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.

Step by step solution

01

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.

02

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: \(E_{cell} = E_{red(R)} - E_{red(O)}\) \(E_{cell} = (+0.49) - (-0.86)\) \(E_{cell} = 1.35 V\)

03

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.

04

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.

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