Question #c9882

1 Answer
Jul 30, 2016

Answer:

WARNING! Long answer.

Explanation:

Here's how I would approach the problem.

You don’t state what kind of hydrogen fuel cell, so I shall assume that you are using a solid oxide fuel cell (SOFC).

The reactions for an SOFC are:

#"Anode:"color(white)(ml) "H"_2 + color(red)(cancel(color(black)("O"^"2-"))) ⇌ "H"_2"O" + color(red)(cancel(color(black)("2e"^"-")))#
#"Cathode:"color(white)(l) "½O"_2 + color(red)(cancel(color(black)("2e"^"-"))) ⇌ color(red)(cancel(color(black)("O"^"2-")))#
#stackrel(——————————————)("Overall:"color(white)(m) "H"_2 + "½O"_2 ⇌ "H"_2"O")#

The Nernst equation for the cell potential is

#color(blue)(|bar(ul(color(white)(a/a) E = E^° - (RT)/(zF)lnQcolor(white)(a/a)|)))" "#

where

#E^°# is the standard cell potential
#R# is the Universal Gas Constant
#T# is the temperature
#z# is the moles of electrons transferred per mole of hydrogen
#F# is the Faraday constant
#Q# is the reaction quotient

#Q = 1/(P_"H₂"P_"O₂"^½)#

Note that water is a liquid at room temperature, so its concentration (activity) does not occur in the reaction quotient.

The Nernst equation then becomes:

#color(blue)(|bar(ul(color(white)(a/a) E = E^° + (RT)/(zF)ln(P_"H₂"P_"O₂"^½)color(white)(a/a)|)))" "#

Calculation of #E^°#

Assume that room temperature is 25 °C.

#E^°# for a hydrogen fuel cell is not the same as #E^°# for a hydrogen half-cell, because the half-reactions are different.

However, the overall reaction is the equilibrium formation of water.

#"H"_2"(g)" + "½O"_2"(g)" ⇌ "H"_2"O(l)"#;

#Δ_"f"G^° "= -237.14 kJ/mol"#

#ΔG^° = -zFE^°#

#E^° = (-ΔG^°)/(zF) = ("-237 140" color(red)(cancel(color(black)("J·mol"^"-1"))))/("2 × 96 485" color(red)(cancel(color(black)("C·mol"^"-1")))) × ("1 V"·color(red)(cancel(color(black)("C"))))/(1 color(red)(cancel(color(black)("J")))) = "1.229 V"#

#E = "1.229 V" + (RT)/(zF)ln(P_"H₂"P_"O₂"^½)#

Calculate the cell potential

An SOFC normally operates at temperatures up to 1000 °C and pressures up to 14 atm.

But we have to calculate the cell potential at 25 °C and pressures of 1 atm.

Then

#E = E^° + (RT)/(zF)ln(P_"H₂"P_"O₂"^½) = "1.229 V" + ("8.314 V"·color(red)(cancel(color(black)("C·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K"))))/("2 × 96 485" color(red)(cancel(color(black)("C·mol"^"-1"))))ln(1sqrt1) = "1.229 V + 0 V" = "1.23 V"#

If you are using a different type of hydrogen fuel cell, you will have to adjust the formula and equations accordingly.