# 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 ?

Jul 15, 2017

$1929 \text{Amperes}$

#### Explanation:

We are given:

$\left(67.2 \text{L H"_2"@S.T.P.")/(5"min}\right)$

Use the factor that converts Liters to Moles:

$\left(67.2 \text{L H"_2"@S.T.P.")/(5"min")(1"mol H"_2)/(22.4"L H"_2"@S.T.P.}\right)$

Please observe how the units cancel:

$\left(67.2 \cancel{\text{L H"_2"@S.T.P."))/(5"min")(1"mol H"_2)/(22.4cancel("L H"_2"@S.T.P.}}\right)$

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

$\left(67.2 \cancel{{\text{L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2"mol e"^-)/(1cancel("mol H}}_{2}}\right)$

$\left(67.2 \cancel{{\text{L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23"e"^-)/(1cancel("mol e}}^{-}}\right)$

Next we use the definition of a Coulomb :

$\left(67.2 \cancel{\text{L H"_2"@S.T.P."))/(5"min")(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P.}}\right)$
$\left(2 \cancel{{\text{mol e"^-))/(1cancel("mol H"_2))(6.02xx10^23cancel("e"^-))/(1cancel("mol e"^-))(1"C")/(6.242xx10^18cancel("e}}^{-}}\right)$

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

$\left(67.2 \cancel{{\text{L H"_2"@S.T.P."))/(5cancel("min"))(1cancel("mol H"_2))/(22.4cancel("L H"_2"@S.T.P."))(2cancel("mol e"^-))/(1cancel("mol H}}_{2}}\right)$(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:

$1929 \text{Amperes}$

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

$\textsf{1930 \textcolor{w h i t e}{x} A}$

#### Explanation:

Hydrogen is oxidised at the anode:

$\textsf{{H}_{2 \left(g\right)} \rightarrow 2 {H}_{\left(a q\right)}^{+} + 2 e}$

This tells us that 1 mole of $\textsf{{H}_{2}}$ produces 2 moles of electrons.

So $\textsf{22.4}$ litres produce 2 moles of electrons.

So $\textsf{67.2}$ litres produce $\textsf{2 \times \frac{67.2}{22.4} = 6}$ moles of electrons.

The charge on 1 mole of electrons is given by the Faraday Constant and is equal to $\textsf{9.65 \times {10}^{4} \textcolor{w h i t e}{x} \text{C/mol}}$.

So the total charge Q produced is given by:

$\textsf{Q = 6 \times 9.65 \times {10}^{4} = 57.9 \times {10}^{4} \textcolor{w h i t e}{x} C}$

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:

$\textsf{I = \frac{Q}{t}}$

$\therefore$$\textsf{I = \frac{57.9 \times {10}^{4}}{300} = 1930 \textcolor{w h i t e}{x} A}$