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Standard free energies of formation (in kJ/ mol) at 298 K are – 237.2, – 394.4 and – 8.2 for H₂O(l), CO₂(g) and pentane (g), respectively. The value of $E_{cell}^{0}$ for the pentane-oxygen fuel cell is :

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2.0968V

1.0968V

1.968V

0.0968V

answer is B.

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Detailed Solution

{C_5}{H_{12}}_{\left( g \right)} + 8{O_{2\left( g \right)}} \to 5C{O_{2\left( g \right)}} + 6{H_2}{O_{\left( l \right)}}

\Delta {G^o} = G_{pro}^o - G_{rea}^o = \left[ {6\left( { - 273.2} \right) + 5\left( { - 394.4} \right)} \right] - \left[ { - 8.2} \right] = - 3387kJ

Calculation of 'n' value

decrease in oxidation state of Oxygen atom = 2

Total decrease in oxidation state of Oxygen atom = 2 x 16 =32

\Delta {G^o} = - nFE_{cell}^o \Rightarrow E_{cell}^o = \frac{{ - \Delta {G^o}}}{{nF}} = \frac{{3387}}{{32 \times 96.5}} = + 1.09V

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