Kindly provide me with the solution for the following question: 82. Given the following half-cell reactions:2H₂O() +2e → H2(g) + 2OH(aq) (10-7 mol/L)= -0.41 VK* (29) + e- → K(s)= -2.92 V2H₂O (1)= -0.815 VE = -0.54 V25 (29) → 12(3)+2e-What is the minimum voltage necessary for the electrolysis of 1.0 mol/L solution of potassium iodide usingplatinum electrodes?a.b.C.→O2(g) + 4H* (ag) (10-7 mol/L) + 4e¯0.13 V0.99 V1.23 VEd.e.3.46 V3.74 V

When an aqueous solution of potassium iodide (KI) is electrolyzed on a platinum electrode, then…

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

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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