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32. Given the following half-cell reactions: 2H₂O (1) + 2e → H₂(g) + 2OH(aq) (10¹ mol/L) E° = -0.41 V E = -2.92 V j+e² → K(s) K* (29) E a. 0.13 V b. 0.99 V C. 1.23 V 2H₂O(1)→ O2(g) + 4H* ► O2(g) + 4H* (aq) (10~¹ mol/L) + 4e¯ 25 (29) → 12(3)+2e= What is the minimum voltage necessary for the electrolysis of 1.0 mol/L solution of potassium iodide using platinum electrodes? E = -0.815 V d. e. = -0.54 V 3.46 V 3.74 V

32. Given the following half-cell reactions: 2H₂O (1) + 2e → H₂(g) + 2OH(aq) (10¹ mol/L) E° = -0.41 V E = -2.92 V j+e² → K(s) K* (29) E a. 0.13 V b. 0.99 V C. 1.23 V 2H₂O(1)→ O2(g) + 4H* ► O2(g) + 4H* (aq) (10~¹ mol/L) + 4e¯ 25 (29) → 12(3)+2e= What is the minimum voltage necessary for the electrolysis of 1.0 mol/L solution of potassium iodide using platinum electrodes? E = -0.815 V d. e. = -0.54 V 3.46 V 3.74 V

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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82. Given the following half-cell reactions: 2H₂O() +2e → H2(g) + 2OH(aq) (10-7 mol/L) = -0.41 V K* (29) + e- → K(s) = -2.92 V 2H₂O (1) = -0.815 V E = -0.54 V 25 (29) → 12(3)+2e- What is the minimum voltage necessary for the electrolysis of 1.0 mol/L solution of potassium iodide using platinum electrodes? a. b. C. →O2(g) + 4H* (ag) (10-7 mol/L) + 4e¯ 0.13 V 0.99 V 1.23 V E d. e. 3.46 V 3.74 V
Transcribed Image Text:82. Given the following half-cell reactions: 2H₂O() +2e → H2(g) + 2OH(aq) (10-7 mol/L) = -0.41 V K* (29) + e- → K(s) = -2.92 V 2H₂O (1) = -0.815 V E = -0.54 V 25 (29) → 12(3)+2e- What is the minimum voltage necessary for the electrolysis of 1.0 mol/L solution of potassium iodide using platinum electrodes? a. b. C. →O2(g) + 4H* (ag) (10-7 mol/L) + 4e¯ 0.13 V 0.99 V 1.23 V E d. e. 3.46 V 3.74 V
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