Chapter 18, Problem 18.39QP

The hydrogen-oxygen fuel cell is described in Section 18.6. (a) What volume of H₂(g), stored at 25°C at a pressure of 155 atm, would be needed to run an electric motor drawing a current of 8.5 A for 3.0 h? (b) What volume (liters) of air at 25°C and 1.00 atm will have to pass into the cell per minute to run the motor? Assume that air is 20 percent O₂ by volume and that all the O₂ is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior.

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.

$\begin{array}{l}\text{Anode (Oxidation)}{\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+4OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{4H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}\\ \text{Cathode (Reduction)}{\text{O}}_{\text{2}}{}_{\text{(g)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}\stackrel{}{\to }{\text{4OH}}^{\text{-}}{}_{\text{(aq)}}\\ \text{Overall}\text{Reaction}{\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\end{array}$

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.

$\text{charges(in}\text{coulombs)=current(in}\text{amperes)×time(s)}$

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

$\text{Number}\text{of}\text{moles}\text{of}\text{electron=charges/Faraday constant (96500 C/mole of electron)}$

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.

$\text{Anode (Oxidation)}{\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+4OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{4H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}$

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

$numberofmolesof\mathrm{hydrogen}=moleof{e}^{-}\times \frac{2mo\text{le}ofHydrogen}{4moleof{e}^{-}}$

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.

$\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

P = Pressure of the gas

V = Volume of the gas

R = Universal gas constant $\text{(8}{\text{.31JK}}^{\text{-1}}{\text{mol}}^{-1}\text{ or 0}{\text{.08206 LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{) }$

T = Temperature in kelvin

n = Number of moles of the gas

Answer to Problem 18.39QP

For the anode reaction  $\text{(Oxidation)}{\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+4OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{4H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}$

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

$\text{charges(in}\text{coulombs)=current(in}\text{amperes)×time(s)}$

Current = 8.5A

Time = 3 h or 10800s

So

$\begin{array}{l}\text{charges(in}\text{coulombs)=8}\text{.5A×10800s}\\ \text{=9}\text{.18}\times {\text{10}}^{\text{4}}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{electron=}\frac{\text{9}\text{.18}\times {\text{10}}^{\text{4}}\text{C}}{{\text{96500 C/mole of e}}^{\text{-}}}\\ =\text{0}\text{.9513}\text{mole}\end{array}$

$\begin{array}{l}\text{Number of}molesof\mathrm{hydrogen}=0.9513\times \frac{2\text{mole}ofHydrogen}{4moleof{e}^{-}}\\ =0.4757\text{mole}\end{array}$

Volume of hydrogen can calculated from ideal gas equation

$\begin{array}{l}\text{PV=nRT}\\ \text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{\text{(0}\text{.4757}\text{mol)(}\text{0}\text{.08206L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)(298K)}}{\text{155atm}}\\ \text{=0}\text{.075L}\end{array}$

Explanation of Solution

For the anode reaction  $\text{(Oxidation)}{\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+4OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{4H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}$

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

$\text{charges(in}\text{coulombs)=current(in}\text{amperes)×time(s)}$

Current = 8.5A

Time = 3 h or 10800s

So

$\begin{array}{l}\text{charges(in}\text{coulombs)=8}\text{.5A×10800s}\\ \text{=9}\text{.18}\times {\text{10}}^{\text{4}}\end{array}$

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

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{electron=}\frac{\text{9}\text{.18}\times {\text{10}}^{\text{4}}\text{C}}{{\text{96500 C/mole of e}}^{\text{-}}}\\ =\text{0}\text{.9513}\text{mole}\end{array}$

$\begin{array}{l}\text{Number of}molesof\mathrm{hydrogen}=0.9513\times \frac{2\text{mole}ofHydrogen}{4moleof{e}^{-}}\\ =0.4757\text{mole}\end{array}$

Volume of hydrogen can calculated from ideal gas equation

$\begin{array}{l}\text{PV=nRT}\\ \text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{\text{(0}\text{.4757}\text{mol)(}\text{0}\text{.08206L}\text{atm}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{)(298K)}}{\text{155atm}}\\ \text{=0}\text{.075L}\end{array}$

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

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.

$\begin{array}{l}\text{Anode (Oxidation)}{\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+4OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{4H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}\\ \text{Cathode (Reduction)}{\text{O}}_{\text{2}}{}_{\text{(g)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}\stackrel{}{\to }{\text{4OH}}^{\text{-}}{}_{\text{(aq)}}\\ \text{Overall}\text{Reaction}{\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2H}}_{\text{2}}{\text{O}}_{\text{(l)}}\end{array}$

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.

