Question

Standard free energies of formation (in kj/mol) at 298 K are -237.2, -394.4 and -8.2 for $H_2O(l),CO_2(g) $ and pentane (g), respectively. The value of $E^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

The Correct Option is C

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

In pentane-oxygen fuel cell following reaction takes place 

\(C_5H_{12}+10H_2O(l) \rightarrow5CO_2+32 H^+ \, +32e^-\)

\(\frac {8O_2+32H^+\, +32e^- \rightarrow 16H_2O(l)} {C_5H_{12}+8O_2 \rightarrow5CO^2+6H_2O(l), E^0=? }\)

\(\Delta G_{reaction}=\boldsymbol{\Sigma} \Delta G_{product}-\boldsymbol{\Sigma} \Delta G_{reactant}\)
\(=5 \times \Delta G_{(CO_2)}+6\Delta G_{(H_2O)}-[\Delta G_{(C_5H_{12})}+8 \times \Delta G_{O_2}]\)
\(=5 \times (-394.4)+6 \times (-237.2)-(-8.2 +0)\) 
= - 1972- 1423.2+ 8.2 
= -3 3 8 7 kJ/mol 
\(\, \, \, \, \, \, \, \, =-3387 \times 10^3 J/mol\)

\(\Delta G=-nFE^0_{cell}\)

\(-3387 \times 10^3=-32 \times 96500 \times E^0_{cell}\)

\(\, \, \, \, \, E^0_{cell}= \frac {-3387 \times 10^3}{-32 \times 96500}=1.0968 V\)