Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
Question
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Chapter 19, Problem 19.57QP

(a)

Interpretation Introduction

To determine: The standard cell potential (Eο) for the given overall reaction.

Interpretation: The standard cell potential (Eο) for the given overall reaction is to be calculated.

Concept introduction: The potential in an electrochemical cell is defined as a measure of energy per unit charge that is available from the oxidation and reductions reactions to carry out the reaction. The values standard electrode potential is obtained with reference to a standard hydrogen electrode.

(a)

Expert Solution
Check Mark

Answer to Problem 19.57QP

The value of standard cell potential (Eο) for the overall reaction is 0.27V_.

Explanation of Solution

The given reaction is,

NH4+(aq)+2O2(g)NO3(aq)+2H+(aq)+H2O(l)

The reduction half cell reaction for the given overall reaction is,

2O2(g)+8H+(aq)+8e4H2O(l)Ecathodeο=1.23V (1)

The oxidation half cell reaction for the given overall reaction is,

NH4+(aq)+3H2O(l)NO3(aq)+10H+(aq)+8eEanodeο=1.50V (2)

The standard cell potential (Eο) for the given overall reaction is calculated by the formula,

Ecellο=EcathodeοEanodeο

Substitute the values of Ecathodeο and Eanodeο from equations (1) and (2) in the above formula to calculate the value of standard cell potential (Eο) for the overall reaction.

Ecellο=1.23V1.50V=0.27V_

Hence, the value of standard cell potential (Eο) for the overall reaction is 0.27V_.

(b)

Interpretation Introduction

To determine: The ratio of [NO3] to [NH4+] when the reaction is at equilibrium at 298K.

Interpretation: The ratio of [NO3] to [NH4+] when the reaction is at equilibrium at 298K is to be calculated.

Concept introduction: Nernst equation is defined as an equation that shows the relationship between the standard electrode potential of an electrochemical cell to the reduction potential, temperature and the concentrations of the chemical substances that undergoes oxidation and reduction.

The Nernst equation in electrochemistry is,

E=EοRTnFln[Red][Ox]

(b)

Expert Solution
Check Mark

Answer to Problem 19.57QP

The ratio of [NO3] to [NH4+] is 1.27×1024:1_.

Explanation of Solution

The pH of the reaction is 7.00.

The partial pressure of oxygen gas is 0.21atm.

The temperature of the reaction is 298K.

The relation between Ecellο and equilibrium constant is,

nFEcellο=RTlnK

Where,

  • n is the number of electrons involved in a chemical reaction.
  • F is the faradays constant (96485Cmol1).
  • R is the universal gas constant.
  • T is the temperature.
  • Ecellο is the standard cell potential.
  • K is the equilibrium constant.

The above equation is rearranged to calculate the value of equilibrium constant at 298K as,

logK=n×Ecellο0.0591

Substitute the values n and Ecellο in the above equation to calculate the value of equilibrium constant.

logK=8×0.27V0.0591logK=2.160.0591logK=36.54K=1036.54

Simplify the above equation,

K=2.88×1037

The concentration of H+ ions is calculated by the formula,

pH=log[H+]

Substitute the value of pH in the above formula to calculate the concentration of H+ ions.

7.00=log[H+][H+]=107.00=1×107M

The ratio of [NO3] to [NH4+] is calculated by the formula,

K=[NO3][H+]2[NH4+][PO2]2

Substitute the values of K, concentration of H+ ions and partial pressure of oxygen gas in the above formula to calculate the ratio of [NO3] to [NH4+].

2.88×1037=[NO3](1×107M)2[NH4+](0.21atm)2[NO3][NH4+]=2.88×1037×(0.21atm)2(1×107M)2[NO3][NH4+]=1.270×10381×1014[NO3][NH4+]=1.27×1024:1_

Hence, the ratio of [NO3] to [NH4+] is 1.27×1024:1_.

Conclusion
  1. a. The value of standard cell potential (Eο) for the overall reaction is 0.27V_.
  2. b. The ratio of [NO3] to [NH4+] is 1.27×1024:1_.

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Chapter 19 Solutions

Chemistry

Ch. 19 - Prob. 19.3VPCh. 19 - Prob. 19.4VPCh. 19 - Prob. 19.5VPCh. 19 - Prob. 19.6VPCh. 19 - Prob. 19.7VPCh. 19 - Prob. 19.8VPCh. 19 - Prob. 19.9VPCh. 19 - Prob. 19.10VPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - Prob. 19.14QPCh. 19 - Prob. 19.15QPCh. 19 - Prob. 19.16QPCh. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.32QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - Prob. 19.35QPCh. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - Prob. 19.43QPCh. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - Prob. 19.51QPCh. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - Prob. 19.55QPCh. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - Prob. 19.66QPCh. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.85QPCh. 19 - Prob. 19.86QPCh. 19 - Prob. 19.87QPCh. 19 - Prob. 19.88QPCh. 19 - Prob. 19.89APCh. 19 - Prob. 19.90APCh. 19 - Prob. 19.91APCh. 19 - Prob. 19.92AP
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