$\text{charges(in}\text{coulombs)=current(in}\text{amperes)×time(s)}$

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

$\text{Number}\text{of}\text{moles}\text{of}\text{electron=charges/Faraday constant (96500 C/mole of electron)}$

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.

$\text{Anode (Oxidation)}{\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+4OH}}^{\text{-}}{}_{\text{(aq)}}\stackrel{}{\to }{\text{4H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}$

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

$numberofmolesof\mathrm{hydrogen}=moleof{e}^{-}\times \frac{2mo\text{le}ofHydrogen}{4moleof{e}^{-}}$

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.

$\begin{array}{l}PV=nRT\\ V=\frac{nRT}{P}\end{array}$

P = Pressure of the gas

V = Volume of the gas

R = Universal gas constant $\text{(8}{\text{.31JK}}^{\text{-1}}{\text{mol}}^{-1}\text{ or 0}{\text{.08206 LatmK}}^{\text{-1}}{\text{mol}}^{\text{-1}}\text{) }$

T = Temperature in kelvin

n = Number of moles of the gas

Answer to Problem 18.39QP

For the cathode half reaction $\text{(Reduction)}{\text{O}}_{\text{2}}{}_{\text{(g)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}\stackrel{}{\to }{\text{4OH}}^{\text{-}}{}_{\text{(aq)}}$

Charges flowing through the circuit for 1 minute

$\begin{array}{l}\text{charges(in}\text{coulombs)=8}\text{.5A×60s}\\ \text{=510C/min}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{electron=}\frac{\text{510C}}{{\text{96500 C/mole of e}}^{\text{-}}}\\ =5.285\times {10}^{-3}\text{mole}\end{array}$

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

$\begin{array}{l}\text{Number}ofmolesof\mathrm{hydrogen}=5.285\times {10}^{-3}\times \frac{1\text{mole}of\text{Oxygen}}{4moleof{e}^{-}}\\ =1.3212\times {10}^{-3}\text{mole}\end{array}$

Volume of oxygen can be calculated by using ideal gas equation

$\begin{array}{l}\text{PV=nRT}\\ \text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{\text{(1}{\text{.3212×10}}^{\text{-3}}\text{mol)(}\text{0}\text{.08206L-atm/K-mol)(298K)}}{\text{1atm}}\\ \text{=0}{\text{.032L O}}_{\text{2}}\text{/min}\end{array}$

Then volume of air flown can be calculated as follows

$\begin{array}{l}=0.032{\text{L O}}_{\text{2}}/\min \times \frac{1.0\text{L}\text{air}}{0.20{\text{LO}}_{\text{2}}}\\ =0.16\text{L of air/min}\end{array}$

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 $\text{(Reduction)}{\text{O}}_{\text{2}}{}_{\text{(g)}}{\text{+ 2H}}_{\text{2}}{\text{O}}_{\text{(l)}}{\text{+4e}}^{\text{-}}\stackrel{}{\to }{\text{4OH}}^{\text{-}}{}_{\text{(aq)}}$

 charges flowing through the circuit for 1 minute

 $\begin{array}{l}\text{charges(in}\text{coulombs)=8}\text{.5A×60s}\\ \text{=510C/min}\end{array}$

$\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}\text{electron=}\frac{\text{510C}}{{\text{96500 C/mole of e}}^{\text{-}}}\\ =5.285\times {10}^{-3}\text{mole}\end{array}$

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

$\begin{array}{l}\text{Number}ofmolesof\mathrm{hydrogen}=5.285\times {10}^{-3}\times \frac{1\text{mole}of\text{Oxygen}}{4moleof{e}^{-}}\\ =1.3212\times {10}^{-3}\text{mole}\end{array}$

Volume of oxygen can be calculated by using ideal gas equation

$\begin{array}{l}\text{PV=nRT}\\ \text{V}\text{=}\frac{\text{nRT}}{\text{P}}\\ \text{=}\frac{\text{(1}{\text{.3212×10}}^{\text{-3}}\text{mol)(}\text{0}\text{.08206L-atm/K-mol)(298K)}}{\text{1atm}}\\ \text{=0}{\text{.032L O}}_{\text{2}}\text{/min}\end{array}$

Then volume of air flown can be calculated as follows

$\begin{array}{l}=0.032{\text{L O}}_{\text{2}}/\min \times \frac{1.0\text{L}\text{air}}{0.20{\text{LO}}_{\text{2}}}\\ =0.16\text{L of air/min}\end{array}$

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.