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Question

Standard free energies of formation (in kJ/mol) at 298 K are -237.2, -394.4 and -8.2 for $$H_{2}O(l),CO_{2}(g)$$ and pentane (g), respectively. The value of $$E^{0}_{cell}$$ for the pentane-oxygen fuel cell is:


A
1.0968V
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B
0.0968V
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C
1.968V
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D
2.0968V
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Solution

The correct option is A $$1.0968 V$$
$$ C_{5}H_{12}+8O_{2}⟶5CO_{2}+6H_{2}O $$

$$ ΔG=G(products)−G(reactants)=5 \times (-394.4) + 6 \times (-237.2) - (-8.2) = -3387 Kjmol^{-1}$$

To get $$E_{cell}$$ use the equation:

$$ ΔG=−nFE_{cell} $$

As O is moving from 0 to -2 oxidation state, n = 32 for this reaction

$$-3387\times1000$$$$= -32 \times 96500 \times E_{cell}$$

$$ E_{cell} = 1.0968 V $$

Chemistry

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