Question

Standard free energies of formation (in $kJ/mol$ ) at $298K$ are $-237.2,-394.4$ and $-8.2$ for $H_{2}O(I),C{O}_2(g)$ 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)

$C_5{H}_{12}+8O_2⟶5C{O}_2+6H_{2}O$
$\Delta G=G(\text{ products })-G(\text{ reactants })=5\times (-394.4)+6\times (-237.2)-(-8.2)=-3387Kjmo{l}^{-1}$
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
$-3387\times 1000=-32\times 96500\times E_{\text{cell }}$
$E_{\text{cell }}=1.0968V$