1 answer

10. Given that the AG for the hydrogen oxygen fuel cell reaction is -474.4 kJ/mol, calculate...

Question:

10. Given that the AG for the hydrogen oxygen fuel cell reaction is -474.4 kJ/mol, calculate the E and K for this fuel cell
10. Given that the AG for the hydrogen oxygen fuel cell reaction is -474.4 kJ/mol, calculate the E' and K for this fuel cell at 25°C: 2 H2(g) + O2(g) → 2 H20 (1)

Answers

10. We know that elements in their free state have 0 oxidation state so H in H2 and O in O2 have 0 oxidation state. In H2O, each H has +1 oxidation state and O has -2 oxidation state. But in the balanced redox reaction, we have 2x2=4 H involved and 2x1=2 O involved. So there is a transfer of 4 electrons from 4 H to 2 O to form 4 H in +1 oxidation state each and 2 O in -2 oxidation state each i.e.

2 H2O molecules. So number of moles of electrons transferred in balanced redox reaction=4

Also we know that free energy change for a reaction AG = -FE

Where n=number of moles of electrons transferred

F=Faraday's constant=96485 C/mol

E°=Electromotive force at standard state

Substitute values for free energy change, F, n in above equation.

-474.4 kJ/mol=-4 mol x 96485 C/mol x E°

(-474.4 kJ/mol x 1000 J/kJ)/(-4 mol x 96485 C/mol)=E°=1.23 V (1 kJ=1000 J)

So E°=1.23 V

Also we know that free energy change for a reaction AG = -RIINK

Where R=Gas constant=8.314 J/Kmol

T=Temperature in K=25°C=25+273 K=298 K (9°C=273 K)

K=Equilibrium constant

Substitute values for free energy change, R and T in above equation we get

-474.4 kJ/mol=-8.314 J/Kmol x 298 K ln K

ln K=(-474.4 kJ/mol x 1000 J/kJ)/(-8.314 J/Kmol x 298 K)=191.48

K=e191.48=1.44 x 1083

So K=1.44 x 1083

.

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