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Pt (s) | H2 (g, ƒ= 1) | HCI (aq, m = 0.1) | AgCl (s) | Ag (s) changes as follows with the temperature: Epila (V) = 0.35510 – 0.3422x104t - 3.2347x10-8² + 6.314x10-f³ (t being the temperature in °C) Write the reaction and calculate AG, AH and AS at 90 °C. Result: -34388 J/mol, 10.94 J/Kmol, -30.42 KJ/mol

Pt (s) | H2 (g, ƒ= 1) | HCI (aq, m = 0.1) | AgCl (s) | Ag (s) changes as follows with the temperature: Epila (V) = 0.35510 – 0.3422x104t - 3.2347x10-8² + 6.314x10-f³ (t being the temperature in °C) Write the reaction and calculate AG, AH and AS at 90 °C. Result: -34388 J/mol, 10.94 J/Kmol, -30.42 KJ/mol

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 the temperature range 0-90 ºC , the potential difference of the following cell:
Pt (s) | H2 (g, f = 1) | HCl (aq, m = 0.1) | AgCl (s) | Ag (s)
changes as follows with the temperature:
Ecell (V) = 0.35510 – 0.3422x10-4t - 3.2347x10-8t2 + 6.314x10-9t3 (t being the temperature in ºC) Write the reaction and calculate ΔG, ΔH and ΔS at 90 ºC.
Result: -34388 J/mol, 10.94 J/Kmol, -30.42 KJ/mol 

Pt (s) | H2 (g, f= 1)| HCl (aq, m = 0.1)| AgCl (s) | Ag (s) changes as follows with the temperature: Epila (V) = 0.35510 – 0.3422x104t - 3.2347x10-82 + 6.314x10-f³ (t being the temperature in °C) Write the reaction and calculate AG, AH and AS at 90 °C. Result: -34388 J/mol, 10.94 J/Kmol, -30.42 KJ/mol
Transcribed Image Text:Pt (s) | H2 (g, f= 1)| HCl (aq, m = 0.1)| AgCl (s) | Ag (s) changes as follows with the temperature: Epila (V) = 0.35510 – 0.3422x104t - 3.2347x10-82 + 6.314x10-f³ (t being the temperature in °C) Write the reaction and calculate AG, AH and AS at 90 °C. Result: -34388 J/mol, 10.94 J/Kmol, -30.42 KJ/mol
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