Question

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. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{C}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.60 \mathrm{~V}\)

Short answer

In galvanic cell (a), the electron flow is from Br₂ (anode) to Cl₂ (cathode). Br⁻ ions will migrate through the salt bridge towards the cathode, while Cl⁻ ions will move towards the anode. The overall balanced equation is: 2 Br⁻ + Cl₂ → Br₂ + 2 Cl⁻ The cell potential for this cell is 0.27 V. For galvanic cell (b), the overall balanced equation is: 10 MnO₄⁻ + 16 H⁺ + 6 IO₄⁻ → 10 Mn²⁺ + 8 IO₃⁻ + 16 H₂O The cell potential for this cell is 0.09 V.

Step-by-step solution

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###

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###

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###

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###

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

Key concepts

Half-Reaction

In galvanic cells, understanding half-reactions is crucial. A half-reaction is a part of the overall reaction, focusing on either the oxidation or reduction process. Each galvanic cell consists of two half-reactions: one for oxidation and one for reduction. Oxidation involves the loss of electrons, while reduction entails the gain of electrons. For instance, in the chloride-bromide cell scenario, the half-reaction for chlorine is: \(\mathrm{Cl}_{2} + 2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}\)It shows chlorine gaining electrons (reduction), whereas bromine loses electrons (oxidation).Each half-reaction provides essential information on electron flow and helps identify which species undergoes oxidation or reduction.

Standard Reduction Potential

The standard reduction potential, denoted as \(\mathscr{E}^{\circ}\), reveals how likely a substance is to gain electrons, thus becoming reduced under standard conditions. A positive value indicates a greater tendency for reduction. For example, chlorine has a \(\mathscr{E}^{\circ}\) of 1.36 V, making it more electrophilic compared to bromine, which has a \(\mathscr{E}^{\circ}\) of 1.09 V. In a galvanic cell, the half-reaction with the higher \(\mathscr{E}^{\circ}\) occurs at the cathode where reduction takes place. By comparing \(\mathscr{E}^{\circ}\) values, we can predict the flow of electrons and determine which substance will become oxidized or reduced.

Cathode and Anode Identification

Identifying the cathode and anode in a galvanic cell is indispensable for understanding electron flow. The cathode is the site of reduction, where positive ions gain electrons, while the anode is the site of oxidation, where negative ions lose electrons. In the chlorine-bromine example:- The cathode reaction is \(\mathrm{Cl}_{2}\rightarrow 2\mathrm{Cl}^-\) because chlorine is reduced.- The anode reaction is \(2\mathrm{Br}^- \rightarrow \mathrm{Br}_{2}\) as bromine is oxidized.Always remember:- **Reduction** occurs at the **cathode**.- **Oxidation** occurs at the **anode**.By understanding these roles, we can effectively analyze galvanic cell functions.

Direction of Electron Flow

In galvanic cells, electrons flow from the anode to the cathode, driven by the difference in reduction potential between the two substances. This flow is crucial for maintaining the electric current necessary for the cell's operation. In a typical cell setup: - Electrons are released at the anode during oxidation. - They travel through the external circuit toward the cathode. For the chlorine-bromine cell, electrons flow from bromine to chlorine. This is because bromine, with its lower standard reduction potential, loses electrons that travel to chlorine, which gains them. The electron flow direction reflects the spontaneous nature of the redox reactions, ensuring electricity generation in galvanic cells.

Ion Migration

Ion migration in a galvanic cell is essential for charge balance and uninterrupted flow of current. Ions move through a salt bridge, which connects the two half-cells. This salt bridge allows ions to migrate to maintain electrical neutrality, preventing charge buildup.- **Anions** move toward the **anode**. For example, \(\mathrm{Br}^-\) ions migrate towards the anode to balance positive charges generated from electron loss.- **Cations** move towards the **cathode**. Similarly, \(\mathrm{Cl}^-\) ions move toward the cathode to neutralize charge from electron gain.By facilitating ion flow, the salt bridge prevents polarization of the cell, ensuring sustained electrical current.