Problem 98 - Chemistry: An Atoms-Focused Approach

Question

Corrosion of Copper Pipes The copper pipes frequently used in household plumbing may corrode and eventually leak. The corrosion reaction is believed to involve the formation of copper( 1 ) chloride: $$ 2 \mathrm{Cu}(s)+\mathrm{Cl}_{2}(a q) \rightarrow 2 \mathrm{CuCl}(s) $$ a. Write balanced cquations for the half-reactions in this redox reaction. b. Calculate \(E_{\text {ran }}^{\circ}\) and \(\Delta G_{\text {rxn }}^{a}\) for the reaction.

Step-by-Step Solution

Answer

Question: Write the balanced half-reactions and calculate the standard cell potential and standard Gibbs free energy change for the corrosion of copper pipes. Answer: The balanced half-reactions are: 1. Oxidation: Cu(s) → CuCl(s) + e^- 2. Reduction: Cl₂(aq) + 2e^- → 2CuCl(s) The standard cell potential (E°rxn) is -1.88V, and the standard Gibbs free energy change (ΔG°rxn) is 361.68 kJ/mol.

Step 1: Identify the half-reactions

First, we need to identify the oxidation and reduction half-reactions from the overall balanced equation: $$ 2 \mathrm{Cu}(s)+\mathrm{Cl}_{2}(a q) \rightarrow 2 \mathrm{CuCl}(s) $$ Oxidation half-reaction: Copper (Cu) loses electrons and transforms into copper(I) chloride (CuCl). To balance the charges, we need to add one electron to the right side of the equation: $$ \mathrm{Cu}(s) \rightarrow \mathrm{CuCl}(s) + e^- $$ Reduction half-reaction: Chlorine (Cl\(_2\)) gains electrons and also transforms into copper(I) chloride (CuCl). To balance the charges, we need to add two electrons to the left side of the equation: $$ \mathrm{Cl}_{2}(aq) + 2e^- \rightarrow 2\mathrm{CuCl}(s) $$ These are the balanced half-reactions for the given overall reaction.

Step 2: Find the standard cell potential (\(E_{\text{rxn}}^{\circ}\))

To calculate the standard cell potential, we need to find the standard electrode potentials for both Cu and Cl. Standard electrode potential for copper oxidation (Cu \(|\) Cu\(^{+}\)) is \(E_{\text{Cu}^{+}| \text{Cu}}^{\circ} = +0.52V\) Standard electrode potential for chlorine reduction (Cl\(_2\) \(|\) Cl\(^-\)) is \(E_{\text{Cl}^-|\text{Cl}_2}^{\circ}=-1.36V\) Now we can calculate the standard cell potential using the following formula: \(E_{\text{rxn}}^{\circ} = E_{\text{red}}^{\circ} - E_{\text{ox}}^{\circ}\) \(E_{\text{rxn}}^{\circ} = (-1.36) - (0.52) = -1.88V\)

Step 3: Calculate the standard Gibbs free energy change

Now that we have the standard cell potential, we can calculate the standard Gibbs free energy change using the equation: \(\Delta G_{\text{rxn}}^{\circ}=-nFE_{\text{rxn}}^{\circ}\) Where \(n\) is the number of moles of electrons transferred during the reaction, \(F\) is Faraday's constant (\(F\approx 96500\ C/mol\)), and \(E_{\text{rxn}}^{\circ}\) is the standard cell potential. In this reaction, two moles of electrons are transferred (from the balanced equation), and we previously calculated \(E_{\text{rxn}}^{\circ} = -1.88V\). Therefore, we can calculate the standard Gibbs free energy change: $$ \Delta G_{\text{rxn}}^{\circ}=-2(96500)(-1.88)\ \text{J/mol} $$ After calculating the result, we get: $$ \Delta G_{\text{rxn}}^{\circ}=361.68\ \text{kJ/mol} $$ In conclusion: a. The balanced half-reactions are: $$ \mathrm{Cu}(s) \rightarrow \mathrm{CuCl}(s) + e^- $$ $$ \mathrm{Cl}_{2}(aq) + 2e^- \rightarrow 2\mathrm{CuCl}(s) $$ b. The standard cell potential \(E_{\text{rxn}}^{\circ} = -1.88V\), and the standard Gibbs free energy change is \(\Delta G_{\text{rxn}}^{\circ}=361.68\ \text{kJ/mol}\).

