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Chapter 19: Problem 37

The hydrogen-oxygen fuel cell is described in Section 19.6 . (a) What volume of \(\mathrm{H}_{2}(g)\), stored at \(25^{\circ} \mathrm{C}\) at a pressure of 155 atm, would be needed to run an electric motor drawing a current of \(8.5 \mathrm{~A}\) for \(3.0 \mathrm{~h} ?\) (b) What volume (liters) of air at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) will have to pass into the cell per minute to run the motor? Assume that air is 20 percent \(\mathrm{O}_{2}\) by volume and that all the \(\mathrm{O}_{2}\) is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior.

Short Answer

Expert verified
The volume of \(H_{2}\) needed to run the motor is __ liters and the volume of air required per minute is __ liters.

Step by step solution

01

Calculate the number of moles of \(H_{2}\)

To find the volume of \(H_{2}\) needed, we need to first know the number of moles of \(H_{2}\) required. This will be calculated using the relationship between electric current and the amount of substance. According to Faraday’s laws of electrolysis, 1 mole of a substance is equivalent to a 1 Faraday charge (96,485 coulombs). Since we are given the current and time, we can find the charge (Q) using the formula \(Q = It\) where I is the current in amperes and t is the time in seconds (\(3.0h = 10800s)\). Then by dividing the calculated charge by one Faraday, we find the moles of \(H_{2}\).
02

Calculate the volume of \(H_{2}\)

With the number of moles of \(H_{2}\) obtained from Step 1, we can use the ideal gas law to find the volume. The ideal gas law is \(PV=nRT\). In this case, we know \(n\), \(P\), \(T\), and \(R\) (the ideal gas constant). Solving for \(V\), we find the volume.
03

Calculate the number of moles of \(O_2\)

From the stoichiometry of the redox reaction of hydrogen and oxygen in a fuel cell, it is seen that one mole of \(O_2\) will need two moles of \(H_2\). So the number of moles of \(O_2\) needed can be calculated as half the number of moles of \(H_{2}\) calculated in Step 1.
04

Calculate the volume of air

Assuming air is 20 percent \(O_2\) by volume, the volume of air needed will be five times the volume of \(O_2\) needed. To find the volume of \(O_2\) needed, we again use the ideal gas law on the number of moles of \(O_2\) computed in Step 3. Finally, we divide by the time given to find the volume of air in liters per minute that will have to pass into the cell.

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Most popular questions from this chapter

A galvanic cell is constructed as follows. One halfcell consists of a platinum wire immersed in a solution containing \(1.0 M \mathrm{Sn}^{2+}\) and \(1.0 M \mathrm{Sn}^{4+} ;\) the other half-cell has a thallium rod immersed in a solution of \(1.0 M \mathrm{TI}^{+}\). (a) Write the half-cell reactions and the overall reaction. (b) What is the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (c) What is the cell voltage if the \(\mathrm{TI}^{+}\) concentration is increased tenfold? \(\left(E_{\mathrm{T} 1^{+} / \mathrm{T} 1}^{\circ}=-0.34 \mathrm{~V} .\right)\)

Balance the following redox equations by the halfreaction method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution)

Describe the basic features of a galvanic cell. Why are the two components of the cell separated from each other?

Predict whether \(\mathrm{Fe}^{3+}\) can oxidize \(\mathrm{I}^{-}\) to \(\mathrm{I}_{2}\) under standard-state conditions.

Calculate the pressure of \(\mathrm{H}_{2}\) (in atm) required to maintain equilibrium with respect to the following reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pb}(s)+2 \mathrm{H}^{+}(a q) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{H}_{2}(g) $$ Given that \(\left[\mathrm{Pb}^{2+}\right]=0.035 \mathrm{M}\) and the solution is buffered at \(\mathrm{pH} 1.60\).
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