Problem 55 - Chemistry: The Central Science in SI Units, Global Edition

Question

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : $$ \begin{array}{l} \text { (a) } \mathrm{Fe}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{Ni}(s) \\ \text { (b) } \mathrm{Co}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Co}^{2+}(a q)+\mathrm{H}_{2}(g) \\ \text { (c) } 10 \mathrm{Br}^{-}(a q)+2 \mathrm{MnO}_{4}^{-}(a q)+16 \mathrm{H}^{+}(a q) \longrightarrow 2 \mathrm{Mn}^{2+}(a q)+8 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Br}_{2}(l) \end{array} $$

Step-by-Step Solution

Answer

The equilibrium constants for the given redox reactions at 298 K are: (a) \(K = 1.21 \times 10^{5}\) (b) \(K = 4.14 \times 10^{7}\) (c) \(K = 8.36 \times 10^{26}\)

Step 1: Determine half-reactions and obtain standard reduction potentials

First, we need to identify the half-reactions for each given redox reaction. Then, we will find their standard reduction potentials (E°) from Appendix E. (a) Oxidation half-reaction: \(\mathrm{Fe}(s) \longrightarrow \mathrm{Fe}^{2+}(aq) + 2e^-\) Reduction half-reaction: \(\mathrm{Ni}^{2+}(aq) + 2e^- \longrightarrow \mathrm{Ni}(s)\) (b) Oxidation half-reaction: \(\mathrm{Co}(s) \longrightarrow \mathrm{Co}^{2+}(aq) + 2e^-\) Reduction half-reaction: \(2\mathrm{H}^+(aq) + 2e^- \longrightarrow \mathrm{H}_{2}(g)\) (c) Oxidation half-reaction: \(2\mathrm{Br}^-(aq) \longrightarrow \mathrm{Br}_{2}(l) + 2e^-\) Reduction half-reaction: \(\mathrm{MnO}_{4}^-(aq) + 8\mathrm{H}^+(aq) + 5e^- \longrightarrow \mathrm{Mn}^{2+}(aq) + 4\mathrm{H}_{2}\mathrm{O}(l)\) Now, look up the standard reduction potentials in Appendix E: - For (a): \(E°_{Fe^{2+}/Fe} = -0.44 \mathrm{V}\) \(E°_{Ni^{2+}/Ni} = -0.26 \mathrm{V}\) - For (b): \(E°_{Co^{2+}/Co} = -0.28 \mathrm{V}\) \(E°_{H_2/H^+} = 0.00 \mathrm{V}\) - For (c): \(E°_{Br_2/Br^-} = 1.09 \mathrm{V}\) \(E°_{MnO_4^-/Mn^{2+}} = 1.51 \mathrm{V}\)

Step 2: Calculate the standard cell potential (E°) for each reaction

Calculate the standard cell potential for each reaction using the standard reduction potentials we found in Step 1, using the formula: \(E°_{cell} = E°_{reduction} - E°_{oxidation}\) (a) \(E°_{cell} = (-0.26 \mathrm{V}) - (-0.44 \mathrm{V}) = 0.18 \mathrm{V}\) (b) \(E°_{cell} = (0.00 \mathrm{V}) - (-0.28 \mathrm{V}) = 0.28 \mathrm{V}\) (c) \(E°_{cell} = (1.51 \mathrm{V}) - (1.09 \mathrm{V}) = 0.42 \mathrm{V}\)

