SATHEE: Electrochemistry - Result Question 59

Electrochemistry - Result Question 59

####61. Standard free energies of formation (in $k J / m o l$ ) at $298 K$ are $- 237.2, - 394.4$ and -8.2 for $H_{2} O (l)$ , $C O_{2} (g)$ and pentane $(g)$ , respectively. The value of $E^{\circ}$ cell for the pentane-oxygen fuel cell is :

(a) $1.968 V$

(b) $2.0968 V$

(d) $0.0968 V$

[2008]

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Solution:

At Anode:

$C_{5} H_{12} + 10 H_{2} O \to 5 C O_{2} + 32 H^{+} + 32 e^{-}$ (pentane)

At Cathode:

$\frac{8 O_{2} + 32 H^{+} + 32 e^{-} \to 16 H_{2} O}{Over all : C_{5} H_{12} + 8 O_{2} \to 5 C O_{2} + 6 H_{2} O}$

Calculation of $Δ G^{\circ}$ for the above reaction $Δ G^{\circ} = [5 \times (- 394.4) + 6 \times (- 237.2)] - [- 8.2]$ $= - 1972.0 - 1423.2 + 8.2 = - 3387.0 k J$ $= - 3387000$ Joules.

From the overall equation we find $n = 32$

Using the relation, $Δ G^{\circ} = - n F E_{cell}^{0}$ and substituting various values, we get

$- 3387000 = - 32 \times 96500 \times E_{cell}^{0} (F = 96500 C)$ or $E_{cell}^{0} = \frac{3387000}{32 \times 96500} = \frac{3387000}{3088000}$

or $\frac{3387}{3088} V = 1.0968 V$

Electrochemistry - Result Question 59

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Electrochemistry - Result Question 59

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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