Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

We look up the reduction potentials for the following half-reactions: 1. \(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\) 2. \(\mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}\) 3. \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}\) The corresponding reduction potentials are: 1. \(E_1^\circ = 0.771\, \text{V}\) 2. \(E_2^\circ = -0.440\, \text{V}\) 3. \(E_3^\circ = -0.036\, \text{V}\)

Adding the reduction potentials for the first two reactions, \(E_1^\circ\) and \(E_2^\circ\), directly doesn't give the potential for the third reaction, \(E_3^\circ\), because the stoichiometry of the electrons in these reactions is not equal. The first reaction involves 1 electron, while the second reaction involves 2 electrons.

To use the first two potentials to calculate the third potential, we need to ensure that the number of electrons in both reactions is the same to satisfy the stoichiometry. We can achieve this by multiplying the first reaction by 2, so the number of electrons in both reactions is the same (3 electrons): 2(\(\mathrm{Fe}^{3+} + e^- \rightarrow \mathrm{Fe}^{2+}\)) = \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}^{2+} + 2\mathrm{Fe}^{2+}\) Now, we can add the above reaction and the second reaction: \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}^{2+} + 2\mathrm{Fe}^{2+} + \mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe}\) The resulting reaction is: \(\mathrm{Fe}^{3+} + 3e^- \rightarrow \mathrm{Fe}\) For the potential of this reaction, we can use the Nernst equation: \(E^\circ = E_1^\circ - \frac{RT}{nF} \ln{\frac{a_\mathrm{Fe^{2+}}}{a_\mathrm{Fe^{3+}}}}\) Since the Nernst equation has logarithmic terms, we cannot just add the two potentials directly. We need to take the logarithmic term into account to obtain the third potential accurately.