Problem 53 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

For the given reactions, equilibrium constants are: (a) \(K = 1.64 \times 10^{13}\), (b) \(K = 1.16 \times 10^{14}\), and (c) \(K = 2.45 \times 10^{61}\).

Step 1: Identify the reduction and oxidation half-reactions

In this step, we will split the given redox reaction into oxidation and reduction half-reactions. Oxidation half-reaction: Fe(s) → Fe²⁺(aq) + 2e⁻ Reduction half-reaction: Ni²⁺(aq) + 2e⁻ → Ni(s)

Step 2: Determine the standard cell potential

Now, we will calculate the standard cell potential (E°) using the reduction potentials given in Appendix E. For Fe²⁺(aq)/Fe(s): E° = -0.44V For Ni²⁺(aq)/Ni(s): E° = -0.23V Since we are subtracting E° values, E° (cell) = E° (cathode) - E° (anode) = -0.23 - (-0.44) = 0.21V

Step 3: Calculate the equilibrium constant

Using the standard cell potential, we can calculate the standard Gibbs Free Energy change (ΔG°) using the equation: ΔG° = -nFE° Where n is the number of moles of electrons transferred and F is the Faraday's constant (96485 C/mol). ΔG° = -2 * 96485 * 0.21 = -40525.7 J/mol Now we can calculate the equilibrium constant (K) using the given formula: K = e^{-\frac{\Delta{G°}}{RT}} = e^{-\frac{-40525.7}{8.314 \times 298}} = 1.64 × 10¹³ (b) Co(s) + 2 H⁺(aq) → Co²⁺(aq) + H₂(g)

Step 1: Identify the reduction and oxidation half-reactions

In this step, we will split the given redox reaction into oxidation and reduction half-reactions. Oxidation half-reaction: Co(s) → Co²⁺(aq) + 2e⁻ Reduction half-reaction: 2 H⁺(aq) + 2e⁻ → H₂(g)

Step 2: Determine the standard cell potential

Now, we will calculate the standard cell potential (E°) using the reduction potentials given in Appendix E. For Co²⁺(aq)/Co(s): E° = -0.28V For H₂(g)/H⁺(aq): E° = 0.00V E° (cell) = E° (cathode) - E° (anode) = 0.00 - (-0.28) = 0.28V

Step 3: Calculate the equilibrium constant

Using the standard cell potential, we can calculate the standard Gibbs Free Energy change (ΔG°) using the equation: ΔG° = -nFE° ΔG° = -2 * 96485 * 0.28 = -54021.2 J/mol Now we can calculate the equilibrium constant (K) using the given formula: K = e^{-\frac{\Delta{G°}}{RT}} = e^{-\frac{-54021.2}{8.314 \times 298}} = 1.16 × 10¹⁴ (c) 10 Br⁻(aq) + 2 MnO₄⁻(aq) + 16 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

Step 1: Identify the reduction and oxidation half-reactions

In this step, we will split the given redox reaction into oxidation and reduction half-reactions. Oxidation half-reaction: 10 Br⁻(aq) → 5 Br₂(l) + 10e⁻ Reduction half-reaction: 2 MnO₄⁻(aq) + 10e⁻ + 16 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l)

Step 2: Determine the standard cell potential

Now, we will calculate the standard cell potential (E°) using the reduction potentials given in Appendix E. For 5 Br₂(l)/10 Br⁻(aq): E° = +1.07V For 2 MnO₄⁻(aq)/2 Mn²⁺(aq): E° = +1.51V E° (cell) = E° (cathode) - E° (anode) = 1.51 - 1.07 = 0.44V

Step 3: Calculate the equilibrium constant

Using the standard cell potential, we can calculate the standard Gibbs Free Energy change (ΔG°) using the equation: ΔG° = -nFE° ΔG° = -10 * 96485 * 0.44 = -424171 J/mol Now we can calculate the equilibrium constant (K) using the given formula: K = e^{-\frac{\Delta{G°}}{RT}} = e^{-\frac{-424171}{8.314 \times 298}} = 2.45 × 10⁶¹ Equilibrium constants for reactions (a), (b), and (c) are 1.64 × 10¹³, 1.16 × 10¹⁴, and 2.45 × 10⁶¹, respectively.

Key Concepts

Equilibrium Constant

In an electrochemical reaction, the equilibrium constant \((K)\) is a crucial factor that links the reaction quotient at equilibrium to the Gibbs Free Energy change. An equilibrium constant provides insights into where the equilibrium of a redox reaction lies, suggesting whether the products or reactants are favored at equilibrium.

The formula to connect the equilibrium constant and Gibbs Free Energy change is:

• \( K = e^{-rac{\Delta G^\circ}{RT}} \)

Here:

• \( \Delta G^\circ \) is the standard Gibbs Free Energy change of the reaction.
• \( R \) is the universal gas constant (8.314 J/(mol·K)).
• \( T \) is the temperature in Kelvin.

If \( K \) is significantly greater than 1, it implies the reaction favors the formation of products at equilibrium. Conversely, if \( K \) is much less than 1, the reactants are favored. A moderate value of \( K \) suggests a more balanced state at equilibrium.

Through calculating \( K \) for reactions, it becomes easier to predict the extent to which a redox reaction will proceed.

Reduction Potentials

Reduction potentials, often denoted as \( E^\circ \), play a vital role in electrochemistry, acting as a measure of the tendency of a chemical species to gain electrons and undergo reduction. Each half-reaction is associated with a specific reduction potential. It is vital to note:

• Positive \( E^\circ \) values suggest a substance readily gains electrons (strong oxidizing agent).
• Negative \( E^\circ \) values indicate a substance less willing to receive electrons.

To calculate the cell potential \( E^\circ_{cell} \) for a complete redox reaction, we utilize the reduction potentials of both the oxidation and reduction reactions. This is done using:

• \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \)

By determining \( E^\circ_{cell} \), you can infer the spontaneity of a reaction. A positive \( E^\circ_{cell} \) indicates that the reaction is spontaneous under standard conditions.

Gibbs Free Energy Change

Gibbs Free Energy change, represented as \( \Delta G \), forms a cornerstone concept in understanding reaction spontaneity. This thermodynamic quantity helps determine whether a reaction can proceed without needing external energy.

• A negative \( \Delta G^\circ \) signifies a spontaneous process.
• A positive \( \Delta G^\circ \) means the process is not spontaneous.
• When \( \Delta G^\circ \) is zero, the system is at equilibrium.

For electrochemical reactions, work or energy changes can be calculated:

• \( \Delta G^\circ = -nFE^\circ_{cell} \)

Where:

• \( n \) is the number of moles of electrons exchanged.
• \( F \) is Faraday's constant (96485 C/mol).
• \( E^\circ_{cell} \) is the standard cell potential.

This relationship helps us connect Gibbs Free Energy to electrochemical cell potential, providing a deeper insight into the feasibility and direction of a reaction.