General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 19, Problem 19.37QP

(a)

Interpretation Introduction

Interpretation:

Calculate the volume of hydrogen with pressure 155 atm that run an electric motor for 3 h and volume of air with 20%oxygen needed to run the electric motor per minute.

Concept introduction:

Hydrogen-Oxygen fuel cell works on the principle of oxidation of hydrogen and reduction of oxygen, it was made up of potassium hydroxide as an electrolyte solution and two inert electrodes. Hydrogen and oxygen gases were bubbled through the anode and cathode compartments. The cell reaction of hydrogen-oxygen fuel cell was shown below.

  Anode (Oxidation) 2H2(g)+4OH-(aq) 4H2O(l)+4e-Cathode (Reduction) O2(g)+ 2H2O(l)+4e- 4OH-(aq)OverallReaction 2H2(g)+O2(g) 2H2O(l)

Calculation of the volume of hydrogen gas used for generating of electricity involves multistep

1) Calculation of total number of charges that flow through the circuit, since coulomb is the amount of electric charge flowing in a circuit in 1s, when current is 1A. So the above statement can represented by the following equation.

  charges(incoulombs)=current(inamperes)×time(s)

On dividing the number of charges with Faraday constant we can attain the number of moles of electron

Numberofmolesofelectron=charges/Faraday constant (96500 C/mole of electron)

From knowing the number of mole of electrons and using the stoichiometry of the reaction, the number of moles of the substance reduced or oxidized can be determined. This can be explained by the representative reaction as shown below.

  Anode (Oxidation) 2H2(g)+4OH-(aq) 4H2O(l)+4e-

2 mole of hydrogen releases 4 mole of electron, so the number of moles of hydrogen oxidized can calculated by the following equation.

  numberofmolesofhydrogen=moleofe×2moleofHydrogen4moleofe

Finally on substituting the number of moles of the product into the ideal gas equation the volume of the gas needed for the cell reaction can be achieved.

  PV=nRTV=nRTP

P = Pressure of the gas

V = Volume of the gas

R = Universal gas constant (8.31JK-1mol-1 or 0.08206 LatmK-1mol-1)   

T = Temperature in kelvin

n = Number of moles of the gas

(a)

Expert Solution
Check Mark

Answer to Problem 19.37QP

For the anode reaction  (Oxidation) 2H2(g)+4OH-(aq) 4H2O(l)+4e-

Number of charges passing through the circuit can be calculated using the formula

charges(incoulombs)=current(inamperes)×time(s)

Current = 8.5A

Time = 3 h or 10800s

So

  charges(incoulombs)=8.5A×10800s =9.18×104

  Numberofmolesofelectron=9.18×104C 96500 C/mole of e- =0.9513mole

  Number ofmolesofhydrogen=0.9513×2moleofHydrogen4moleofe =0.4757mole

Volume of hydrogen can calculated from ideal gas equation

PV=nRTV =nRTP =(0.4757mol)(0.08206LatmK-1mol-1)(298K)155atm =0.075L

Explanation of Solution

For the anode reaction  (Oxidation) 2H2(g)+4OH-(aq) 4H2O(l)+4e-

Number of charges passing through the circuit can be calculated using the formula

charges(incoulombs)=current(inamperes)×time(s)

Current = 8.5A

Time = 3 h or 10800s

So

  charges(incoulombs)=8.5A×10800s =9.18×104

On dividing the number of charges by faraday constant number of moles of electrons passing the circuit can be calculated as shown below

  Numberofmolesofelectron=9.18×104C 96500 C/mole of e- =0.9513mole

  Number ofmolesofhydrogen=0.9513×2moleofHydrogen4moleofe =0.4757mole

Volume of hydrogen can calculated from ideal gas equation

PV=nRTV =nRTP =(0.4757mol)(0.08206LatmK-1mol-1)(298K)155atm =0.075L

The volume of hydrogen with pressure 155atm, needed to run a motor of 8.5A for 3 hrs was calculated to be 0.075L.

(b)

Interpretation Introduction

Interpretation:

Calculate the volume of hydrogen with pressure 155 atm that run an electric motor for 3 h and volume of air with 20%oxygen needed to run the electric motor per minute.

Concept introduction:

Hydrogen-Oxygen fuel cell works on the principle of oxidation of hydrogen and reduction of oxygen, it was made up of potassium hydroxide as an electrolyte solution and two inert electrodes. Hydrogen and oxygen gases were bubbled through the anode and cathode compartments. The cell reaction of hydrogen-oxygen fuel cell was shown below.

  Anode (Oxidation) 2H2(g)+4OH-(aq) 4H2O(l)+4e-Cathode (Reduction) O2(g)+ 2H2O(l)+4e- 4OH-(aq)OverallReaction 2H2(g)+O2(g) 2H2O(l)

Calculation of the volume of hydrogen gas used for generating of electricity involves multistep

1) Calculation of total number of charges that flow through the circuit, since coulomb is the amount of electric charge flowing in a circuit in 1s, when current is 1A. So the above statement can represented by the following equation.

  charges(incoulombs)=current(inamperes)×time(s)

On dividing the number of charges with Faraday constant we can attain the number of moles of electron

Numberofmolesofelectron=charges/Faraday constant (96500 C/mole of electron)

From knowing the number of mole of electrons and using the stoichiometry of the reaction, the number of moles of the substance reduced or oxidized can be determined. This can be explained by the representative reaction as shown below.

  Anode (Oxidation) 2H2(g)+4OH-(aq) 4H2O(l)+4e-

2 mole of hydrogen releases 4 mole of electron, so the number of moles of hydrogen oxidized can calculated by the following equation.

  numberofmolesofhydrogen=moleofe×2moleofHydrogen4moleofe

Finally on substituting the number of moles of the product into the ideal gas equation the volume of the gas needed for the cell reaction can be achieved.

