Consider a galvanic cell based on the following half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\ \hline {\mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La}} & {-2.37} \\\ {\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}} & {-0.44}\end{array}$$ a. What is the expected cell potential with all components in their standard states? b. What is the oxidizing agent in the overall cell reaction? c. What substances make up the anode compartment? d. In the standard cell, in which direction do the electrons flow? e. How many electrons are transferred per unit of cell reaction? f. If this cell is set up at \(25^{\circ} \mathrm{C}\) with \(\left[\mathrm{Fe}^{2+}\right]=2.00 \times 10^{-4} M\) and \(\left[\mathrm{La}^{3+}\right]=3.00 \times 10^{-3} M,\) what is the expected cell potential?

To calculate the expected cell potential, we must first identify the two half-reactions as reduction and oxidation reactions. The half-reaction with more positive standard reduction potential will proceed as reduction and the other will undergo oxidation. In this case, \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}\) has a higher standard \(\mathscr{E}^{\circ}\) than \(\mathrm{La}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{La}\), so Fe is undergoing reduction, and La is undergoing oxidation. By reversing the La half-reaction, we can obtain the oxidation half-reaction and its standard potential. Now, the oxidation half-reaction is \(\mathrm{La} \longrightarrow \mathrm{La}^{3+} + 3 \mathrm{e}^{-}\), with standard potential \(\mathscr{E}^{\circ}=+2.37\) V. The net cell potential can be calculated as: \(\mathscr{E}_{cell}^{\circ} = \mathscr{E}_{cathode}^{\circ} - \mathscr{E}_{anode}^{\circ}\)

Now, using the given standard potentials, we can calculate the expected cell potential under standard conditions. \(\mathscr{E}_{cell}^{\circ} = (-0.44) - (+2.37)=-2.81\) V.

The oxidizing agent is a substance that gains electrons in a redox reaction. In this case, the oxidizing agent in the overall cell reaction is the reduced form of the reduction half-reaction: \(\mathrm{Fe}^{2+}\).

The anode compartment consists of the substances that are oxidized. In this case, it is \(\mathrm{La}\) and \(\mathrm{La}^{3+}\).

In a galvanic cell, electrons flow from the anode to the cathode. In this reaction, electrons flow from the La compartment (the anode) to the Fe compartment (the cathode).

To determine the number of electrons transferred per unit of cell reaction, we need to balance the redox equation: \(2\mathrm{La} + 3\mathrm{Fe}^{2+} \longrightarrow 2\mathrm{La}^{3+} + 3\mathrm{Fe}\) In this balanced equation, 6 electrons are being transferred.

To calculate the cell potential under non-standard conditions, we can use the Nernst equation: \(\mathscr{E}_{cell} = \mathscr{E}_{cell}^{\circ} - \frac{0.05916}{n} \log(Q)\), where n is the number of electrons transferred (in our case, n=6) and Q is the reaction quotient. For our reaction, \(Q = \frac{[\mathrm{La}^{3+}]^{2}}{[\mathrm{Fe}^{2+}]^{3}} = \frac{(3.00\times10^{-3})^{2}}{(2.00\times10^{-4})^{3}}\) Now, substituting the values into the equation, we get: \(\mathscr{E}_{cell} = -2.81 - \frac{0.05916}{6} \log\left( \frac{(3.00\times10^{-3})^{2}}{(2.00\times10^{-4})^{3}} \right) \) After calculating, we find the non-standard cell potential to be \(\mathscr{E}_{cell} = -2.66\) V.