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The oxidation of NH4* to NO3¯ in acid solution (pH = 5.600) is described by the following equation: NH(ag)+20,(g) –→ NO,(ag)+2H*(aq) + H,0(1) The overall reaction for the oxidation has an E°cell = -0.2710 V. E°= 1.50 V for the half-reaction NO, (ag) + 10H* (aq) + 8e¯ → NH(aq) + 3H,0(1) i See Periodic Table Be sure to solve the equation using a base-10 logarithm (log) rather than the natural logarithm (In). What is the ratio of [NO3¯] to [NH4*] at 298 K if Po2 = 0.180 atm? Assume that the reaction is at equilibrium.

The oxidation of NH4* to NO3¯ in acid solution (pH = 5.600) is described by the following equation: NH(ag)+20,(g) –→ NO,(ag)+2H*(aq) + H,0(1) The overall reaction for the oxidation has an E°cell = -0.2710 V. E°= 1.50 V for the half-reaction NO, (ag) + 10H* (aq) + 8e¯ → NH(aq) + 3H,0(1) i See Periodic Table Be sure to solve the equation using a base-10 logarithm (log) rather than the natural logarithm (In). What is the ratio of [NO3¯] to [NH4*] at 298 K if Po2 = 0.180 atm? Assume that the reaction is at equilibrium.

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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The oxidation of NH4* to NO3 in acid solution (pH = 5.600) is described by the following equation: %3D NH(ag)+20,(g) → NO, (ag) +2H*(aq)+H,0(1) The overall reaction for the oxidation has an E°cell = -0.2710 V. E° = 1.50 V for the half-reaction NO, (ag) + 10H* (aq) + 8e → NH(aq) +3H,0(1) i See Periodic Table Be sure to solve the equation using a base-10 logarithm (log) rather than the natural logarithm (In). What is the ratio of [NO3] to [NH4*] at 298 K if Po2 = 0.180 atm? Assume that the reaction is at equilibrium.
Transcribed Image Text:The oxidation of NH4* to NO3 in acid solution (pH = 5.600) is described by the following equation: %3D NH(ag)+20,(g) → NO, (ag) +2H*(aq)+H,0(1) The overall reaction for the oxidation has an E°cell = -0.2710 V. E° = 1.50 V for the half-reaction NO, (ag) + 10H* (aq) + 8e → NH(aq) +3H,0(1) i See Periodic Table Be sure to solve the equation using a base-10 logarithm (log) rather than the natural logarithm (In). What is the ratio of [NO3] to [NH4*] at 298 K if Po2 = 0.180 atm? Assume that the reaction is at equilibrium.
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