Problem 40 - Chemistry The Central Science In Si Units

Question

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ} &=-0.86 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\mathrm{red}}^{\circ} &=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(I)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ} &=+0.49 \mathrm{~V} \end{aligned} $$ (a) Write the equation for the combination of these halfcell 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

Answer

Largest positive emf: \(0.92\) V, reaction includes IO and Eu. Smallest positive emf: \(1.35\) V, reaction includes IO and Au.

Step 1: Determine the Potential of Possible Reactions

For part (a), find the combination of half-reactions that gives the largest positive emf. The overall reaction will only be favorable if the cell potential is positive. Examine combinations by reversing the least negative components since reduction potential becomes oxidation potential when reversed.\[E_{cell} = E_{red} + E_{ox}\]For the largest positive emf:1. Reverse \(\text{Eu}^{3+} + \text{e}^{-} \rightarrow \text{Eu}^{2+} \), \(E_{ox}^{ ext{Eu}} = +0.43 \text{ V}\).

Step 2: Combine with Iodate Reduction

Combine the iodate reduction with europium oxidation:-\(\text{IO}^- + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{I}^- + 2\text{OH}^-\), \(E_{red}^{\text{IO}^-} = +0.49 \text{ V}\)The emf for the combined cell becomes:\[E_{cell} = E_{red}^{\text{IO}^-} + E_{ox}^{\text{Eu}} = 0.49 + 0.43 = 0.92 \text{ V}\]The overall reaction is:\[\text{IO}^- + \text{H}_2\text{O} + \text{Eu}^{3+} \rightarrow \text{I}^- + 2 \text{OH}^- + \text{Eu}^{2+}\]

Step 3: For the Smallest Positive EMF

For part (b), find the combination that leads to the smallest positive emf. Set the less favorable (less positive) reactions in combination.Reverse the gold half-reaction:-\(\text{Au}(s) + 4 \text{Br}^- \rightarrow \text{AuBr}_4^- + 3\text{e}^-\), \(E_{ox}^{\text{Au}} = +0.86 \text{ V}\).Use the iodate half-reaction unchanged:- \(\text{IO}^- + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{I}^- + 2\text{OH}^-\), \(E_{red}^{\text{IO}^-} = +0.49 \text{ V}\).

Step 4: Combine and Calculate Smallest Positive EMF

Calculate the cell potential:\[E_{cell} = E_{red}^{\text{IO}^-} + E_{ox}^{\text{Au}} = 0.49 + 0.86 = 1.35 \text{ V}\]The overall balanced reaction, which must include balancing electrons, is:\[3(\text{IO}^- + \text{H}_2\text{O} + 2 \text{e}^- \rightarrow \text{I}^- + 2 \text{OH}^- ) \\text{Au}(s) + 4 \text{Br}^- \rightarrow \text{AuBr}_4^- + 3 \text{e}^-\]This results in:\[3 \text{IO}^- + 3\text{H}_2\text{O} + \text{Au}(s) + 4 \text{Br}^- \rightarrow 3 \text{I}^- + 6 \text{OH}^- + \text{AuBr}_4^-\]

Key Concepts

Standard Reduction Potential

The standard reduction potential is a measure that tells us how likely a particular chemical species is to gain electrons, meaning how easily it gets reduced. It’s expressed in volts (V). Each half-reaction in electrochemistry has its own standard reduction potential.

• A higher positive value indicates a substance that readily gains electrons and gets reduced.
• A more negative value indicates a substance that does not easily get reduced and would rather lose electrons and be oxidized.

These potentials are determined under standard conditions: 25°C, 1M concentrations for solutions, 1 atm pressure for gases, and pure solids or liquids. They are essential in predicting the direction and feasibility of a redox reaction.
In this context, comparing the given standard reduction potentials helps in deciding which half-reaction should undergo oxidation and which should undergo reduction in order to achieve the desired cell potential.

Cell Potential

Cell potential, denoted as \(E_{cell}\), represents the driving force behind an electrochemical reaction and is calculated by combining the potentials of the two half-cells in a redox reaction. It’s the voltage difference between two half-cells.

The formula to calculate cell potential is:

• \(E_{cell} = E_{red} + E_{ox}\)

Here, \(E_{red}\) is the reduction potential of the half-reaction that gains electrons and \(E_{ox}\) is the oxidation potential of the half-reaction that loses electrons.

To find the oxidation potential, you invert the sign of the standard reduction potential of the reversed half-reaction. A positive \(E_{cell}\) means the reaction is spontaneous and will proceed as written.

In the given exercise, combinations of standard reduction potentials are calculated to find the largest positive \(E_{cell}\), making it easier to decide which reactions yield the most favorable outcome.

Half-Reactions

Half-reactions are essential components of redox reactions that describe either the oxidation or reduction process separately. Each one shows either the gain (reduction) or loss (oxidation) of electrons.

In writing half-reactions:

• Reduction half-reaction shows electrons being gained.
• Oxidation half-reaction shows electrons being lost.

Combining these two processes forms a complete redox reaction. Balancing half-reactions includes making sure the number of electrons lost in the oxidation half equals the number gained in the reduction half.

In the original exercise, the half-reactions given include both with positive and negative standard reduction potentials, representing their tendencies for reduction or oxidation.

By manipulating the reactions depending on these potentials, you can determine which reactions will lead to the greatest or smallest cell potential when combined. This method is crucial for solving problems related to electrochemical cells, predicting reaction outcomes, and understanding redox processes.