  PV=nRTV=nRTP

P = Pressure of the gas

V = Volume of the gas

R = Universal gas constant (8.31JK-1mol-1 or 0.08206 LatmK-1mol-1)   

T = Temperature in kelvin

n = Number of moles of the gas

(b)

Expert Solution
Check Mark

Answer to Problem 19.37QP

For the cathode half reaction (Reduction) O2(g)+ 2H2O(l)+4e-4OH-(aq)

Charges flowing through the circuit for 1 minute

charges(incoulombs)=8.5A×60s =510C/min

Numberofmolesofelectron=510C 96500 C/mole of e- =5.285×103mole

Thus obtained number of moles of electron can be used to determine the number of moles of hydrogen

  Numberofmolesofhydrogen=5.285×103×1moleofOxygen4moleofe =1.3212×103mole

Volume of oxygen can be calculated by using ideal gas equation

  PV=nRTV =nRTP =(1.3212×10-3mol)(0.08206L-atm/K-mol)(298K)1atm =0.032L O2/min

Then volume of air flown can be calculated as follows

  =0.032L O2/min×1.0Lair0.20LO2=0.16L of air/min

Explanation of Solution

The volume of air flowing through the fuel cell can calculated in a step by step manner

For the cathode half reaction (Reduction) O2(g)+ 2H2O(l)+4e-4OH-(aq)

 charges flowing through the circuit for 1 minute

charges(incoulombs)=8.5A×60s =510C/min

Numberofmolesofelectron=510C 96500 C/mole of e- =5.285×103mole

Thus obtained number of moles of electron can be used to determine the number of moles of hydrogen

  Numberofmolesofhydrogen=5.285×103×1moleofOxygen4moleofe =1.3212×103mole

Volume of oxygen can be calculated by using ideal gas equation

  PV=nRTV =nRTP =(1.3212×10-3mol)(0.08206L-atm/K-mol)(298K)1atm =0.032L O2/min

Then volume of air flown can be calculated as follows

=0.032L O2/min×1.0Lair0.20LO2=0.16L of air/min

The volume of air with 20% oxygen and pressure 1atm, needed to run a motor of 8.5A for 1 hr was determined as 0.16L.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 19 Solutions

General Chemistry

Ch. 19.5 - Prob. 2PECh. 19.5 - Prob. 1RCCh. 19.7 - Prob. 1RCCh. 19.8 - Prob. 1RCCh. 19.8 - An aqueous solution of Mg(NO3)2 is electrolyzed....Ch. 19.8 - Prob. 2RCCh. 19.8 - Prob. 2PECh. 19.8 - Prob. 3RCCh. 19 - Prob. 19.1QPCh. 19 - Prob. 19.2QPCh. 19 - Prob. 19.3QPCh. 19 - Prob. 19.4QPCh. 19 - Prob. 19.5QPCh. 19 - Prob. 19.6QPCh. 19 - 19.7 What is the difference between the...Ch. 19 - Prob. 19.8QPCh. 19 - Prob. 19.9QPCh. 19 - Prob. 19.10QPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - 19.14 Which of the following reagents can oxidize...Ch. 19 - 19.15 Consider the following half-reactions: (aq)...Ch. 19 - 19.16 Predict whether the following reactions...Ch. 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. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - 19.35 Explain the differences between a primary...Ch. 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 - 19.43 What is the difference between a galvanic...Ch. 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 - 19.51 Calculate the amounts of Cu and Br2 produced...Ch. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - 19.55 What is the hourly production rate of...Ch. 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 - 19.66 A sample of iron ore weighing 0.2792 g was...Ch. 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. 19.86QPCh. 19 - Prob. 19.87QPCh. 19 - Prob. 19.88QPCh. 19 - Prob. 19.89QPCh. 19 - Prob. 19.90QPCh. 19 - Prob. 19.91QPCh. 19 - Prob. 19.92QPCh. 19 - Prob. 19.93QPCh. 19 - Prob. 19.94QPCh. 19 - Prob. 19.95QPCh. 19 - Prob. 19.96QPCh. 19 - Prob. 19.97QPCh. 19 - Prob. 19.98QPCh. 19 - Prob. 19.99QPCh. 19 - Prob. 19.100QPCh. 19 - Prob. 19.101QPCh. 19 - 19.102 The magnitudes (but not the signs) of the...Ch. 19 - Prob. 19.103QPCh. 19 - Prob. 19.104QPCh. 19 - Prob. 19.105QPCh. 19 - Prob. 19.106QPCh. 19 - Prob. 19.107QPCh. 19 - Prob. 19.108QPCh. 19 - Prob. 19.109QPCh. 19 - Prob. 19.110QPCh. 19 - 19.111 A spoon was silver-plated electro lyrically...Ch. 19 - Prob. 19.112QPCh. 19 - Prob. 19.113QPCh. 19 - Prob. 19.114QPCh. 19 - Prob. 19.115QPCh. 19 - Prob. 19.116QPCh. 19 - Prob. 19.117QPCh. 19 - Prob. 19.118QPCh. 19 - Prob. 19.119QPCh. 19 - Prob. 19.120QPCh. 19 - Prob. 19.121SPCh. 19 - Prob. 19.122SPCh. 19 - Prob. 19.123SPCh. 19 - Prob. 19.124SPCh. 19 - Prob. 19.125SPCh. 19 - Prob. 19.126SPCh. 19 - Prob. 19.128SPCh. 19 - Prob. 19.129SPCh. 19 - Prob. 19.130SP
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Electrolysis; Author: Tyler DeWitt;https://www.youtube.com/watch?v=dRtSjJCKkIo;License: Standard YouTube License, CC-BY