Question

Given the following data: \(\mathrm{PtCl}_{4}^{2-}(a q)+2 e^{-} \longrightarrow \operatorname{Pt}(s)+4 \mathrm{Cl}^{-}(a q)\) $$ \begin{aligned} & E_{\mathrm{red}}^{\circ} &=0.73 \mathrm{~V} \\ \mathrm{Pt}^{2+}(a q)+2 e^{-} \longrightarrow \operatorname{Pt}(s) & E_{\mathrm{red}}^{\circ} &=1.20 \mathrm{~V} \end{aligned} $$ Find \(K_{\mathrm{f}}\) for \(\mathrm{PtCl}_{4}^{2-}\) at \(25^{\circ} \mathrm{C}\).

Short answer

The formation constant for PtCl₄²⁻ complex at 25°C is approximately 3.09 x 10¹⁶.

Step-by-step solution

Combine the Half-Reactions

First, we need to combine the given half-reactions to form the overall redox equation: $$\mathrm{PtCl}_{4}^{2-}(aq) + 2e^- \longrightarrow \mathrm{Pt}(s) + 4\mathrm{Cl}^-(aq)$$ $$\mathrm{Pt}^{2+}(aq) + 2e^- \longrightarrow \mathrm{Pt}(s)$$ Subtract the second equation from the first equation to get the overall equation: $$\mathrm{Pt}^{2+}(aq) + 4\mathrm{Cl}^-(aq) \longleftrightarrow \mathrm{PtCl}_{4}^{2-}(aq)$$

Calculate the overall standard reduction potential

Now, we can calculate the overall standard reduction potential \(E^\circ\) for the complex formation reaction. This can be found by subtracting the standard reduction potential of \(\mathrm{Pt}^{2+}\) from the standard reduction potential of \(\mathrm{PtCl}_{4}^{2-}\) complex: $$E^\circ_\text{overall} = E^\circ_\text{complex} - E^\circ_\text{Pt^{2+}} = 0.73 \,\text{V} - 1.20 \,\text{V} = -0.47 \,\text{V}$$

Use Nernst Equation to find K\(_\mathrm{f}\)

Now that we have the overall standard reduction potential, we can use the Nernst equation to find \(K_\mathrm{f}\). The Nernst equation is: $$E = E^\circ - \frac{RT}{nF} \ln Q$$ Where \(E\) is the cell potential, \(E^\circ\) is the standard cell potential, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred, \(F\) is the Faraday's constant, and \(Q\) is the reaction quotient. Since the complex is in its equilibrium state, the overall cell potential \(E\) will be 0 (\(E_\text{cell} = 0\)) and the reaction quotient \(Q = K_\mathrm{f}\). The number of electrons transferred \(n\) is 2. Rewriting the Nernst equation for the equilibrium constant, we get: $$0 = -0.47\,\text{V} -\frac{(8.314\,\text{J K}^{-1}\,\text{mol}^{-1})(298\,\text{K})}{(2)(96485\,\text{C mol}^{-1})} \ln K_\mathrm{f}$$

Solve for K\(_\mathrm{f}\)

Now we can solve the equation for \(K_\mathrm{f}\): $$\frac{(8.314\,\text{J K}^{-1}\,\text{mol}^{-1})(298\,\text{K})}{(2)(96485\,\text{C mol}^{-1})} \ln K_\mathrm{f} = 0.47\,\text{V}$$ Divide both sides by \(\frac{(8.314\,\text{J K}^{-1}\,\text{mol}^{-1})(298\,\text{K})}{(2)(96485\,\text{C mol}^{-1})}\): $$\ln K_\mathrm{f} = \frac{0.47\,\text{V}(2)(96485\,\text{C mol}^{-1})}{(8.314\,\text{J K}^{-1}\,\text{mol}^{-1})(298\,\text{K})}$$ Evaluate the right-hand side: $$\ln K_\mathrm{f} \approx 38.02$$ Now we can take the exponential to find \(K_\mathrm{f}\): $$K_\mathrm{f} = e^{38.02} \approx 3.09 \times10^{16}$$ Hence, the formation constant for \(\mathrm{PtCl}_{4}^{2-}\) at 25°C is \(K_\mathrm{f} \approx 3.09 \times10^{16}\).