Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{8}^{\circ}=1.09 \mathrm{~V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{C}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.60 \mathrm{~V}\)

Step by step solution

01

Compare the standard reduction potentials

We know that the half-reaction with the highest standard reduction potential \(\mathscr{E}^{\circ}\) will be reduced, whereas the one with the lower \(\mathscr{E}^{\circ}\) will be oxidized. We compare the given potentials: \(\mathrm{Cl}_{2} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ} = 1.36 \mathrm{~V}\) \(\mathrm{Br}_{2} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ} = 1.09 \mathrm{~V}\) Since the reduction potential of Cl\(_2\) is higher than that of Br\(_2\), the reduction of Cl\(_2\) will occur. Thus, the electron flow will be from Br\(_2\) (oxidation) to Cl\(_2\) (reduction). ###Step 2: Determine the direction of ion migration###

02

Anions move towards the anode; cations move towards the cathode

As Br\(_2\) is oxidized to Br\(^-\), the concentration of Br\(^-\) will increase near the anode (Br\(_2\)). Therefore, Br\(^-\) anions will migrate through the salt bridge towards the other side to keep the charge balanced. Similarly, as Cl\(_2\) is reduced to Cl\(^-\), Cl\(^-\) will migrate towards the side where reduction takes place to maintain charge neutrality. ###Step 3: Identify the cathode and anode###

03

Reduction occurs at the cathode; oxidation occurs at the anode

In this cell, Cl\(_2\) is reduced (gains electrons), and this reaction occurs at the cathode. Br\(_2\) is oxidized (loses electrons), and this reaction occurs at the anode. Therefore, Cl\(_2\) is at the cathode and Br\(_2\) is at the anode. ###Step 4: Obtain the overall balanced equation###

04

Combine the half-reactions and balance

To obtain the overall balanced equation, we need to combine the half-reactions and balance them. Since both half-reactions involve the transfer of 2 electrons, we can directly add them without any further adjustment: 2 Br\(^-\) \( \rightarrow \) Br\(_2\) + 2 e\(^-\) (Balancing by multiplying the original half-reaction by 2) Cl\(_2\) + 2 e\(^-\) \( \rightarrow \) 2 Cl\(^-\) Adding the two half-reactions: 2 Br\(^-\) + Cl\(_2\) \( \rightarrow \) Br\(_2\) + 2 Cl\(^-\) ###Step 5: Determine the cell potential###

05

Calculate the cell potential by subtracting potentials

To determine the cell potential (\(\mathscr{E}^{\circ}\)), subtract the standard reduction potential of the anode from the standard reduction potential of the cathode: Cell Potential, \(\mathscr{E}^{\circ}\) = \(\mathscr{E}^{\circ}_{\text{cathode}} - \mathscr{E}^{\circ}_\text{anode}\) \(\mathscr{E}^{\circ}\) = 1.36 V - 1.09 V = 0.27 V ##Galvanic cell (b): Following the same process as for cell (a), we can obtain the required information for cell (b). Note that for cell (b), the half-reactions need to be balanced by multiplying the first half-reaction by 2 and the second by 5 to obtain equivalent electron transfer numbers: Overall equation: 10 MnO\(_4^{-}\) + 16 H\(^+\) + 6 IO\(_4^{-}\) \( \rightarrow \) 10 Mn\(^{2+}\) + 8 IO\(_3^{-}\) + 16 H\(_2\)O Cell Potential, \(\mathscr{E}^{\circ}\) = 1.60 V - 1.51 V = 0.09 V

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