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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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
Half-reaction E° (V) Cr3+ (aq) + 3e- → Cr (s) -0.74 Fe2+ (aq) + 2e- → Fe (s) -0.440 Fe3+ (aq) + e- → Fe2+ (s) +0.771 Sn4+ (aq) + 2e- → Sn2+ (aq) +0.154 The standard cell potential (E°cell) for the voltaic cell based on the reaction below is ________ V.Sn2+ (aq) + 2Fe3+ (aq) → 2Fe2+ (aq) + Sn4+ (aq) a). +1.21 b). +0.617 c). -0.46 d). +1.39 e). +0.46
Consider the standard reduction potentials and redox couples used in a galvanic cell. Unbalanced Half-Reaction E°(V) MnO4 (aq) + 3e MnO2(s) (at basic pH) 0.59 Fe3+(aq) + e- Fe2*(aq) 0.77 What is the balanced reaction taking place in the cell (in basic solution)? 2 H20(1) + Mn04 (aq) + 3 Fe3+(aq) → 3 Fe2*(aq) + MnO2(s) + 4 OH- MnO2(s) + 4 OH¯ + 3 Fe3+(aq) 3 Fe2+(aq) + 2 H20(1) + MnO4¯(aq) MO 3 Fe2*(aq) + 2 H2O(1) + MnO4¯(aq) MnO2(s) + 4 OH¯ + 3 Fe3+(aq) M 3 Fe2+(aq) + MnO2(s) + 4 OH → 2 H20(1) + MnO4 (aq) + 3 Fe3+(aq)
If you said it was possible to calculate the cell voltage, do so and enter your answer here. Round your Ov answer to 2 significant digits.
For the galvanic cell at 298 K 2X(s) + 3Y2*(aq) –→ 2x3*(aq) + 3Y(s) E° cell = 0.39 V What is the measured cell potential (Ecell) when [Y2*] = 1.0 M and [X3+] = 0.01 M? Periodic Table and Datasheet O 0.39 V O 0.35 V O 0.43 V O 0.51 V
Determine the cell potential of an electrochemical cell based on the following two half-reactions: Oxidation: Ni (s) ⟶⟶ Ni2+ (aq, 2.0 M) + 2e- Reduction: VO2+ (aq, 0.010 M) + 2H+ (aq, 1.0 M) + e- →→ VO2+ (aq, 2.0 M) + H2O (l)
Given the following two half-reactions, write the overall reaction in the direction in which it is product- favored, and calculate the standard cell potential. Pb2*(aq) + 2 e- - → Pb(s) E° = -0.126 V Fe3*(aq) + e¯ → Fe2*(s) E° = +0.771 V Pb2*(aq) + 2 Fe²*(s) → Pb(s) + 2 Fe³*(aq); = +0.897 V cell Pb2*(aq) + Fe²*(s) → Pb(s) + Fe³*(aq); E, cell = +0.645 V Pb(s) + 2 Fe³*(aq) → Pb²*(aq) + 2 Fe²*(s); F° cell = +1.416 V Pb(s) + 2 Fe*(aq) → Pb²*(aq) + 2 Fe²*(s); E. 'cell = +0.897 V Pb(s) + Fe³*(aq) → Pb²*(aq) + Fe²*(s); = +0.645 V
Which of the following reactions occurs at the anode of an electrochemical cell composed of Zn│Zn2+(aq) and Fe│Fe2+(aq) half-cells? Their reduction potentials are as follows: Zn2+ + 2e– --> Zn; Eo = -0.763 V Fe2+ + 2e– --> Fe; Eo = -0.447 V (A) Zn2+ + 2e– --> Cr; (B) Zn --> Zn2+ + 2e–; (C) Fe --> Fe2+ + 2e–; (D) Fe2+ + 2e– --> Fe;

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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