In a hydrogen/oxygen fuel cell 67.2 litres of hydrogen is consumed in 5 minutes. What electric current will be produced at #sf(0^@C)# and 1 atmosphere pressure ?

2 Answers
Jul 15, 2017

Answer:

#1929"Amperes"#

Explanation:

We are given:

#(67.2"L H"_2"@S.T.P.")/(5"min")#

Use the factor that converts Liters to Moles:

#(67.2"L H"_2"@S.T.P.")/(5"min")(1"mol H"_2)/(22.4"L H"_2"@S.T.P.")#

Please observe how the units cancel:

#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1"mol H"_2)/(22.4cancel("L H"_2"@S.T.P."))#

We know that 1 mole of H#"_2# produces 2 moles of electrons:

#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2"mol e"^-)/(1cancel("mol H"_2))#

Use Avagadro's Constant :

#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23"e"^-)/(1cancel("mol e"^-))#

Next we use the definition of a Coulomb :

#(67.2cancel("L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))#
#(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e"^-))#

Because we know that an Ampere is 1 Coulomb per second we must convert minutes to seconds:

#(67.2cancel("L H"_2"@S.T.P."))/(5cancel("min"))(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))##(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e"^-))(1cancel("min"))/(60"s")#

Do the multiplication and division and the answer will be in Amperes:

#1929"Amperes"#

This seems reasonable, because the consumption rate of Hydrogen is quite high and the potential of a single cell is between 0.6 and 0.7 volts, thereby, making the output power about 1.3 kW. To practically do this, one would stack the cells in a series-parallel configuration.

Jul 15, 2017

Answer:

#sf(1930color(white)(x)A)#

Explanation:

Hydrogen is oxidised at the anode:

#sf(H_(2(g))rarr2H_((aq))^(+)+2e)#

This tells us that 1 mole of #sf(H_2)# produces 2 moles of electrons.

So #sf(22.4)# litres produce 2 moles of electrons.

So #sf(67.2)# litres produce #sf(2xx67.2/22.4=6)# moles of electrons.

The charge on 1 mole of electrons is given by the Faraday Constant and is equal to #sf(9.65xx10^4color(white)(x)"C/mol")#.

So the total charge Q produced is given by:

#sf(Q=6xx9.65xx10^(4)=57.9xx10^(4)color(white)(x)C)#

This is produced in 5 minutes which = 5 x 60 = 300 s.

Electric current I is the rate of flow of charge as given by:

#sf(I=Q/t)#

#:.##sf(I=(57.9xx10^(4))/(300)=1930color(white)(x)A)#