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In a certain experiment, a Galvanic cell has a cell emf = 0.54 V at 25 °C. Given that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate [H+], AND PH. Zn (s) | Zn2+ (1.0 M) || H+ (?? M) | H₂ (1.0 atm) | Pt (s) Oxidation: Reduction: (2e)+2H+ H2 E° = E₁₁+/H2 - Ezn/ZR+ 2+ E° = 0 (-0.76) Zn Zn2+ +(2e- n = 2 E = E°- 0.0257 V In Q n 0.54 = 0.76 0.0257 [Zn2+]PH2 In 2 [H+]2 E° = +0.76 V 0.0257 (1.0)(1.0) -0.22 = 1 In 2 (x)² 17.1 = In (1.0) (x)² e17.1 = (1.0) (x)² x = [H+] = 1.935 x 10-4 M NOTE: use "log" for pH calcs! pH = -log(1.935 x 10 4 M) pH = 3.713

In a certain experiment, a Galvanic cell has a cell emf = 0.54 V at 25 °C. Given that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate [H+], AND PH. Zn (s) | Zn2+ (1.0 M) || H+ (?? M) | H₂ (1.0 atm) | Pt (s) Oxidation: Reduction: (2e)+2H+ H2 E° = E₁₁+/H2 - Ezn/ZR+ 2+ E° = 0 (-0.76) Zn Zn2+ +(2e- n = 2 E = E°- 0.0257 V In Q n 0.54 = 0.76 0.0257 [Zn2+]PH2 In 2 [H+]2 E° = +0.76 V 0.0257 (1.0)(1.0) -0.22 = 1 In 2 (x)² 17.1 = In (1.0) (x)² e17.1 = (1.0) (x)² x = [H+] = 1.935 x 10-4 M NOTE: use "log" for pH calcs! pH = -log(1.935 x 10 4 M) pH = 3.713

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
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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In a certain experiment, a Galvanic cell has a cell emf = 0.54 V at 25 °C. Given that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate [H+], AND PH. Zn (s) | Zn2+ (1.0 M) || H+ (?? M) | H₂ (1.0 atm) | Pt (s) Oxidation: Reduction: (2e)+2H+ H2 E° = E₁₁+/H2 - Ezn/ZR+ 2+ E° = 0 (-0.76) Zn Zn2+ +(2e- n = 2 E = E°- 0.0257 V In Q n 0.54 = 0.76 0.0257 [Zn2+]PH2 In 2 [H+]2 E° = +0.76 V 0.0257 (1.0)(1.0) -0.22 = 1 In 2 (x)² 17.1 = In (1.0) (x)² e17.1 = (1.0) (x)² x = [H+] = 1.935 x 10-4 M NOTE: use "log" for pH calcs! pH = -log(1.935 x 10 4 M) pH = 3.713
Transcribed Image Text:In a certain experiment, a Galvanic cell has a cell emf = 0.54 V at 25 °C. Given that [Zn2+] = 1.0 M and PH2 = 1.0 atm. Calculate [H+], AND PH. Zn (s) | Zn2+ (1.0 M) || H+ (?? M) | H₂ (1.0 atm) | Pt (s) Oxidation: Reduction: (2e)+2H+ H2 E° = E₁₁+/H2 - Ezn/ZR+ 2+ E° = 0 (-0.76) Zn Zn2+ +(2e- n = 2 E = E°- 0.0257 V In Q n 0.54 = 0.76 0.0257 [Zn2+]PH2 In 2 [H+]2 E° = +0.76 V 0.0257 (1.0)(1.0) -0.22 = 1 In 2 (x)² 17.1 = In (1.0) (x)² e17.1 = (1.0) (x)² x = [H+] = 1.935 x 10-4 M NOTE: use "log" for pH calcs! pH = -log(1.935 x 10 4 M) pH = 3.713
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