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Chemistry Question on Electrochemical Cells

Standard free energies of formation (in kj/mol) at 298 K are -237.2, -394.4 and -8.2 for H2O(l),CO2(g)H_2O(l),CO_2(g) and pentane (g), respectively. The value of Ecell0E^0_{cell} for the pentane-oxygen fuel cell is

A

1.968 V

B

2.0968 V

C

1.0968 V

D

0.0968 V

Answer

1.0968 V

Explanation

Solution

Cell potential is the measure of the voltage difference between the anode and cathode.
ΔG of H2O(l)=−237.2KJ/mol\Delta G \, of \, H_2O(l)=-237.2 KJ/mol
ΔG of CO2(g)=−394.4 KJ/mol\Delta G \, of \, CO_2(g)=-394.4 \, KJ/mol
ΔG of pentane(g)=−8.2kJ/mol\Delta G \, of \, pentane (g)=-8.2 kJ/mol

In pentane-oxygen fuel cell following reaction takes place

C5H12+10H2O(l)→5CO2+32H+ +32e−C_5H_{12}+10H_2O(l) \rightarrow5CO_2+32 H^+ \, +32e^-

8O2+32H+ +32e−→16H2O(l)C5H12+8O2→5CO2+6H2O(l),E0=?\frac {8O_2+32H^+\, +32e^- \rightarrow 16H_2O(l)} {C_5H_{12}+8O_2 \rightarrow5CO^2+6H_2O(l), E^0=? }

ΔGreaction=ΣΔGproduct−ΣΔGreactant\Delta G_{reaction}=\boldsymbol{\Sigma} \Delta G_{product}-\boldsymbol{\Sigma} \Delta G_{reactant}
=5×ΔG(CO2)+6ΔG(H2O)−[ΔG(C5H12)+8×ΔGO2]=5 \times \Delta G_{(CO_2)}+6\Delta G_{(H_2O)}-[\Delta G_{(C_5H_{12})}+8 \times \Delta G_{O_2}]
=5×(−394.4)+6×(−237.2)−(−8.2+0)=5 \times (-394.4)+6 \times (-237.2)-(-8.2 +0)
= - 1972- 1423.2+ 8.2
= -3 3 8 7 kJ/mol
        =−3387×103J/mol\, \, \, \, \, \, \, \, =-3387 \times 10^3 J/mol

ΔG=−nFEcell0\Delta G=-nFE^0_{cell}

−3387×103=−32×96500×Ecell0-3387 \times 10^3=-32 \times 96500 \times E^0_{cell}

     Ecell0=−3387×103−32×96500=1.0968V\, \, \, \, \, E^0_{cell}= \frac {-3387 \times 10^3}{-32 \times 96500}=1.0968 V