Question

Standard free energies of formation (in $kJ / mol$ ) at $298 \,K$ are $-237.2,-394.4$ and $-8.2$ for $H _{2} O (I), CO _{2}(g)$ and pentane $(g)$, respectively. The value of $E_{\text {cell }}^{\circ}$ 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

Answer: (C)

$C _{5} H _{12}+8 O _{2} \longrightarrow 5 CO _{2}+6 H _{2} O$
$\Delta G = G (\text { products })- G (\text { reactants })=5 \times(-394.4)+6 \times(-237.2)-(-8.2)=-3387 \,Kj\,mol ^{-1}$
To get $E _{\text {cell }}$ use the equation:
$\Delta G =- nF E _{\text {cell }}$
As $O$ is moving from o to $-2$ oxidation state, $n =32$ for this reaction
$-3387 \times 1000=-32 \times 96500 \times E_{\text {cell }}$
$E _{\text {cell }}=1.0968 \,V$