Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Ni²+ (aq) + Pb(s) → Ni(s) + Pb²+ (aq) Ni2+ Pb²+ (aq) + 2e → Pb(s) Ee = -0.126 V red (aq) + 2e → Ni(s) E = -0.250 V red AG° = kj K for this reaction would be ✓than one.
Use standard reduction potentials to calculate the standard free energy change in kJ for the reaction: Ni²+ (aq) + Pb(s) → Ni(s) + Pb²+ (aq) Ni2+ Pb²+ (aq) + 2e → Pb(s) Ee = -0.126 V red (aq) + 2e → Ni(s) E = -0.250 V red AG° = kj K for this reaction would be ✓than one.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Calculating the Standard Free Energy Change for a Redox Reaction**
To calculate the standard free energy change (ΔG°) in kJ for the given redox reaction, we use standard reduction potentials.
**Reaction:**
\[ \text{Ni}^{2+} (aq) + \text{Pb} (s) \rightarrow \text{Ni} (s) + \text{Pb}^{2+} (aq) \]
Below are the standard reduction potentials (\(E^\circ_{\text{red}}\)) for the relevant half-reactions:
\[
\begin{array}{c|c}
\text{Half-Reaction} & E^\circ_{\text{red}}\\
\hline
\text{Ni}^{2+} (aq) + 2e^- \rightarrow \text{Ni} (s) & -0.250 \, \text{V} \\
\text{Pb}^{2+} (aq) + 2e^- \rightarrow \text{Pb} (s) & -0.126 \, \text{V} \\
\end{array}
\]
**Calculations:**
\[
\Delta G^\circ = \boxed{\phantom{000}} \text{ kJ}
\]
The equilibrium constant \(K\) for this reaction would be:
\[ K \text{ is } \boxed{\text{greater or less}} \text{ than one.} \]
**Explanation:**
The information provided shows the standard reduction potentials for nickel and lead in their respective half-reactions. The differences in these potentials allow us to determine the cell potential (E°) of the overall reaction, which can then be used to calculate the Gibbs free energy change. The value of \(K\) will indicate if the reaction favors products or reactants.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ad64212-e120-4fe2-97f2-e4e87c5c792a%2F0dee2a02-9aff-48f5-9ac4-41ec8ff1ba6d%2Fuhpu65_processed.png&w=3840&q=75)
Transcribed Image Text:**Calculating the Standard Free Energy Change for a Redox Reaction**
To calculate the standard free energy change (ΔG°) in kJ for the given redox reaction, we use standard reduction potentials.
**Reaction:**
\[ \text{Ni}^{2+} (aq) + \text{Pb} (s) \rightarrow \text{Ni} (s) + \text{Pb}^{2+} (aq) \]
Below are the standard reduction potentials (\(E^\circ_{\text{red}}\)) for the relevant half-reactions:
\[
\begin{array}{c|c}
\text{Half-Reaction} & E^\circ_{\text{red}}\\
\hline
\text{Ni}^{2+} (aq) + 2e^- \rightarrow \text{Ni} (s) & -0.250 \, \text{V} \\
\text{Pb}^{2+} (aq) + 2e^- \rightarrow \text{Pb} (s) & -0.126 \, \text{V} \\
\end{array}
\]
**Calculations:**
\[
\Delta G^\circ = \boxed{\phantom{000}} \text{ kJ}
\]
The equilibrium constant \(K\) for this reaction would be:
\[ K \text{ is } \boxed{\text{greater or less}} \text{ than one.} \]
**Explanation:**
The information provided shows the standard reduction potentials for nickel and lead in their respective half-reactions. The differences in these potentials allow us to determine the cell potential (E°) of the overall reaction, which can then be used to calculate the Gibbs free energy change. The value of \(K\) will indicate if the reaction favors products or reactants.
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