Problem 106 - Chemistry The Molecular Science

Question

An electrochemical cell is made by immersing a strip of chromium into a \(1.0-\mathrm{M}\) solution of \(\mathrm{Cr}^{3+}\) and a strip of gold into a \(1.0-\mathrm{M}\) solution of \(\mathrm{Au}^{3+} .\) The half-cells are connected by a salt bridge. A wire and light bulb complete the circuit. (a) Write the balanced chemical equation for the reaction that is product- favored. (b) Calculate the cell potential. (c) Draw a sketch of the cell and indicate the anode, cathode, and direction of electron flow.

Step-by-Step Solution

Answer

(a) \(\text{Cr} + \text{Au}^{3+} \rightarrow \text{Cr}^{3+} + \text{Au}\) (b) 2.24 V (c) Electrons flow from Cr (anode) to Au (cathode).

Step 1: Identify Half-Reactions

Determine the two half-reactions involved in the electrochemical cell. Chromium undergoes oxidation, and Gold undergoes reduction as follows:\[\text{Oxidation (Anode): } \text{Cr} \rightarrow \text{Cr}^{3+} + 3e^- \]\[\text{Reduction (Cathode): } \text{Au}^{3+} + 3e^- \rightarrow \text{Au}\]

Step 2: Write the Balanced Chemical Equation

Combine the oxidation and reduction half-reactions to form the balanced overall chemical equation. Ensure that the electrons cancel out:\[\text{Cr} + \text{Au}^{3+} \rightarrow \text{Cr}^{3+} + \text{Au}\]

Step 3: Calculate Standard Cell Potentials

Look up the standard reduction potentials from a table:\[E^\circ(\text{Cr}^{3+}/\text{Cr}) = -0.74\, V\] \[E^\circ(\text{Au}^{3+}/\text{Au}) = +1.50\, V\]Calculate the standard cell potential using the formula:\[E^\circ_{\text{cell}} = E^\circ(\text{cathode}) - E^\circ(\text{anode}) = 1.50 - (-0.74) = 2.24\, V\]

Step 4: Draw the Electrochemical Cell

Sketch the cell, labeling the anode (Chromium electrode) and cathode (Gold electrode). Indicate the direction of electron flow from the anode to the cathode through the wire, with the light bulb in series. The cell starts with electrodes immersed in their respective ion solutions connected by a salt bridge.

Key Concepts

Oxidation-Reduction Reaction

In an electrochemical cell, oxidation-reduction reactions, commonly known as redox reactions, are the driving force behind the generation of electric current. When these reactions occur, one substance loses electrons (oxidation) while another substance gains electrons (reduction), leading to a flow of electrons through an external circuit.

In the case of the given electrochemical cell, chromium undergoes oxidation by losing electrons, transforming from Cr to Cr³⁺:

• Oxidation (at the Anode): \( \text{Cr} \rightarrow \text{Cr}^{3+} + 3e^- \)
• Reduction (at the Cathode): \( \text{Au}^{3+} + 3e^- \rightarrow \text{Au} \)

Combining these two half-reactions provides the balanced equation: \ \( \text{Cr} + \text{Au}^{3+} \rightarrow \text{Cr}^{3+} + \text{Au} \).

Through understanding these reactions, it’s clear how materials interact and drive the flow of electricity in the cell.

Cell Potential Calculation

The cell potential is a measure of the driving force behind an electrochemical reaction, indicating how much electric potential difference (voltage) the cell can produce. To calculate the standard cell potential, we utilize the standard reduction potentials from a reference table. This involves subtracting the anode’s potential from the cathode’s potential as in the equation:

• \( E^\circ_{\text{cell}} = E^\circ(\text{cathode}) - E^\circ(\text{anode}) \)

For the chromium-gold cell, the given values are:

• \( E^\circ(\text{Cr}^{3+}/\text{Cr}) = -0.74\, V \)
• \( E^\circ(\text{Au}^{3+}/\text{Au}) = +1.50\, V \)

Plugging these into the formula, we find:\[ E^\circ_{\text{cell}} = 1.50 - (-0.74) = 2.24\, V \]This result indicates that the electrochemical cell would exert a potential difference of 2.24 volts under standard conditions.

Cell Diagram and Electron Flow

An electrochemical cell diagram helps visualize the setup and functioning of the redox process. In our specific example, figure out the orientation of components like electrodes, wires, and the salt bridge, and understand how these physical arrangements contribute to the cell's operation.

Key elements of the diagram:

• Anode (Chromium): The site of oxidation, where electrons are released from the chromium strip into the solution and travel through the wire.
• Cathode (Gold): The location where reduction occurs, allowing Au³⁺ ions to pick up electrons and form solid gold.
• Electron flow: Arrows in diagrams illustrate electron movement through the wire from the anode towards the cathode, powering a connected device such as a light bulb.
• Salt bridge: Necessary for maintaining ionic balance and completing the circuit, it lets ions freely move between the two half-cells, but without direct mixing of different solutions.

Such an arrangement ensures efficient electron transfer, sustaining the current flow through the circuit.

Standard Reduction Potentials

Standard reduction potentials (E°) play a crucial role in determining the behavior and efficiency of electrochemical cells. These values, obtained from standardized tables, represent how strongly a substance tends to gain electrons (or be reduced) under standard conditions.

Each half-reaction has an associated standard reduction potential, allowing us to predict which species will undergo reduction or oxidation.

• Electrode with higher reduction potential will act as the cathode as it attracts electrons.
• Conversely, the electrode with the lower potential will function as the anode by releasing electrons.

For our cell:

• Au³⁺/Au has a higher E° value (+1.50 V), making it the cathode.
• Cr³⁺/Cr, with a lower E° (-0.74 V), serves as the anode.

Knowing these potentials allows us to design efficient cells and calculate the potential difference, indicating the energy available to do work.