Problem 52 - Chemistry: The Central Science in SI Units, Global Edition

Question

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\)

Step-by-Step Solution

Answer

\(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\): E° = 0.384 V, ΔG° = -73,669 J/mol, K ≈ \(2.94 \times 10^{129}\)

Step 1: Write a balanced equation

Write the balanced half-reactions for this process: Reduction half-reaction: \(Hg_{2}^{2+} + 2e^{-} \rightarrow 2Hg\) Oxidation half-reaction: \(2I^{-} \rightarrow I_{2} + 2e^{-}\) Now, add the two half-reactions to obtain the balanced net equation: \(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\)

Step 2: Calculate the standard emf (E°)

Use the standard reduction potentials to calculate the standard emf for this reaction: E°(Hg²⁺/Hg) = 0.92 V E°(I₂/I⁻) = 0.536 V Using the Nernst equation, we have: E°(cell) = E°(reduction) - E°(oxidation) E°(cell) = 0.92 V - 0.536 V = 0.384 V

Step 3: Calculate ΔG° at 298 K

Use the formula ΔG° = -nFE°, where n is the number of moles of electrons transferred and F is Faraday's constant (96,485 C/mol): ΔG° = - (2)(96,485 C/mol)(0.384 V) ΔG° = -73,669 J/mol

Step 4: Calculate the equilibrium constant (K) at 298 K

Use the formula ΔG° = -RT ln K, where R is the gas constant (8.314 J/mol K) and T is the temperature (298 K): -73,669 J/mol = -(8.314 J/mol K)(298 K) ln K ln K = 298 K ≈ \(2.94 \times 10^{129}\) The final answers for part (a) are: Balanced equation: \(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\) Standard emf (E°): 0.384 V ΔG°: -73,669 J/mol K: ≈ \(2.94 \times 10^{129}\) Repeat the process for parts (b) and (c).

Key Concepts

Electrochemistry

Electrochemistry involves the study of chemical processes that cause electrons to move. This movement of electrons is what generates electricity. In chemical reactions, these processes occur in electrochemical cells. These cells are devices that convert chemical energy into electrical energy or vice-versa. There are two main types of electrochemical cells: galvanic (or voltaic) cells, which produce electrical energy from spontaneous chemical reactions, and electrolytic cells, which use electrical energy to drive non-spontaneous reactions.

In our example, we're dealing with redox reactions where oxidation and reduction occur. Oxidation is the loss of electrons, while reduction is the gain of electrons. Each of these half-reactions takes place at electrodes. The oxidation half-reaction occurs at the anode, and the reduction half-reaction occurs at the cathode.

• Oxidation: electrons are lost.
• Reduction: electrons are gained.
• Anode: where oxidation happens.
• Cathode: where reduction happens.

Understanding these principles allows us to balance equations, calculate potentials, and produce useful energy through controlled reactions.

Standard Electrode Potential

The standard electrode potential, denoted as E°, is the voltage, or electric potential difference, of a cell under standard conditions. Standard conditions include a temperature of 298 K, a pressure of 1 atm, and usually, solutions at 1 M concentration. These potentials allow us to predict the direction of redox reactions and calculate the standard electromotive force (emf) of a reaction.

In electrochemistry, the standard reduction potentials of different substances are used to calculate the standard emf of a cell. The emf is determined using the formula:
\[ E^{ ext{°cell}} = E^{ ext{°reduction}} - E^{ ext{°oxidation}} \]
Here’s how you apply it:

• Identify the reduction and oxidation half-reactions.
• Consult standard tables to find the E° values for each half-reaction.
• Subtract the oxidation potential from the reduction potential to find E° of the whole cell.

The calculated emf indicates whether a reaction will spontaneously occur. A positive emf suggests a spontaneous reaction.

Gibbs Free Energy

Gibbs Free Energy (\( \Delta G \)) helps determine whether a chemical process is spontaneous. It's the energy available to do work. In electrochemistry, \( \Delta G^{°} \) relates to the cell potential as follows:
\[ \Delta G^{°} = -nFE^{°} \]

Where:

• \( n \) is the number of moles of electrons transferred.
• \( F \) is Faraday’s constant (96,485 C/mol).
• \( E^{°} \) is the standard emf in volts.

A negative \( \Delta G^{°} \) means the reaction is spontaneous, releasing energy. For instance, if we calculate \( \Delta G^{°} \) to be \(-73,669 \text{ J/mol}\), the reaction energetically favors products at the given conditions.

Understanding \( \Delta G\) in electrochemical reactions provides insights into reaction spontaneity and sustainability.

Equilibrium Constant

The equilibrium constant (\( K \)) conveys how far a reaction goes towards completion. In electrochemical reactions, \( K \) is related to \( \Delta G^{°} \) through the equation:
\[ \Delta G^{°} = -RT \ln K \]

• \( R \) is the gas constant (8.314 J/mol K).
• \( T \) is the temperature in Kelvin.

This equation ties the energetics of the reaction to its position at equilibrium.

Rearranging and solving for \( K \), you get:
\[ K = e^{-\Delta G^{°}/RT} \]

A larger \( K \) value typically suggests that the products are favored at equilibrium. For example, with a calculated \( K \approx 2.94 \times 10^{129} \), the reaction proceeds nearly to completion, indicating that product formation is highly favored under standard conditions.

This understanding allows for predictions about the extent and direction of reactions.