_{red}\): IO⁻(aq)+H₂O(l)+2 e⁻ → I⁻(aq)+2 OH⁻(aq) with \(E_{red}^{\circ}=+0.49V\) - Second-lowest \(E_{red}\): Eu³⁺(aq)+ e⁻ → Eu²⁺(aq) with \(E_{red}^{\circ}=-0.43V\) Now, we reverse the oxidation half-reaction because we need it to act as an oxidation.

Writing the smallest positive emf reaction and calculating the emf

Smallest positive emf reaction: IO⁻(aq)+H₂O(l)+2 e⁻ → I⁻(aq)+2 OH⁻(aq) (Oxidation) Eu³⁺(aq)+ e⁻ → Eu²⁺(aq) (Reduction) Calculate the emf: \(E_{cell} = E_{red(R)} - E_{red(O)}\) \(E_{cell} = (-0.43) - (+0.49)\) \(E_{cell} = -0.92 V\) However, we are asked for the smallest positive emf, so we need to reverse this reaction, giving us: I⁻(aq)+2 OH⁻(aq) → IO⁻(aq)+H₂O(l)+2 e⁻ (Reduction) Eu²⁺(aq) → Eu³⁺(aq)+ e⁻ (Oxidation) Recalculate the emf: \(E_{cell} = +0.92 V\) (a) The combination of half-cell reactions that leads to the largest positive emf is Au(s)+4 Br⁻(aq) → AuBr₄⁻(aq)+3 e⁻ and IO⁻(aq)+H₂O(l)+2 e⁻ → I⁻(aq)+2 OH⁻(aq), with an emf of 1.35 V. (b) The combination of half-cell reactions that leads to the smallest positive emf is I⁻(aq)+2 OH⁻(aq) → IO⁻(aq)+H₂O(l)+2 e⁻ and Eu²⁺(aq) → Eu³⁺(aq)+ e⁻, with an emf of 0.92 V.

Key concepts

Standard Reduction Potentials

The standard reduction potential, denoted as \(E_{\text{red}}^{\circ}\), is a measure of the tendency of a chemical species to gain electrons and become reduced. These potentials are crucial in predicting the direction of redox reactions and the voltage of an electrochemical cell. Each substance has a unique standard reduction potential under standard conditions of 25°C, 1 atm pressure, and 1 M concentration.

To understand reduction potentials, it's essential to realize that the more positive the \(E_{\text{red}}^{\circ}\), the greater the species’ affinity for electrons. Conversely, a more negative \(E_{\text{red}}^{\circ}\) indicates a lower affinity. For example, in the given problem:

• \(IO^-\) has a positive potential of \(+0.49 \, V\), making it a strong oxidizing agent.
• \(AuBr_4^-\) with \(-0.86 \, V\) acts as a reducing agent since it's more negative.

This balance between reduction potentials guides which reactions will occur and how they can be combined in electrochemical cells.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two chemical species. In these reactions, one substance is oxidized (loses electrons), and another is reduced (gains electrons). This electron exchange is what makes electrochemical cells function.

To construct a redox reaction, you identify two half-reactions: oxidation and reduction. For instance:

• **Oxidation Half-Reaction:** A typical example from the exercise is the reaction: \(Au(s)+4 \text{Br}^- \rightarrow AuBr_4^- + 3e^-\), where gold is oxidized.
• **Reduction Half-Reaction:** Another part from the exercise is: \(IO^-+H_2O+2e^- \rightarrow I^-+2 \text{OH}^-\), where \(IO^-\) is reduced.

The overall redox reaction is balanced in terms of both mass and charge, ensuring that electrons lost in oxidation are equal to electrons gained in reduction. Understanding redox reactions is key to predicting cell reactions and potentials.

Emf Calculation

The electromotive force (emf) of an electrochemical cell is the potential difference between two half-cells under standard conditions. It tells us how much work a cell can do due to the flow of electrons from the anodic to cathodic compartments.

The emf can be calculated using the difference between the reduction potentials of the cathode and anode. The formula is:
\[E_{\text{cell}} = E_{\text{red (cathode)}} - E_{\text{red (anode)}}\]
For the largest positive emf in the exercise, you look for the combination of half-reactions that maximizes this difference:

• Cathodic Reaction: \(IO^- (aq)+H_2O (l)+2e^- \rightarrow I^- (aq)+2 OH^- (aq)\) with \(+0.49 \, V\)
• Anodic Reaction (oxidized): \(Au(s)+4 \text{Br}^- \rightarrow AuBr_4^- (aq)+3e^-\) with \(-0.86 \, V\)
• The emf: \(E_{\text{cell}} = 0.49 - (-0.86) = 1.35 \, V\)

The emf value determines not only the feasibility of the reaction but also the potential energy the cell can provide. Understanding emf calculations helps predict cell performance and efficiency.