Question

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at $298\mathrm{K}:$ $$\begin{array}{l}\text{ (a) }\mathrm{Fe}(s)+{\mathrm{Ni}}^{2+}(aq)⟶{\mathrm{Fe}}^{2+}(aq)+\mathrm{Ni}(s)\\ \text{ (b) }\mathrm{Co}(s)+2{\mathrm{H}}^{+}(aq)⟶{\mathrm{Co}}^{2+}(aq)+{\mathrm{H}}_2(g)\\ \text{ (c) }10{\mathrm{Br}}^{-}(aq)+2{\mathrm{MnO}}_4^{-}(aq)+16{\mathrm{H}}^{+}(aq)\to \\ 2{\mathrm{Mn}}^{2+}(aq)+8{\mathrm{H}}_2\mathrm{O}(l)+5{\mathrm{Br}}_2(l)\end{array}$$

Short answer

The equilibrium constants for the given reactions at 298 K are: a) Fe(s) + Ni²⁺(aq) → Fe²⁺(aq) + Ni(s): $K\approx 5.17\times {10}^{-15}$ b) Co(s) + 2 H⁺(aq) → Co²⁺(aq) + H₂(g): $K\approx 2.57\times {10}^{-5}$ c) 10 Br⁻(aq) + 2 MnO₄⁻(aq) + 16 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l): $K\approx 2.73\times {10}^{-5}$

Step-by-step solution

Reaction (a): Fe(s) + Ni²⁺(aq) → Fe²⁺(aq) + Ni(s)

First, let's identify the half-reactions and their standard reduction potentials from Appendix E. The half-reactions and their potentials will be: - Fe²⁺(aq) + 2e⁻ → Fe(s), E° = -0.44 V - Ni²⁺(aq) + 2e⁻ → Ni(s), E° = -0.25 V Step 1: Calculate the standard cell potential The given reaction is essentially the reverse of the Fe half-reaction, making that half-reaction an oxidation process. Therefore, the standard reduction potentials should be calculated by reversing the sign of the Fe half-reaction and summing both potentials: $$E^0=(-0.44)+(-0.25)=-0.69V$$ Step 2: Calculate the equilibrium constant Now, we can use the Nernst equation to calculate K: $$E^0=\frac{RT}{nF}\ln K$$ Using the values at 298 K: R = 8.314 J/(mol·K), T = 298 K, n = 2 (two electrons are exchanged in both half-reactions), and F = 96485 C/mol, we get: $$-0.69=\frac{(8.314)(298)}{(2)(96485)}\ln K$$ Solve for K: $$\ln K=-32.96$$ $$K=e^{-32.96}\approx 5.17\times {10}^{-15}$$

Reaction (b): Co(s) + 2 H⁺(aq) → Co²⁺(aq) + H₂(g)

First, let's identify the half-reactions and their standard reduction potentials from Appendix E. The half-reactions and their potentials will be: - Co²⁺(aq) + 2e⁻ → Co(s), E° = -0.28 V - 2H⁺(aq) + 2e⁻ → H₂(g), E° = 0.00 V Step 1: Calculate the standard cell potential The given reaction is essentially the reverse of the Co half-reaction, making that half-reaction an oxidation process. Therefore, the standard reduction potentials should be calculated by reversing the sign of the Co half-reaction and summing both potentials: $$E^0=(-0.28)+(0.00)=-0.28V$$ Step 2: Calculate the equilibrium constant Now, using the Nernst equation to calculate K: $$E^0=\frac{RT}{nF}\ln K$$ Using the values at 298 K: R = 8.314 J/(mol·K), T = 298 K, n = 2 (two electrons are exchanged in both half-reactions), and F = 96485 C/mol, we get: $$-0.28=\frac{(8.314)(298)}{(2)(96485)}\ln K$$ Solve for K: $$\ln K=-10.56$$ $$K=e^{-10.56}\approx 2.57\times {10}^{-5}$$

