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

Short answer

For part a, the cathode is \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}\) and the anode is \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}\). The overall balanced equation is \(2\mathrm{Br}_{2} \rightarrow 4\mathrm{Br}^{-}+\mathrm{Cl}_{2}+2 \mathrm{e}^{-}\), and the standard cell potential is \(0.27 \mathrm{V}\). For part b, the cathode is \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O}\) and the anode is \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}\). The overall balanced equation is \(10\mathrm{IO}_{4}^{-}+16 \mathrm{H}^{+}+2\mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+} \rightarrow 5\mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O}+2\mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}\), and the standard cell potential is \(0.09 \mathrm{V}\).

Step-by-step solution

The half-reaction with the higher reduction potential will occur at the cathode and the one with the lower reduction potential will occur at the anode. Cathode: \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{V}\) Anode: \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{E}^{\circ}=1.09 \mathrm{V}\) 2. Determine the direction of electron flow and ion migration

Electrons flow from anode to cathode. In the salt bridge, anions will move towards the anode and cations will move towards the cathode. 3. Write the overall balanced equation

Multiply the anode half-reaction by 2 to balance the number of electrons, and then add the two half-reactions together. Anode: \(2(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-})\) Cathode: \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-}\) Overall: \(2\mathrm{Br}_{2} \rightarrow 4\mathrm{Br}^{-}+\mathrm{Cl}_{2}+2 \mathrm{e}^{-}\) 4. Calculate the standard cell potential

\(\mathscr{E}^{\circ}_{cell}=\mathscr{E}^{\circ}_{cathode}-\mathscr{E}^{\circ}_{anode}=1.36 \mathrm{V}-1.09 \mathrm{V}=0.27 \mathrm{V}\) #Part b# 1. Determine the cathode and anode

The half-reaction with the higher reduction potential will occur at the cathode, and the one with the lower reduction potential will occur at the anode. Cathode: \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.60 \mathrm{V}\) Anode: \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.51 \mathrm{V}\) 2. Determine the direction of electron flow and ion migration

Electrons flow from anode to cathode. In the salt bridge, anions will move towards the anode and cations will move towards the cathode. 3. Write the overall balanced equation

Multiply the anode half-reaction by 2 and the cathode half-reaction by 5 to balance the number of electrons, and then add the two half-reactions together. Anode: \(2(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O})\) Cathode: \(5(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O})\) Overall: \(10\mathrm{IO}_{4}^{-}+16 \mathrm{H}^{+}+2\mathrm{MnO}_{4}^{-}+16 \mathrm{H}^{+} \rightarrow 5\mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O}+2\mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}\) 4. Calculate the standard cell potential

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

Key concepts

Standard Cell Potential

The standard cell potential, denoted as \( \mathscr{E}^{\circ}_{cell} \), is a vital concept in electrochemistry that measures the voltage difference between two half-cells in a galvanic cell when all components are in their standard states, generally at 1M concentration for solutions and 1 atm pressure for gases. This value is indicative of the cell's ability to do electrical work, where a higher potential means a greater capability to drive an electron flow through an external circuit.

To calculate the standard cell potential of a galvanic cell, we subtract the standard reduction potential of the anode \( \mathscr{E}^{\circ}_{anode} \) from that of the cathode \( \mathscr{E}^{\circ}_{cathode} \), which is \( \mathscr{E}^{\circ}_{cell} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} \). This calculation presupposes that the more positive half-reaction will occur at the cathode, undergoing reduction, and the less positive (or more negative) half-reaction will occur at the anode, undergoing oxidation.

For instance, in the exercise provided, the standard cell potential for the reaction involving chlorine and bromine is calculated as \( 1.36 \mathrm{V} - 1.09 \mathrm{V} = 0.27 \mathrm{V} \). This numerical value not only reveals the spontaneity of the reaction but also reflects the efficiency of the galvanic cell in question.

Electron Flow

In any electrochemical cell, electrons flow from the anode to the cathode through the external circuit. This movement generates an electric current that can be harnessed to do work. The anode is where oxidation takes place, meaning the loss of electrons, which then travel through the wire toward the cathode where reduction occurs, or where the electrons are gained.

It's essential to understand that electron flow direction determines the cell's polarity. For instance, in the given exercise, the standard cell potentials suggest that electrons will move from bromine (Br2) to chlorine (Cl2) in the first scenario, because bromine has the lower reduction potential and is therefore the anode. Additionally, cations (positively charged ions) move towards the cathode within the cell, and anions (negatively charged ions) move towards the anode, maintaining charge balance, a concept that is closely linked to the electron flow in the cell.

Cathode and Anode Reactions

The cathode and anode roles in a galvanic cell are pivotal; the cathode is where the reduction takes place, and the anode is where oxidation occurs. This relationship is best remembered by the acronym 'Red Cat, An Ox', standing for reduction at the cathode and oxidation at the anode.

In the exercise's galvanic cells, for the first case, chlorine gas \( \mathrm{Cl}_{2} \) is being reduced to chloride ions \( \mathrm{Cl}^{-} \) at the cathode, while bromine gas \( \mathrm{Br}_{2} \) is being oxidized to bromide ions \( \mathrm{Br}^{-} \) at the anode. The reactions at these electrodes are crucial for the operation of the galvanic cell as they encapsulate the essential electron transfer processes that enable the flow of electric current. Understanding each half-reaction and identifying which occurs at the cathode or anode is crucial to master the dynamics of galvanic cells.

Overall Balanced Equation

Synthesizing the half-reactions at the anode and cathode results in the overall balanced equation, which portrays a complete view of the electrochemical process. This involves ensuring that the number of electrons lost in the oxidation process is equal to the number gained in the reduction, which may require multiplying one or both half-reactions by appropriate factors.

In the given exercise, we observe a consistent electron exchange that must be balanced. For the chlorine-bromine cell, the overall balanced equation combines doubled bromine anode reaction with the chlorine cathode reaction, resulting in a complete electron circuit. For the exercise's second case involving permanganate \( \mathrm{MnO}_4^{-} \) and periodate \( \mathrm{IO}_{4}^{-} \), a 2:5 ratio is required to balance the electron transfer. Properly balancing these reactions is crucial because it reflects the stoichiometry of the electrochemical process, ensuring that mass and charge are conserved throughout the reaction.