Key Concepts

Redox Reactions

Redox reactions are fascinating because they involve the transformation of substances by the exchange of electrons. In the corrosion of copper pipes, this type of reaction takes place. Redox stands for "reduction-oxidation." This means that in one part of the reaction, a substance loses electrons (oxidation), and in another part, a different substance gains those electrons (reduction).

For example, let's look at the corrosion reaction of copper pipes:

• Oxidation: Copper ( \(\text{Cu}\)) transforms into copper(I) chloride ( \(\text{CuCl}\)) by losing electrons.
• Reduction: Chlorine ( \(\text{Cl}_2\)) gains electrons and becomes chloride ions, which then react with copper to form copper(I) chloride.

Oxidation and reduction always happen together in redox reactions. That's why the complete equation shows both processes occurring simultaneously. When balancing these reactions, we ensure the same number of electrons is gained and lost, preserving the principle of charge conservation.

Standard Cell Potential

The standard cell potential (\(E_{\text{rxn}}^{\circ}\)) is a critical concept in electrochemistry. It tells us the voltage potential or the electrical force of a cell under standard conditions when redox reactions take place. Essentially, it measures how eagerly the electrons will transfer in a given reaction.

To calculate the standard cell potential, we rely on standard electrode potentials of half-reactions. These are known values for specific ions or molecules:

• Standard electrode potential for copper oxidation ( \(\text{Cu} \mid \text{Cu}^{+}\)) is \(+0.52 \text{V}\)
• Standard electrode potential for chlorine reduction ( \(\text{Cl}_2 \mid \text{Cl}^-\)) is \(-1.36 \text{V}\)

The standard cell potential is calculated by subtracting the oxidation potential from the reduction potential:\[E_{\text{rxn}}^{\circ} = E_{\text{red}}^{\circ} - E_{\text{ox}}^{\circ}\]

In our example, this leads to:\[E_{\text{rxn}}^{\circ} = (-1.36 \text{V}) - (0.52 \text{V}) = -1.88 \text{V}\]

A negative standard cell potential indicates that the reaction is nonspontaneous under standard conditions, meaning additional energy would be required for the reaction to occur.

Gibbs Free Energy Change

Gibbs free energy change (\(\Delta G\)) is crucial for understanding whether a reaction can occur spontaneously. It tells us if a reaction does or does not need additional energy to proceed. When it comes to electrochemical cells, we can relate Gibbs free energy change to the standard cell potential using this formula:\[\Delta G_{\text{rxn}}^{\circ} = -nFE_{\text{rxn}}^{\circ}\]

Here:

• \(n\) is the number of moles of electrons transferred during the reaction.
• \(F\) is Faraday's constant, approximately \(96500\ C/mol\).
• \(E_{\text{rxn}}^{\circ}\) is the standard cell potential.

In our copper pipe corrosion example, \(n = 2\), and we calculated \(E_{\text{rxn}}^{\circ} = -1.88 \text{V}\). Substituting into the formula:\[\Delta G_{\text{rxn}}^{\circ} = -2 \times 96500 \times (-1.88)\ \text{J/mol}\]

After solving this, we find:\[\Delta G_{\text{rxn}}^{\circ} = 361.68\ \text{kJ/mol}\]

A positive Gibbs free energy change also indicates that the reaction is nonspontaneous under standard conditions. More energy from an external source would be required for the reaction to take place spontaneously.