Reaction (c): 10 Br⁻(aq) + 2 MnO₄⁻(aq) + 16 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

First, let's identify the half-reactions and their standard reduction potentials from Appendix E. The half-reactions and their potentials will be: - 2MnO₄⁻(aq) + 8H₂O(l) + 10e⁻ → 2Mn²⁺(aq) + 16H⁺(aq) + (-1.51) V - 5Br₂(l) + 10e⁻ → 10Br⁻(aq), E° = 1.07 V Step 1: Calculate the standard cell potential For this reaction, we simply have to sum both potentials: $$E^0=(-1.51)+(1.07)=-0.44V$$ Step 2: Calculate the equilibrium constant Now, using the Nernst equation to calculate K: $$E^0=\frac{RT}{nF}\ln K$$ Using the values at 298 K: R = 8.314 J/(mol·K), T = 298 K, n = 10 (ten electrons are exchanged in both half-reactions), and F = 96485 C/mol, we get: $$-0.44=\frac{(8.314)(298)}{(10)(96485)}\ln K$$ Solve for K: $$\ln K=-10.52$$ $$K=e^{-10.52}\approx 2.73\times {10}^{-5}$$

Key concepts

Equilibrium Constant Calculation

Understanding how to calculate an equilibrium constant (K) for a chemical reaction is critical in the field of chemistry, particularly when studying reactions in an electrochemical context. The value of K provides insight into the extent to which a reaction will proceed under standard conditions.

In the case of redox reactions, the equilibrium constant can be calculated using cell potentials and the Nernst equation. As shown in the textbook solution, after obtaining the standard reduction potentials for the half-reactions, we calculate the standard cell potential ($E^0$) which is the sum of the potentials for each half-reaction involved in the overall reaction.

However, when a half-reaction is reversed (as in oxidation), its potential's sign is also reversed. Once we have our standard cell potential, we can use the equation derived from the Nernst equation for the reaction at standard conditions:$$E^0=\frac{RT}{nF}\ln K$$where R is the ideal gas constant, T is temperature in Kelvin, n is the number of moles of electrons exchanged, and F is the Faraday constant. The calculated value of $K$ gives us the equilibrium constant, indicating how far the reaction goes towards products or reactants.

Nernst Equation

The Nernst equation is a fundamental equation in electrochemistry, providing a connection between the electrochemical potential of a cell and the concentrations of reactants and products. It is expressed as:$$E=E^0-\frac{RT}{nF}\ln Q$$where $E$ is the cell potential, $E^0$ is the standard cell potential, $R$ is the ideal gas constant, $T$ is the temperature in Kelvin, $n$ is the number of moles of electrons transferred in the reaction, $F$ is the Faraday constant, and $Q$ is the reaction quotient.

At equilibrium, $E$ becomes zero, leading to the simplified equation used for equilibrium constant calculation, as $Q$ becomes $K$. The textbook solutions highlight the application of the Nernst equation in obtaining the equilibrium constant. In simple terms, the Nernst equation adjusts the standard cell potential to reflect the actual conditions of a reaction taking place at concentrations different from standard conditions.

Electrochemistry

Electrochemistry is the branch of chemistry that deals with the relationship between electrical energy and chemical changes. It's an interdisciplinary realm that crosses over into physics and materials science, underpinning technologies such as batteries, fuel cells, and electroplating.

The standard reduction potentials of various half-reactions are cornerstone elements in electrochemistry. These potentials, measured in volts (V), indicate a substance's tendency to gain electrons and be reduced. They are determined under standard conditions (1 M concentration, 1 atm pressure, and 298 K), and provide a scale against which all redox reactions can be compared.

In the given solutions, the reactions' equilibrium constants were calculated using standard reduction potentials, highlighting the intricacies of electrochemical reactions. These constants reflect the thermodynamic feasibility of reactions, and emphasizing their concept enhances understanding in fundamental electrochemistry.