Problem 40 - Chemistry: The Central Science

Question

Given the following half-reactions and associated standard reduction potentials: \(\mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)\) \(\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) \quad E_{\text {red }}^{\mathrm{o}}=-0.858 \mathrm{~V}\) \(\mathrm{HO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{2}(a q)+\mathrm{e}^{\circ} \longrightarrow \mathrm{Eu}^{2}(a q)=-0.43 \mathrm{~V}\) \(1 \mathrm{O}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)\) (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate

Step-by-Step Solution

Answer

The combination of half-cell reactions that leads to the largest positive emf is given by: 2I⁻(aq) + Eu³⁺(aq) → I₂(s) + Eu²⁺(aq) and the emf value is 0.858 V.

Step 1: Identify the most positive and negative potentials

It has been given in the exercise that the reduction potentials are as follows: AuBr4- + 3e- → Au(s) + 4Br- Eu3+(aq) + e- → Eu2+(aq), Eredº = -0.858V HO- + H2O(l) + 2e- → Eu2(aq) + eº, Eredº = -0.430V The last half-reaction is missing the reduction potential, but by observing the given reactions, we can see that the reaction with the most positive potential is indeed the one with the missing value: I2 + 2e- → 2I-(aq) The most negative potential is -0.858V (Eu3+ to Eu2+), and the least negative potential for reduction is -0.43V (HO- to Eu2). Now, we must combine the two half-reactions that received these values.

Step 2: Balance the half-reactions

Write the most positive (I2) and the most negative (Eu3+) half-reactions. Remember that the most positive reaction must be reversed to become the oxidation half-reaction. X is the missing reduction potential of the missing reaction. Oxidation Half-Reaction: 2I-(aq) → I2(s) + 2e-, Eº = -X V Reduction Half-Reaction: Eu3+(aq) + e- → Eu2+(aq), Eredº = -0.858 V Since both reactions already have the same number of electrons transferred, they don't need any further balancing.

Step 3: Add the half-reactions and determine the emf

Add the oxidation and reduction half-reactions and write the total equation for the redox process: 2I-(aq) + Eu3+(aq) → I2(s) + Eu2+(aq) Now, use the formula for the total emf: Eº = Eº(oxidation) + Eº(reduction) = -X - (-0.858) In this case, we obtain the largest positive emf when X is at its minimum value, so the value of this emf is 0.858 V.

Key Concepts

Standard Reduction Potential

Standard reduction potential is a key concept in electrochemistry. It tells us how likely a chemical species will gain electrons, also known as a reduction. Measured in volts, this potential is a comparative measure, which means it shows how one reaction stands in relation to others.

In general, more positive values indicate a greater tendency to gain electrons and get reduced, while more negative values suggest a stronger likelihood to lose electrons, hence get oxidized when reversed in reactions. For example, among the given half-reactions, the one involving \(\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)\) has a very negative potential of \(-0.858 \mathrm{~V}\), indicating that it has a lower tendency to undergo reduction under standard conditions.

As a student, always remember to compare potentials to identify the reaction that acts as an oxidizing agent and a reducing agent. The more positive the standard reduction potential, the stronger the oxidizing agent.

Redox Reactions

Redox reactions are fundamental chemical processes that involve the transfer of electrons between two chemical species. One species undergoes oxidation, losing electrons, while the other undergoes reduction, gaining those electrons. These reactions are vital in many biological and chemical systems, enabling energy transfer and chemical changes.

In the exercise, two half-reactions are combined: the reduction of \(\mathrm{Eu}^{3+}\) and the oxidation of \(\mathrm{I}^{-}\). The electrons gained by \(\mathrm{Eu}^{3+}\) in the reduction half are exactly balanced by the electrons lost by \(\mathrm{I}^{-}\) in its oxidation half, promoting electron conservation. Together, these form a spontaneous redox pair leading to a positive emf, reflecting a favorable energy change.

Key to mastering redox reactions is understanding the roles of reducing and oxidizing agents. The reducing agent donates electrons and gets oxidized, while the oxidizing agent accepts electrons and gets reduced.

Half-Reaction Balancing

Balancing half-reactions is a critical step in solving redox problems. This process ensures the electrons exchanged between the reducing and oxidizing agents are equal, allowing the overall reaction to proceed without any discrepancies.

Here’s a friendly walkthrough for balancing:

• Identify the oxidation and reduction reactions by comparing standard potentials.
• Balance the atoms undergoing oxidation or reduction first. As in the case of our example reactions, make sure the electrons gained and lost are equal.
• Adjust electron transfer so the reactions can be added together to form an overall equation.

In the example provided, the balancing requires reversing the oxidation half-reaction to ensure electron flow continuity between the reactions. The played out formula \(2I^- + Eu^{3+} \rightarrow I_2 + Eu^{2+}\) demonstrates this, emphasizing balanced electron exchange leading to appropriate emf calculations. Balancing, therefore, confirms the integrity and feasibility of the electrochemical reaction.