Using the standard reduction potentials listed in Appendix E, ...

Question

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K:

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

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

(c) 10 Br^-(aq) + 2 MnO₄^-(aq) + 16 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

Accepted Answer

Identify the half-reactions involved in the redox process. For the given reaction, the half-reactions are: $ \text{Br}^- \rightarrow \text{Br}_2 $ and $ \text{MnO}_4^- \rightarrow \text{Mn}^{2+} $.. Look up the standard reduction potentials (E°) for each half-reaction from Appendix E. The standard reduction potential for $ \text{Br}_2 + 2e^- \rightarrow 2\text{Br}^- $ and $ \text{MnO}_4^- + 8H^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4H_2O $ are needed.. Calculate the standard cell potential (E°_cell) using the formula: $ E°_\text{cell} = E°_\text{cathode} - E°_\text{anode} $. Identify which half-reaction is the reduction (cathode) and which is the oxidation (anode).. Use the Nernst equation to relate the standard cell potential to the equilibrium constant (K): $ E°_\text{cell} = \frac{RT}{nF} \ln K $, where $ R $ is the gas constant, $ T $ is the temperature in Kelvin, $ n $ is the number of moles of electrons transferred, and $ F $ is Faraday's constant.. Rearrange the Nernst equation to solve for the equilibrium constant $ K $: $ \ln K = \frac{nFE°_\text{cell}}{RT} $. Substitute the known values to find $ K $.

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Key Concepts

Standard Reduction Potentials

Nernst Equation

Equilibrium Constant (K)