Problem 2 - Chemistry Matter and Change

Question

For each of these pairs of half-reactions, write the balanced equation for the overall cell reaction, and calculate the standard cell potential. Describe the reaction using cell notation. Refer to Chapter 19 to review writing and balancing redox equations. $$\mathrm{Co}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Co}(\mathrm{s}) \text { and } \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s})$$

Step-by-Step Solution

Answer

The balanced reaction is \(3\mathrm{Co}^{2+} + 2\mathrm{Cr}^{3+} \rightarrow 3\mathrm{Co} + 2\mathrm{Cr}\) and the standard cell potential is 0.46 V.

Step 1: Write the Reduction Half-Reactions

The half-reactions given are:1. \(\mathrm{Co}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Co}(\mathrm{s})\)2. \(\mathrm{Cr}^{3+}(\mathrm{aq}) + 3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s})\)

Step 2: Balance Electrons in Each Half-Reaction

The half-reactions need to have the same number of electrons. Co needs 2 electrons, while Cr requires 3 electrons. Therefore, multiply the cobalt half-reaction by 3 and the chromium half-reaction by 2:1. \(3(\mathrm{Co}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Co}(\mathrm{s}))\) becomes \(3\mathrm{Co}^{2+}(\mathrm{aq}) + 6 \mathrm{e}^{-} \rightarrow 3 \mathrm{Co}(\mathrm{s})\)2. \(2(\mathrm{Cr}^{3+}(\mathrm{aq}) + 3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s}))\) becomes \(2\mathrm{Cr}^{3+}(\mathrm{aq}) + 6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}(\mathrm{s})\)

Step 3: Write the Overall Redox Reaction

Add the balanced half-reactions together:\(3\mathrm{Co}^{2+}(\mathrm{aq}) + 6 \mathrm{e}^{-} + 2\mathrm{Cr}^{3+}(\mathrm{aq}) + 6 \mathrm{e}^{-} \rightarrow 3 \mathrm{Co}(\mathrm{s}) + 2 \mathrm{Cr}(\mathrm{s})\)The electrons cancel out, so the overall balanced cell reaction is:\(3\mathrm{Co}^{2+}(\mathrm{aq}) + 2\mathrm{Cr}^{3+}(\mathrm{aq}) \rightarrow 3 \mathrm{Co}(\mathrm{s}) + 2 \mathrm{Cr}(\mathrm{s})\)

Step 4: Calculate Standard Cell Potential

The standard reduction potentials are:\(E^⦵(\mathrm{Co}^{2+}/\mathrm{Co}) = -0.28\, \mathrm{V}\)\(E^⦵(\mathrm{Cr}^{3+}/\mathrm{Cr}) = -0.74\, \mathrm{V}\)Since cobalt is reduced, using the formula for cell potential, select the higher potential as the reduction and the lower as oxidation:\[E^⦵_{\mathrm{cell}} = E^⦵_{\mathrm{reduction}} - E^⦵_{\mathrm{oxidation}} = -0.28\, \mathrm{V} - (-0.74\, \mathrm{V}) = 0.46\, \mathrm{V}\]

Step 5: Write the Cell Notation

The cell notation is written as:\(\mathrm{Cr}(\mathrm{s}) | \mathrm{Cr}^{3+}(\mathrm{aq}) || \mathrm{Co}^{2+}(\mathrm{aq}) | \mathrm{Co}(\mathrm{s})\)

Key Concepts

Balancing Redox Equations

Balancing redox equations involves ensuring that the number of electrons lost in oxidation equals the number gained in reduction. This technique is fundamental because, in a chemical reaction, mass and charge must be conserved. For the given half-reactions:

• Cobalt: \( \mathrm{Co}^{2+}(\mathrm{aq}) + 2 \mathrm{e}^{-} \rightarrow \mathrm{Co}(\mathrm{s}) \)
• Chromium: \( \mathrm{Cr}^{3+}(\mathrm{aq}) + 3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s}) \)

The cobalt half-reaction needs to be multiplied by 3 to provide a total of 6 electrons, while the chromium half-reaction is multiplied by 2 for the same reason. This leads to an equal electron exchange, allowing the two half-reactions to be combined:3\( \mathrm{Co}^{2+}(\mathrm{aq}) \)+ 6\( \mathrm{e}^{-} \)+ 2\( \mathrm{Cr}^{3+}(\mathrm{aq}) \)+ 6\( \mathrm{e}^{-} \rightarrow 3 \mathrm{Co}(\mathrm{s}) + 2 \mathrm{Cr}(\mathrm{s}) \)
Remember, the electrons should cancel out when combined correctly.

Electrochemical Cells

An electrochemical cell is a device that converts chemical energy into electrical energy through redox reactions. It consists of two electrodes: an anode and a cathode. In an electrochemical cell:

• Oxidation occurs at the anode.
• Reduction takes place at the cathode.

In the context of our half-reactions, the setup involves the chromium acting as the anode where it undergoes oxidation, while cobalt serves as the cathode getting reduced. The flow of electrons from the reducing agent to the oxidizing agent through an external circuit generates electric current, making these cells crucial in batteries and various industrial processes.

Standard Cell Potential

The standard cell potential \( (E^⦵_{\mathrm{cell}}) \) portrays the ability of a cell to produce electricity. It is calculated by taking the difference between the reduction potentials of the cathode and the anode. For the half-reactions entry data:

• Cobalt reduction potential \( E^⦵(\mathrm{Co}^{2+}/\mathrm{Co}) = -0.28 \, \mathrm{V} \)
• Chromium reduction potential \( E^⦵(\mathrm{Cr}^{3+}/\mathrm{Cr}) = -0.74 \, \mathrm{V} \)

The cell potential is positive \(E^⦵_{\mathrm{cell}} = -0.28 \, \mathrm{V} - (-0.74 \, \mathrm{V}) = 0.46 \, \mathrm{V} \), which indicates a spontaneous reaction, making it feasible for practical applications like chemical batteries.

Half-Reaction Method

The half-reaction method is a systematic way to balance redox reactions, especially for complex reactions in aqueous solutions. This approach involves the following key steps:

• Identify and write the two half-reactions.
• Balance the atoms other than O and H first.
• Add water \((\mathrm{H}_2\mathrm{O})\) to balance O atoms and hydrogen ions \((\mathrm{H}^+)\) to balance H atoms if the reactions occur in acidic solutions. For basic solutions, add \(\mathrm{OH}^-\).
• Balance the resulting electrical charge by adding electrons.
• Scale the half-reactions, so they have equal electrons, allowing them to cancel out upon addition.

In the reaction provided, the half-reactions were already shown. Balancing steps involve matching the number of electrons in each half-reaction before they were combined to form an overall balanced redox equation showing the flow of electrons from one species to another.