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
B
0.0968V
C
1.968V
D
2.0968V

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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