Step 3: Calculate the equilibrium constants

Now, we will use the Nernst equation to calculate the equilibrium constants for each redox reaction. The Nernst equation is given by: \[E°_{cell}=\frac{RT}{nF} \ln K\] Where \(E°_{cell}\) is the standard cell potential, \(R\) is the gas constant (8.314 J/mol·K), \(T\) is the temperature in Kelvin (298 K in this case), \(n\) is the number of electrons transferred in the reaction, \(F\) is Faraday's constant (96500 C/mol), and \(K\) is the equilibrium constant. Rearrange the Nernst equation to solve for \(K\). \[K = \exp{\left(\frac{nFE°_{cell}}{RT}\right)}\] (a) \(n = 2\) \(K = \exp{\left(\frac{2 \times 96500 \times 0.18}{8.314 \times 298}\right)} = 1.21 \times 10^{5}\) (b) \(n = 2\) \(K = \exp{\left(\frac{2 \times 96500 \times 0.28}{8.314 \times 298}\right)} = 4.14 \times 10^{7}\) (c) \(n = 5\) \(K = \exp{\left(\frac{5 \times 96500 \times 0.42}{8.314 \times 298}\right)} = 8.36 \times 10^{26}\) In conclusion, the equilibrium constants for the given redox reactions at 298 K are: (a) \(K = 1.21 \times 10^{5}\) (b) \(K = 4.14 \times 10^{7}\) (c) \(K = 8.36 \times 10^{26}\)

Key Concepts

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. In these reactions, one substance undergoes oxidation, losing electrons, while another undergoes reduction, gaining those electrons. This flow of electrons can be harnessed to produce electrical energy in electrochemical cells.

To understand these reactions better, it's important to break them down into two parts:

• **Oxidation Half-Reaction**: This is where the substance loses electrons. For instance, in the oxidation of iron: \[ \text{Fe} (s) \rightarrow \text{Fe}^{2+} (aq) + 2e^- \]. Here, iron is oxidized by losing electrons.
• **Reduction Half-Reaction**: This involves the gain of electrons, such as the reduction of nickel ions: \[ \text{Ni}^{2+} (aq) + 2e^- \rightarrow \text{Ni} (s) \]. Nickel ions gain electrons and are reduced.

The beauty of redox reactions lies in their ability to combine these two half-reactions to form an overall reaction that balances both mass and charge. This is essential for calculating the equilibrium constant, which tells us how the reactants and products are distributed at equilibrium.

Nernst Equation

The Nernst Equation is a fundamental tool in electrochemistry that allows us to calculate the potential of an electrochemical cell under non-standard conditions. It's a bridge between an electrochemical cell's standard conditions and its actual operating conditions.

The Nernst Equation is given by:\[ E = E° - \frac{RT}{nF} \ln Q \]Where:

• **\(E\)** is the cell potential at non-standard conditions.
• **\(E°\)** is the standard cell potential.
• **\(R\)** is the universal gas constant (8.314 J/mol·K).
• **\(T\)** is the temperature in Kelvin.
• **\(n\)** is the number of moles of electrons transferred in the reaction.
• **\(F\)** is Faraday’s constant (96500 C/mol).
• **\(Q\)** is the reaction quotient, representing the ratio of concentrations of products to reactants.

For reactions at equilibrium, \(E\) becomes zero, and \(Q\) turns into the equilibrium constant \(K\). Rearranging the Nernst Equation for this scenario allows us to calculate \(K\), providing insights into the reaction's favorability under specific conditions.

Standard Reduction Potentials

Standard Reduction Potentials are essential in understanding the likelihood of a reduction half-reaction occurring. They are a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. These potentials are measured under standard conditions: 1 M concentration, 1 atm pressure, and 25°C (298 K).

Each half-reaction has a specific standard reduction potential, denoted as \(E°\). A more positive \(E°\) value means a greater tendency for the substance to gain electrons and be reduced. Conversely, a more negative \(E°\) indicates that the substance is less likely to be reduced and might instead be oxidized.

For example, in the half-reaction \(\text{Ni}^{2+} (aq) + 2e^- \rightarrow \text{Ni} (s)\), the standard reduction potential is \(-0.26 \text{ V}\). When comparing two half-reactions, the one with the higher \(E°\) will proceed in the forward direction as the reduction half-reaction. This principle is used to calculate the standard cell potential \(E°_{cell}\) for a complete redox reaction: \[ E°_{cell} = E°_{red} - E°_{ox} \]. Calculating \(E°_{cell}\) helps us predict the reaction's spontaneity and strength via the equilibrium constant \(K\).