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

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): \(E°_{Br_2/Br^-} = 1.09 \mathrm{V}\) \(E°_{MnO_4^-/Mn^{2+}} = 1.51 \mathrm{V}\)

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) \(E°_{cell} = (1.51 \mathrm{V}) - (1.09 \mathrm{V}) = 0.42 \mathrm{V}\)

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) \(K = 8.36 \times 10^{26}\)