91Ó°ÊÓ

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. \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{~V}\) \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2}\) \(\mathscr{C}^{\circ}=0.68 \mathrm{~V}\) \(\begin{array}{ll}\text { b. } \mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} & \mathscr{E}^{\circ}=-1.18 \mathrm{~V} \\ \mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} & \mathscr{E}^{\circ}=-0.036 \mathrm{~V}\end{array}\)

Step by step solution

01

1. Identify the anode and cathode based on the standard potentials

Reduction potentials are given, and the reaction with the highest standard reduction potential will occur at the cathode, while the reaction with the lower reduction potential will happen at the anode. Cathode reaction: \(\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78\mathrm{~V}\) Anode reaction: \(\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\mathrm{O}_{2}\quad\mathscr{E}^{\circ}=0.68 \mathrm{~V}\)

02

2. Determine the direction of electron flow and ion migration

Electrons flow from the anode (negative side) to the cathode (positive side). For ion migration, anions move from the cathode to the anode, and cations move from the anode to the cathode through the salt bridge.

03

3. Write the overall balanced equation for the reactions

To get the overall balanced equation, we need to combine the anode and cathode reactions: \(\mathrm{H}_{2} \mathrm{O}_{2} + 2 \mathrm{H}^+ + 2 \mathrm{e}^- + \mathrm{O}_2 + 2 \mathrm{H}^+ + 2 \mathrm{e}^- \rightarrow 2 \mathrm{H}_2 \mathrm{O} + \mathrm{H}_{2}\mathrm{O}_{2}\) After simplifying, the overall balanced equation is: \(\mathrm{O}_{2} + 2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{H}_{2} \mathrm{O}_{2}\)

04

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

Standard cell potential (\(\mathscr{E}^{\circ}\)) is determined by subtracting the cathode's standard potential from the anode's standard potential: \(\mathscr{E}^{\circ} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = 1.78 \mathrm{~V} - 0.68 \mathrm{~V} = 1.1 \mathrm{~V}\) #Case b: Sketching the galvanic cell for given half-reactions#

05

1. Identify the anode and cathode based on the standard potentials

Cathode reaction: \(\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036\mathrm{~V}\) Anode reaction: \(\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \mathscr{E}^{\circ}=-1.18 \mathrm{~V}\)

06

2. Determine the direction of electron flow and ion migration

Electrons flow from the anode (negative side) to the cathode (positive side). Anions move from the cathode to the anode and cations move from the anode to the cathode through the salt bridge.

07

3. Write the overall balanced equation for the reactions

To get the overall balanced equation, we need to combine the anode and cathode reactions and balance the number of electrons. By multiplying the cathode reaction by 2 and the anode reaction by 3, the electrons will be balanced: \(2(\mathrm{Fe}^{3+} + 3 \mathrm{e}^-) \rightarrow 6 \mathrm{e}^- + 6 \mathrm{Fe}^{3+} \rightarrow 3\mathrm{Fe^{2+}} + \mathrm{Mn} \rightarrow \mathrm{Mn}^{3+} + 2 \mathrm{Fe}\) After simplifying, the overall balanced equation is: \(2 \mathrm{Fe}^{3+} + 3 \mathrm{Mn}^{2+} \rightarrow 2 \mathrm{Mn}^{3+} + 3 \mathrm{Fe}\)

08

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

\(\mathscr{E}^{\circ} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = -0.036 \mathrm{~V} - (-1.18 \mathrm{~V}) = 1.144 \mathrm{~V}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-reactions

In the context of a galvanic cell, half-reactions are crucial for understanding how the cell operates. These reactions split the overall cell process into two parts: oxidation and reduction. Each half-reaction happens in a different part of the cell, known as the anode and cathode respectively.

For example, consider the half-reactions given in the exercise:

• Cathode reaction: \[\text{H}_2\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow 2\text{H}_2\text{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{~V}\]
• Anode reaction: \[\text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2\text{O}_2 \quad \mathscr{E}^{\circ}=0.68 \mathrm{~V}\]

Each half-reaction represents a part of the complete electrochemical reaction. Importantly, electrons are transferred, with one half-reaction releasing electrons, while the other requires them. This movement of electrons is what generates the electrical energy in a galvanic cell.

Standard reduction potential

Standard reduction potential, denoted as \(\mathscr{E}^{\circ}\), is a measure of the tendency of a chemical species to be reduced, measured under standard conditions. It is a crucial factor in determining how electrons will flow between substances in a galvanic cell.

The greater the standard reduction potential, the higher the tendency of a species to gain electrons and hence get reduced. For instance, in the cell described in the exercise:

• The cathode has a standard reduction potential of \(1.78\, \mathrm{V}\)
• The anode has a standard reduction potential of \(0.68\, \mathrm{V}\)

The higher value at the cathode indicates that it will readily occur as the reduction half. Comparing these values allows you to predict which will act as the site for reduction and which for oxidation in your galvanic cell.

Anode and cathode identification

Understanding the distinction between anode and cathode is central to mastering galvanic cells. These two types of electrodes play distinct functions in the flow of electrons throughout the cell.

- **The Anode**: This is where oxidation occurs, meaning it is the source of electrons. In our examples, the anode is the reactant in the half-reaction with the lower standard reduction potential (\(0.68\, \mathrm{V}\)).- **The Cathode**: This is where reduction occurs, and electrons flow towards it. Correspondingly, it is assigned the higher standard reduction potential \(1.78\, \mathrm{V}\).

Think of it this way: in a galvanic cell, electrons leave the anode and head to the cathode. This process is driven by the difference in reduction potential between the two electrodes.

Overall balanced equation

Creating an overall balanced equation from two half-reactions is akin to putting together the pieces of a puzzle. The balanced equation represents the entire flow of electrons and ions in the cell.

To start, ensure that the electrons gained from and lost by both half-reactions are equal. In this case, you combine the anode and cathode reactions:\[\text{H}_2\text{O}_2 + 2\text{H}^+ + 2\text{e}^- + \text{O}_2 + 2\text{H}^+ + 2\text{e}^- \rightarrow 2\text{H}_2\text{O} + \text{H}_2\text{O}_2\]After simplification, it becomes:\[\text{O}_2 + 2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{H}_2\text{O}_2\]The overall balanced equation summarizes the overall reaction and confirms charge and mass balance.

Direction of electron flow

In galvanic cells, electron flow is critical to generating electric current. Electrons move through an external circuit from the anode to the cathode, utilizing potential energy from the reaction.

• The electron flow in our example travels from anode's lower potential (\(0.68\, \mathrm{V}\)) to cathode's higher potential (\(1.78\, \mathrm{V}\))
• Simultaneously, within the salt bridge, ions migrate to maintain charge: anions head towards the anode and cations towards the cathode.

This electron journey is the basis for the functionality of the cell as it leads to electron movement in the external circuit, allowing it to do work.