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. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad 8^{\circ}=1.09 \mathrm{~V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\) \(\mathscr{b}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E ^ { \circ }}=1.60 \mathrm{~V}\)

Short answer

a. The overall balanced equation for the galvanic cell with Cl2 and Br2 reactions is \(\mathrm{Cl}_{2} + \mathrm{Br}_{2} \rightarrow 2 \mathrm{Cl}^{-} + 2 \mathrm{Br}^{-}\), with a standard cell potential \(\mathscr{E}^{\circ}\) of \(0.27 \mathrm{~V}\). The anode is the bromine electrode, and the cathode is the chlorine electrode. Electrons flow from the anode to the cathode, and ions migrate through the salt bridge. b. The overall balanced equation for the galvanic cell with MnO4- and IO4- reactions is \(2 (\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}) + 5(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}) \rightarrow 2(\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}) + 5(\mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O})\), with a standard cell potential \(\mathscr{E}^{\circ}\) of \(0.09 \mathrm{~V}\). The anode is the manganese electrode, and the cathode is the iodine electrode. Electrons flow from the anode to the cathode, and ions migrate through the salt bridge.

Step-by-step solution

Identify the half-reactions

Since the half-reactions are given, we just need to identify the anode (oxidation) and the cathode (reduction). \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{~V}\) In a galvanic cell, the half-reaction with the higher reduction potential is the one that will undergo reduction. Here, the \(\mathrm{Cl}_2\) reaction has a higher \(\mathscr{E}^{\circ}\), so it will happen at the cathode.

Write the overall balanced equation

To obtain the overall balanced equation, we multiply each half-reaction with an appropriate factor so that the number of electrons is the same in both reactions. In this case, we don't need to multiply anything. Overall equation: \(\mathrm{Cl}_{2} + \mathrm{Br}_{2} \rightarrow 2 \mathrm{Cl}^{-} + 2 \mathrm{Br}^{-}\)

Determine \(\mathscr{E}^{\circ}\) for the cell

\(\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = 1.36 \mathrm{~V} - 1.09 \mathrm{~V} = 0.27 \mathrm{~V}\)

Sketch the cell

Anode: Bromine electrode (oxidation) Cathode: Chlorine electrode (reduction) Electron flow: From anode to cathode (Br2 to Cl2) Ion migration: Cl- ions move from cathode to anode, Br- ions move from anode to cathode through the salt bridge. b. Galvanic cell with MnO4- and IO4- reactions:

Identify the half-reactions

Since the half-reactions are given, we just need to identify the anode (oxidation) and the cathode (reduction). \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.60 \mathrm{~V}\) Here, the \(\mathrm{IO}_{4}^{-}\) reaction has a higher \(\mathscr{E}^{\circ}\), so it will happen at the cathode.

Write the overall balanced equation

To obtain the overall balanced equation, we multiply each half-reaction with an appropriate factor so that the number of electrons is the same in both reactions. We'll need to multiply the first reaction by 2 and the second by 5. Overall equation: \(2 (\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}) + 5(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}) \rightarrow 2(\mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}) + 5(\mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O})\)

Determine \(\mathscr{E}^{\circ}\) for the cell

\(\mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = 1.60 \mathrm{~V} - 1.51 \mathrm{~V} = 0.09 \mathrm{~V}\)

Sketch the cell

Anode: Manganese electrode (oxidation) Cathode: Iodine electrode (reduction) Electron flow: From anode to cathode (MnO4- to IO4-) Ion migration: Mn2+ ions move from anode to cathode, IO3- ions move from cathode to anode through the salt bridge. H+ ions will also migrate to maintain electric neutrality.

Key concepts

Electrode Potential

Electrode potential is a key concept in understanding galvanic cells. It refers to the voltage that a particular half-cell reaction generates. This is often represented as \( \mathscr{E}^{\circ} \), which indicates the standard electrode potential under standard conditions (1.0 M concentration, 1.0 atm pressure, typically at 25°C). In a galvanic cell, the half-reaction with a higher electrode potential tends to act as the cathode, undergoing reduction. The difference in electrode potential between the two half-cells is crucial, as it defines the voltage of the galvanic cell.
Remember, electrode potential helps to determine which half-reaction will be oxidized and which will be reduced. The half-cell with the higher electrode potential undergoes reduction, making it the cathode, while the half-cell with the lower electrode potential undergoes oxidation at the anode.
Thus, electrode potential gives us the direction and magnitude of the potential change, dictating the overall cell voltage.

Electron Flow

Electron flow is essential in the functioning of a galvanic cell. In any electrochemical cell, electrons travel from the anode to the cathode through an external circuit. This flow of electrons generates electrical energy, which can be harnessed to power devices.
Understanding electron flow begins with identifying the anode and the cathode. The anode is always the site of oxidation, where electrons are released into the external circuit. Conversely, the cathode is where reduction occurs, and electrons are gained from the external circuit.
In summary:

• Electrons flow spontaneously from anode to cathode.
• Anode is the oxidation site; cathode is the reduction site.
• This flow generates current, the useful output of a galvanic cell.

Half-Reactions

Half-reactions are the fundamental building blocks of balancing redox reactions in galvanic cells. Each half-reaction represents either the oxidation or reduction process occurring in a galvanic cell.
Redox reactions are typically split into two half-reactions:

• Oxidation half-reaction occurs at the anode, capturing the loss of electrons.
• Reduction half-reaction occurs at the cathode, capturing the gain of electrons.

Half-reactions reveal how electrons are transferred in a galvanic cell, and they are crucial for calculating overall cell potential. To form a complete reaction, half-reactions must be balanced to ensure the number of electrons gained in reduction equals those lost in oxidation.

Salt Bridge

A salt bridge is an integral component of a galvanic cell that facilitates the movement of ions and maintains electrical neutrality. Without it, the reaction would quickly halt as the solutions in the half-cells become electrically imbalanced.
The primary functions of a salt bridge are:

• Allowing ion flow between the two half-cells to complete the circuit.
• Preventing the solutions in the half-cells from mixing directly.
• Maintaining charge balance by allowing ions to migrate to counteract the formation of excess positive or negative charges in the half-cells.

Salt bridges often contain a gel or solution of an inert salt like KCl, which does not react with the other substances in the cell, thus ensuring sustained ion flow and steady electron transfer.