**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.*

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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

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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