Problem 54 - 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) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l)-\ldots\) \(3 \mathrm{Ce}^{3+}(a q)+\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}{\underline{\phantom{xx}}}^{3-}(a q)-\cdots \rightarrow\) \(\mathrm{N}_{2}(g)+5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4^{-}}(a q)\)

Step-by-Step Solution

Answer

In summary, for reaction (a) $\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)$, the equilibrium constant (K) at $298\mathrm{~K}$ is \(1.31 \times 10^{10}\).

Step 1: The two half-reactions are: Oxidation: \(\mathrm{Cu}(s) \longrightarrow \mathrm{Cu}^{2+}(a q) + 2 e^-\) Reduction: \(2 \mathrm{Ag}^{+}(a q) + 2 e^- \longrightarrow 2 \mathrm{Ag}(s)\) #Step 2: Determine the standard reduction potentials#

Using Appendix E, we find the standard reduction potentials for the half-reactions: Cu²⁺/Cu: E° = +0.337 V Ag⁺/Ag: E° = +0.799 V #Step 3: Calculate the overall cell potential (E°)#

Step 2: The cell potential is equal to the reduction potential of the cathode minus the reduction potential of the anode: E° = E°(cathode) - E°(anode). In our case, the copper half-reaction is the anode, and the silver half-reaction is the cathode. E° = (+0.799 V) - (+0.337 V) = +0.462 V #Step 4: Find the equilibrium constant (K)#

Using the relationship between cell potential and equilibrium constant: \[\Delta G° = -nFE°\] Also, using the relationship between Gibbs free energy and equilibrium constant: \[\Delta G° = -RT \ln K\] Combining these two equations and solving for K, we get: \[K= \exp{ \frac{-nFE°}{RT}}\] where n = number of electrons transferred in the reaction (in this case, n = 2), F is the Faraday's constant (96485 C/mol), R is the gas constant (8.314 J/mol·K), and T is the temperature in Kelvin (298 K). Substituting the values: \[K= \exp{ \frac{-2 \times 96485 \text{C/mol} \times 0.462 \text{V}}{8.314 \text{J/mol·K} \times 298 \text{K}} }\] \[K = 1.31 \times 10^{10}\] The equilibrium constant for reaction (a) is \(1.31 \times 10^{10}\). For reactions (b) and (c), we would follow the same steps of identifying half-reactions, determining standard reduction potentials, calculating the cell potential, and then finding the equilibrium constant.

Key Concepts

Standard Reduction Potential

Understanding standard reduction potential, often denoted as E°, is crucial in the field of electrochemistry. It is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. Each species has an intrinsic standard reduction potential, expressed in volts (V), which is determined under standard conditions: a concentration of 1 M for aqueous species, a pressure of 1 bar for gases, and pure solids or liquids in their standard states at 25°C (298 K).

Standard reduction potentials are available in tables, like the Appendix E mentioned in the exercise, which allow us to compare different substances. For example, we know that if the standard reduction potential is more positive, the substance is more likely to be reduced. In the context of our exercise, copper (Cu) has a standard reduction potential of +0.337 V, while silver (Ag⁺) has a potential of +0.799 V. This indicates that silver ions have a greater tendency to gain electrons and form silver metal compared to copper ions forming copper metal.

• The standard reduction potential is essential for calculating the cell potential of an electrochemical reaction.
• It provides insights into the spontaneity of reduction-oxidation (redox) reactions.

By knowing the standard reduction potential, we can predict the direction of electron flow in an electrochemical cell and determine which species will get oxidized or reduced. This leads to the next concept, the Nernst equation, which allows us to calculate the actual cell potential under non-standard conditions.

Nernst Equation

The Nernst equation comes into play when we need to calculate the potential of an electrochemical cell under any conditions, not just the standard ones. It's derived from thermodynamics and incorporates both the standard reduction potential and the concentrations of the reactants and products involved in the reaction.

For a reaction at standard conditions, the standard cell potential (E°) would suffice. However, conditions often vary, and that's where the Nernst equation becomes a powerful tool. It is mathematically expressed as:\[\begin{equation}E = E° - \frac{RT}{nF} \ln Q\text{,}\end{equation}\]where E is the cell potential, R is the universal gas constant, T is the temperature in kelvin, n is the number of moles of electrons exchanged, F is Faraday's constant, and Q is the reaction quotient, which reflects the actual concentrations of reactants and products. In the logarithmic term, natural logarithms are used, which can also be expressed as a base-10 logarithm by adjusting coefficients appropriately.

• The equation highlights the impact of concentration and temperature on cell potential.
• It is particularly useful when dealing with non-standard conditions.

In our exercise, the concentrations are standard, so Q equals 1, and the ln(Q) term drops out of the equation. However, it's important to understand how deviations from standard conditions could affect the outcome of an electrochemical reaction.

Gibbs Free Energy

The concept of Gibbs free energy, symbolized as ΔG, is pivotal in predicting the spontaneity of a process. In electrochemical terms, it's directly connected to the cell potential and equilibrium constant. Gibbs free energy can tell us whether a reaction will occur on its own, without any added energy. A negative ΔG indicates a spontaneous reaction, while a positive ΔG suggests non-spontaneity.

The link between Gibbs free energy and the equilibrium constant (K) is defined by the equation:\[\begin{equation}ΔG° = -RT\ln K\text{,}\end{equation}\]where R is the gas constant, T is the temperature in Kelvins, and K is the equilibrium constant. This equation reveals that a larger equilibrium constant corresponds to a more negative ΔG°, indicating a stronger tendency for the reaction to proceed to completion.

In the exercise, we used the relationship between ΔG° and the cell potential:\[\begin{equation}ΔG° = -nFE°\text{,}\end{equation}\]where n is the number of electrons transferred and F is Faraday's constant. Combining these equations allows us to link E°, the standard cell potential, directly to the equilibrium constant, enabling us to calculate K from known potentials. Therefore, Gibbs free energy acts as a bridge between thermodynamics and electrochemistry, forming a foundation for understanding and predicting chemical reactions.

• A key player in determining reaction spontaneity.
• Connects electrochemical potentials to equilibrium states.

By understanding Gibbs free energy, students can appreciate how energy considerations dictate the direction and extent of chemical reactions.