Given the half-cell reduction potentials, the overall cell reaction, and cell potential below:Fe3+(aq) +e + Fe2*(aq)E°= 0.77 VVO2*(aq) + 2H*(aq) + e- → Vo2*(aq) + H2O(I)E =1.00 VWhat is the equilibrium constant for this reaction at 298K?9.1x1027.7x1031.3x10-48.7x1029

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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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Consider the following reduction half-reactions: MnO4 (aq) + 8 H*(aq) + 5 e¯ = Mn²+ (aq) + 4H₂O(1) E° = +1.51 V Al³+ (aq) + 3 e 2Al(s) E° = -1.68 V Cl2(g) + 2 e 2 Cl¯(aq) E° = +1.36 V Mg²+ (aq) + 2 e Mg(s) E° = -2.36 V Calculate the standard cell potential for the most spontaneous reaction that can be produced when two of the given half-reactions are combined to form an electrochemical cell. -0.85 V +3.87 V -4.04 V O +2.87 V
Calculate the cell potential given the standard reduction potentials for the half reactions of the following hypothetical elements: X3+(aq) + e- → X2+(aq) SRP: +0.57 V Y2(ℓ) + 2e- → 2Y-(aq) SRP: +0.41 V
Consider the following reduction half-reactions: MnO4 +8 H*( +5e Mn²+ E° = +1.51 V (aq) (aq) + *(aq) + 4 H₂0 (1) Al³+ (aq) + 3 e Al(s) E° = -1.68 V Cl2(g) + 2 e 2 Cl¯ (aq) E° = +1.36 V Mg2+ (aq) + 2 e Mg(s) E° = -2.36 V Which of the following substances is the strongest oxidizing agent? O Mg²+ (aq) MnO4 (aq) O Mg(s) O Mn²+ *(aq)
Using the following standard reduction potentials, 34 Fe *(aq) + e - Fe2*(aq) Ni2*(ag) + 2 e -Ni(s) E° = +0.77 V E° = -0.26 V calculate the standard cell potential for the cell reaction given below, and determine whether or not this reaction is spontaneous under standard conditions 34, Ni 2*( aq) + 2 Fe 2*( aq) → 2 Fe *( aq) + Ni( s) E° = +1.03, nonspontaneous E° = + 1.03, spontaneous E° = -1.03 V, spontaneous E° = -1.03 V, nonspontaneous
When [Hg2+] = 1.16 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 2.583 V. What is the Al3+ concentration in this cell? 3H92*(aq) + 2Al(s) – 3Hg(8) + 2AI3*(aq) [A|3*] = mol/L 6.724x10^-4 Show Approach Hide Tutor Steps Resubmit TUTOR STEP The first step is to separate the overall reaction into oxidation and reduction half-reactions. These are shown here. Hg2*(aq) + 2e" – Hg(e) reduction Al(s) → A3*(aq) + 3e oxidation Next (1 of 5)
When [Pb²+] = 1.10 M, the observed cell potential at 298 K for an electrochemical cell with the reaction shown below is 0.683 V. What is the Cr³+ concentration in this cell? 3Pb²+ (aq) + 2Cr(s) [Cr³+] = → 3Pb(s) + 2Cr³+ (aq) mol/L
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Given the half-cell reduction potentials, the overall cell reaction, and cell potential below: Fe3+(aq) +e + Fe2*(aq) E°= 0.77 V VO2*(aq) + 2H*(aq) + e- → Vo2*(aq) + H2O(I) E =1.00 V What is the equilibrium constant for this reaction at 298K? 9.1x102 7.7x103 1.3x10-4 8.7x1029

Given the half-cell reduction potentials, the overall cell reaction, and cell potential below: Fe3+(aq) +e + Fe2(aq) E°= 0.77 V VO2(aq) + 2H(aq) + e- → Vo2(aq) + H2O(I) E =1.00 V What is the equilibrium constant for this reaction at 298K? 9.1x102 7.7x103 1.3x10-4 8.7x1029