A solution containing 0.13 M Pb2+ , 1.5 x 10-6 M Pb+, 1.5 x 10-6 M Mn²+, 0.13 M MnO, , and 0.88 M HNO, was prepared. For this solution, the balanced reduction half-reactions and overall net reaction shown can occur. 5 [Pb+ +2e = Pb2+] 2[MnO, + 8 H* +5e= Mn2+ + 4 H, O] 5 Pb4+ +2 Mn2+ + 8 H,0=5 Pb²+ + 2 MnO, + 16 H* E; = 1.690 V E = 1.507 V Determine Ell, AG , and K for this reaction. Ecell 30.9 Incorrect AG = -176567.55
A solution containing 0.13 M Pb2+ , 1.5 x 10-6 M Pb+, 1.5 x 10-6 M Mn²+, 0.13 M MnO, , and 0.88 M HNO, was prepared. For this solution, the balanced reduction half-reactions and overall net reaction shown can occur. 5 [Pb+ +2e = Pb2+] 2[MnO, + 8 H* +5e= Mn2+ + 4 H, O] 5 Pb4+ +2 Mn2+ + 8 H,0=5 Pb²+ + 2 MnO, + 16 H* E; = 1.690 V E = 1.507 V Determine Ell, AG , and K for this reaction. Ecell 30.9 Incorrect AG = -176567.55
Chemistry
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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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Corrosion is defined as an activity that transforms refined metals into more chemically stable forms such as oxide, hydroxide, carbonate, or sulfide. It refers to the slow decomposition of things (typically metals); thanks to chemical and/or electrochemical reactions with their surroundings. Corrosion engineering is the science of preventing and controlling corrosion.
Corrosion
Corrosion is defined as an activity that transforms refined metals into more chemically stable forms such as oxide, hydroxide, carbonate, or sulfide. It refers to the slow decomposition of things (typically metals); thanks to chemical and/or electrochemical reactions with their surroundings. Corrosion engineering is the science of preventing and controlling corrosion.
Question
![**Electrochemical Reaction Analysis**
A solution containing:
- 0.13 M \(\text{Pb}^{2+}\)
- \(1.5 \times 10^{-6}\) M \(\text{Pb}^{4+}\)
- \(1.5 \times 10^{-6}\) M \(\text{Mn}^{2+}\)
- 0.13 M \(\text{MnO}_4^-\)
- 0.88 M \(\text{HNO}_3\)
was prepared. For this solution, the balanced reduction half-reactions and overall net reaction shown can occur:
\[
5 \left[\text{Pb}^{4+} + 2e^- \right] \rightleftharpoons \text{Pb}^{2+} \quad E^\circ = 1.690 \text{ V}
\]
\[
2 \left[\text{MnO}_4^- + 8H^+ + 5e^- \right] \rightleftharpoons \text{Mn}^{2+} + 4H_2O \quad E^\circ = 1.507 \text{ V}
\]
\[
\text{Net: } 5\text{Pb}^{4+} + 2\text{Mn}^{2+} + 8H_2O \rightleftharpoons 5\text{Pb}^{2+} + 2\text{MnO}_4^- + 16H^+
\]
**Determine** \(E^\circ_{\text{cell}}, \Delta G^\circ,\) and \(K\) for this reaction.
\[
E^\circ_{\text{cell}} = \boxed{30.9} \text{ V}
\]
*(Note: Answer marked as incorrect.)*
\[
\Delta G^\circ = -176567.55 \text{ J}
\]
In this example, the calculated standard cell potential (\(E^\circ_{\text{cell}}\)) is labeled "Incorrect," suggesting that the result needs further verification. The Gibbs free energy change (\(\Delta G^\circ\)) has been calculated, reflecting the spontaneity of the reaction. Further calculations would be necessary to find the correct \(E^\circ_{\text{cell}}\) and subsequently the equilibrium constant \(K\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc14feaa7-f26c-4630-9d28-afce0e4a19e9%2F48a5d7a7-5b31-4361-89c4-25eb55d42a2a%2Fma4aj9m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Electrochemical Reaction Analysis**
A solution containing:
- 0.13 M \(\text{Pb}^{2+}\)
- \(1.5 \times 10^{-6}\) M \(\text{Pb}^{4+}\)
- \(1.5 \times 10^{-6}\) M \(\text{Mn}^{2+}\)
- 0.13 M \(\text{MnO}_4^-\)
- 0.88 M \(\text{HNO}_3\)
was prepared. For this solution, the balanced reduction half-reactions and overall net reaction shown can occur:
\[
5 \left[\text{Pb}^{4+} + 2e^- \right] \rightleftharpoons \text{Pb}^{2+} \quad E^\circ = 1.690 \text{ V}
\]
\[
2 \left[\text{MnO}_4^- + 8H^+ + 5e^- \right] \rightleftharpoons \text{Mn}^{2+} + 4H_2O \quad E^\circ = 1.507 \text{ V}
\]
\[
\text{Net: } 5\text{Pb}^{4+} + 2\text{Mn}^{2+} + 8H_2O \rightleftharpoons 5\text{Pb}^{2+} + 2\text{MnO}_4^- + 16H^+
\]
**Determine** \(E^\circ_{\text{cell}}, \Delta G^\circ,\) and \(K\) for this reaction.
\[
E^\circ_{\text{cell}} = \boxed{30.9} \text{ V}
\]
*(Note: Answer marked as incorrect.)*
\[
\Delta G^\circ = -176567.55 \text{ J}
\]
In this example, the calculated standard cell potential (\(E^\circ_{\text{cell}}\)) is labeled "Incorrect," suggesting that the result needs further verification. The Gibbs free energy change (\(\Delta G^\circ\)) has been calculated, reflecting the spontaneity of the reaction. Further calculations would be necessary to find the correct \(E^\circ_{\text{cell}}\) and subsequently the equilibrium constant \(K\).
![**Calculate the value for the cell potential, \( E_{\text{cell}} \), and the free energy, \( \Delta G \), for the given conditions.**
\[ E_{\text{cell}} = -0.02 \, \text{V} \]
*Incorrect*
\[ \Delta G = 23272.35 \, \text{J} \]
*Incorrect*
---
**Calculate the value of \( E_{\text{cell}} \) for this system at equilibrium.**
\[ E_{\text{cell}} = 30.9 \, \text{V} \]
*Incorrect*
---
*Note: Always double-check your calculations with standardized formulas and ensure all units are correctly applied. For further assistance, review the theoretical background on cell potential and free energy calculations.*](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc14feaa7-f26c-4630-9d28-afce0e4a19e9%2F48a5d7a7-5b31-4361-89c4-25eb55d42a2a%2F2q74vf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculate the value for the cell potential, \( E_{\text{cell}} \), and the free energy, \( \Delta G \), for the given conditions.**
\[ E_{\text{cell}} = -0.02 \, \text{V} \]
*Incorrect*
\[ \Delta G = 23272.35 \, \text{J} \]
*Incorrect*
---
**Calculate the value of \( E_{\text{cell}} \) for this system at equilibrium.**
\[ E_{\text{cell}} = 30.9 \, \text{V} \]
*Incorrect*
---
*Note: Always double-check your calculations with standardized formulas and ensure all units are correctly applied. For further assistance, review the theoretical background on cell potential and free energy calculations.*
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