Given the following data: $$ \begin{array}{cc} \mathrm{PtCl}_{4}{ }^{2-}(a q)+2 e^{-} \longrightarrow \mathrm{Pt}(s)+4 \mathrm{Cl}^{-}(a q) & E^{o}=+0.73 \mathrm{~V} \\ \mathrm{Pt}^{2+}(a q)+4 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{PtCl}_{4}^{2-}(a q) & K_{f}=1 \times 10^{16} \end{array} $$ find \(E^{\circ}\) for the half-cell $$ \mathrm{Pt}^{2+}(a q)+2 e^{-} \longrightarrow \operatorname{Pt}(s) $$

Write down the overall reaction by combining the given reactions: $$ \begin{array}{c} \mathrm{PtCl}_{4}^{2-}(a q)+2 e^{-} \longrightarrow \mathrm{Pt}(s)+4\mathrm{Cl}^{-}(a q)\\ \hline \mathrm{Pt}^{2+}(a q)+4 \mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{PtCl}_{4}^{2-}(a q) \end{array} $$ Adding the above two reactions: $$ \mathrm{Pt}^{2+}(a q)+2 e^{-} \longrightarrow \operatorname{Pt}(s) $$ This is the half-cell reaction we need to find the standard reduction potential for.

Since we know the formation constant of the overall reaction, we can calculate the equilibrium constant (K_c) for the half-cell reaction using the formula: $$ K_c = \frac{K_f}{E^\circ} $$ Where \(K_f\) is the formation constant, and \(E^\circ\) is the standard reduction potential of the given half-cell reaction. For the half-cell reaction: $$ \mathrm{PtCl}_{4}^{2-}(a q)+2 e^{-} \longrightarrow \mathrm{Pt}(s)+4\mathrm{Cl}^{-}(a q) $$ Using the given values: $$ K_c = \frac{1 \times 10^{16}}{0.73} $$ $$ K_c \approx 1.37 \times 10^{16} $$

Now, we will use the Nernst Equation which relates the standard potential (Eº) of a half-cell reaction, the number of electrons involved (n), the equilibrium constant (K_c), and the standard reduction potential (Eº) of the half-cell reaction we want to find. The equation is: $$ 0 = E^°_{Pt^{2+}/Pt} - \frac{0.0592}{n} \log{K_c} $$ We already have the value of K_c, and in our half-cell reaction, 2 electrons are transferred (n=2). Inserting the values into the equation: $$ 0 = E^°_{Pt^{2+}/Pt} - \frac{0.0592}{2} \log{(1.37 \times 10^{16})} $$ Now solve for \(E^°_{Pt^{2+}/Pt}\): $$ E^°_{Pt^{2+}/Pt} = \frac{0.0592}{2} \log{(1.37 \times 10^{16})} $$ $$ E^°_{Pt^{2+}/Pt} \approx 1.18~V $$ So the standard reduction potential for the half-cell reaction: $$ \mathrm{Pt}^{2+}(a q)+2 e^{-} \longrightarrow \operatorname{Pt}(s) $$ is \(E^°_{Pt^{2+}/Pt} \approx 1.18 V\).