Problem 56 - Chemistry: The Central Science

Question

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) - (a) \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2}+(a q)+2 \mathrm{Ag}(s)\) (b) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+\) \(\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\) \(5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4}(a q)\)

Step-by-Step Solution

Answer

The equilibrium constants for the given reactions at 298 K are as follows: (a) \( K_a = 1.14 \times 10^{15} \) (b) \( K_b = 4.89 \times 10^{-22} \) (c) \( K_c = 7.74 \)

Step 1: Identify the half-reactions for each reaction

First, we need to identify the half-reactions, so we can look up the standard reduction potentials in Appendix E. (a) Cu(s)+2 Ag+(aq) → Cu2+(aq)+2 Ag(s) Half-reactions: Oxidation: Cu(s) → Cu2+(aq)+2e- Reduction: 2 Ag+(aq)+2e- → 2 Ag (s) (b) 3 Ce4+(aq)+Bi(s)+H2O(l) → 3 Ce3+(aq)+BiO+(aq)+2 H+(aq) Half-reactions: Oxidation: Bi(s) → BiO+(aq)+H+(aq) + e- Reduction: 3 Ce4+(aq)+3e- → 3 Ce3+(aq) (c) N2H5+(aq)+4Fe(CN)63-(aq) → N2(g)+5 H+(aq)+4 Fe(CN)64-(aq) Half-reactions: Oxidation: N2H5+(aq) → N2(g)+5 H+(aq)+5e- Reduction: 4 Fe(CN)63-(aq)+4e- → 4 Fe(CN)64-(aq)

Step 2: Calculate standard cell potentials

Now let's find the standard reduction potentials from Appendix E and calculate the cell potentials for each reaction. (a) Oxidation: Cu(s) → Cu2+(aq)+2e-, E° = +0.337 V Reduction: 2 Ag+(aq)+2e- → 2 Ag(s), E° = +0.799 V Cell potential E°(cell) = E°(reduction) - E°(oxidation) = 0.799 - 0.337 = 0.462 V (b) Oxidation: Bi(s) → BiO+(aq)+H+(aq) + e-, E° = -0.308 V Reduction: 3 Ce4+(aq)+3e- → 3 Ce3+(aq), E° = -1.61 V Cell potential E°(cell) = E°(reduction) - E°(oxidation) = (-1.61) - (-0.308) = -1.302 V (c) Oxidation: N2H5+(aq) → N2(g)+5 H+(aq)+5e-, E° = -0.182 V Reduction: 4 Fe(CN)63-(aq)+4e- → 4 Fe(CN)64-(aq), E° = -0.356 V Cell potential E°(cell) = E°(reduction) - E°(oxidation) = (-0.356) - (-0.182) = -0.174 V

Step 3: Calculate the equilibrium constants

Now we can calculate the equilibrium constants using the Nernst equation: \( K = 10^{(\frac{-n \times E°(cell)}{0.05916})} \) (a) For reaction (a), n = 2 because there are 2 moles of electrons transferred, E°(cell) = 0.462 V: \( K_a = 10^{(\frac{-2 \times 0.462}{0.05916})} = 1.14 \times 10^{15} \) (b) For reaction (b), n = 1 because there is 1 mole of electrons transferred, E°(cell) = -1.302 V: \( K_b = 10^{(\frac{-1 \times -1.302}{0.05916})} = 4.89 \times 10^{-22} \) (c) For reaction (c), n = 5 because there are 5 moles of electrons transferred, E°(cell) = -0.174 V: \( K_c = 10^{(\frac{-5 \times -0.174}{0.05916})} = 7.74 \) Therefore, the equilibrium constants for reactions (a), (b), and (c) are: \( K_a = 1.14 \times 10^{15} \) \( K_b = 4.89 \times 10^{-22} \) \( K_c = 7.74 \)

Key Concepts

Standard Reduction Potential

The standard reduction potential is a measure of the tendency of a chemical species to gain electrons, thereby undergoing reduction. It is a critical parameter that helps determine the direction of electron flow in electrochemical cells. The units are volts (V), and these values are typically found in tables, such as Appendix E in your textbooks.

• The higher the standard reduction potential, the greater the tendency for a species to be reduced.
• The lower (or the more negative) the value, the less likely the species will accept electrons.

In simple terms, think of it as a ranking of how strongly different substances pull electrons towards themselves. For example, in reaction (a), silver ions (Ag⁺) have a higher standard reduction potential compared to copper ions (Cu²⁺), making silver ions more likely to gain electrons and be reduced.

Nernst Equation

The Nernst equation is an essential tool in electrochemistry, used to calculate the cell potential at any point other than standard conditions. It links the cell potential to the concentrations of the reacting species, enabling the determination of the equilibrium constant for the reaction.
The equation is given by:
\[ E = E° - \left( \frac{RT}{nF} \right) \ln Q \]
Where:

• \(E\) = cell potential at non-standard conditions (V)
• \(E°\) = standard cell potential (V)
• \(R\) = universal gas constant \( (8.314 \, J/mol \, K)\)
• \(T\) = temperature in Kelvin
• \(n\) = number of moles of electrons exchanged
• \(F\) = Faraday's constant \((96485 \, C/mol)\)
• \(Q\) = reaction quotient

For simplification at standard temperature of 298 K, a common form of the Nernst Equation is:\[ E = E° - \left( \frac{0.05916}{n} \right) \log Q \]This form is often useful in determining the equilibrium constant of a reaction.

Half-Reaction

A half-reaction is part of the redox reaction that shows either oxidation or reduction occurring, including all electrons involved in the transfer. Each half-reaction consists of two parts: the oxidized species and reduced species.

• Oxidation half-reaction: This involves the loss of electrons. For example, in reaction (a), copper is oxidized as it loses electrons to form copper ions.
• Reduction half-reaction: This involves the gain of electrons. Continuing with reaction (a), silver ions gain electrons to form solid silver.

By using the standard reduction potentials of these half-reactions, you can calculate the overall cell potential, which is key to understanding how favorable the reaction is overall.

Cell Potential

Cell potential, also known as electromotive force (emf), is the driving force behind electron flow in an electrochemical cell. It is expressed in volts (V) and calculated from the standard reduction potentials of the half-reactions.
The formula to find the standard cell potential \(E°(\text{cell})\) is:
\[ E°(\text{cell}) = E°(\text{reduction}) - E°(\text{oxidation}) \]

• A positive cell potential indicates a spontaneous redox reaction.
• A negative cell potential suggests a non-spontaneous process under standard conditions.

In the original exercise, calculating these potentials helps you determine the feasibility of the reactions and their equilibrium constants, aiding in predicting the direction of electron flow.

Electron Transfer

Electron transfer is the fundamental chemical process of redox reactions where electrons are shifted from one reactant (the reducing agent) to another (the oxidizing agent). These transfers are what drive the generation of cell potential in electrochemical cells.
Key aspects of electron transfer include:

• It results in the release or absorption of energy, depending on whether the electrons are moving to a lower or higher energy state.
• In balanced equations, both the number of electrons lost in oxidation and gained in reduction are equal.

Understanding electron transfer is crucial because it dictates the direction and spontaneity of a reaction. The number of electrons involved, denoted by \(n\) in the Nernst equation, is directly related to the equilibrium constant calculations sought in the exercise.