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_ cell ^0 for the pentane-oxygen fuel cell is

Correct answer: C. 1.0968 ~V aligned & C_5 H_ 12(g) +8 O_ 2(9) 5 CO_ 2(g) +6 H_2 O_ (e) & Δ G_r=5(Δ G_r)_ CO_2 +6(Δ G_ r )_ H_2 O -(Δ G_ r )_ Pentane & =5(-5394.4)+6(-237.2)-(-8.2) & =-1972-1423.2+8.2 & Δ G_ R =-3387 kJ / mole & C_5^ (-12)/(5) H_ 12 5 4 C_2 aligned ) For 5 carbon atoms, there is a change of electrons from - 12 to 20 aligned & No. of electrons involved =32 e^ - & Δ G_r=-3387 × 10^3=-32 × E^ × 96500 & E^ =1.097 V aligned )

NEET 2008

Exam: NEET
Year: 2008
Subject: Chemistry
Chapter: Hydrocarbons

NEET 2008 · Chemistry · Hydrocarbons

Standard free energies of formation (in kJ / mol ) at 298 ~K are...

Standard free energies of formation (in

\mathrm{kJ} / \mathrm{mol}

) at

298 \mathrm{~K}

are

-237.2,-394.4

and -8.2 for

\mathrm{H}_2 \mathrm{O}(l) \mathrm{CO}_2(g)

and pentane

(g)

, respectively. The value of

E_{\text{cell }}^0

for the pentane-oxygen fuel cell is

Options

A $1.968 \mathrm{~V}$
B $2.0968 \mathrm{~V}$
C $1.0968 \mathrm{~V}$
D $0.0968 \mathrm{~V}$

Correct answer

C. $1.0968 \mathrm{~V}$

Step-by-step solution

\begin{aligned} & \mathrm{C}_5 \mathrm{H}_{12(g)}+8 \mathrm{O}_{2(9)} \longrightarrow 5 \mathrm{CO}_{2(g)}+6 \mathrm{H}_2 \mathrm{O}_{(\mathrm{e})} \\ & \therefore \Delta \mathrm{G}_r=5\left(\Delta \mathrm{G}_r\right)_{\mathrm{CO}_2}+6\left(\Delta \mathrm{G}_{\mathrm{r}}\right)_{\mathrm{H}_2 \mathrm{O}}-\left(\Delta \mathrm{G}_{\mathrm{r}}\right)_{\text{Pentane }} \\ & =5(-5394.4)+6(-237.2)-(-8.2) \\ & =-1972-1423.2+8.2 \\ & \Delta \mathrm{G}_{\mathrm{R}}=-3387 \mathrm{ kJ} / \mathrm{mole} \\ & \mathrm{C}_5^{\frac{-12}{5}} \mathrm{H}_{12} \longrightarrow 5 \stackrel{4}{\mathrm{C}_2} \end{aligned}

)

For 5 carbon atoms, there is a change of electrons from - 12 to 20

\begin{aligned} & \therefore\text{ No.\text{ of }\text{ electrons }\text{ involved } }=32 \mathrm{e}^{-} \\ & \Delta G_r=-3387 \times 10^3=-32 \times E^{\circ} \times 96500 \\ & E^{\circ}=1.097 V \end{aligned}

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