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Electrochemistry question

2009 · Shift 0 · Q20

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Electrochemistry
2009 · Shift 0 · Q20

Electrochemistry question

2009 · Shift 0 · Q20

JEE MainChemistryElectrochemistryMCQ+4 / −1

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is

CH_3OH

(l) + 3/2

O_2

\to

CO_2

(g) + 2

H_2O

(l) At 298K standard Gibb’s energies of formation for

CH_3OH

(l),

H_2O

(l) and

CO_2

(g) are -166.2, -237.2 and -394.4 kJ mol−1 respectively. If standard enthalpy of combustion of methanol is -726 kJ mol−1, efficiency of the fuel cell will be

A
87%
B
90%
C
97%
D
80%

View written solutionFree

Correct answer: C

Use fuel cell efficiency formula

For a fuel cell, the maximum efficiency is

\eta = \frac{\Delta G^\circ}{\Delta H^\circ}

Since both $\Delta G^\circ$ and $\Delta H^\circ$ are negative for the combustion reaction, we use their magnitudes:

\eta = \frac{|\Delta G^\circ|}{|\Delta H^\circ|}

Calculate $\Delta G^\circ$ for the reaction

Given reaction:

CH_3OH(l) + \frac{3}{2}O_2(g) \to CO_2(g) + 2H_2O(l)

Standard Gibbs energies of formation:

$\Delta G_f^\circ[CH_3OH(l)] = -166.2\,\text{kJ mol}^{-1}$
$\Delta G_f^\circ[H_2O(l)] = -237.2\,\text{kJ mol}^{-1}$
$\Delta G_f^\circ[CO_2(g)] = -394.4\,\text{kJ mol}^{-1}$
$\Delta G_f^\circ[O_2(g)] = 0$

Now,

\Delta G^\circ = \sum \nu \Delta G_f^\circ(\text{products}) - \sum \nu \Delta G_f^\circ(\text{reactants})

\Delta G^\circ = \left[(-394.4) + 2(-237.2)\right] - \left[(-166.2) + \frac{3}{2}(0)\right]

= (-394.4 - 474.4) - (-166.2)

= -868.8 + 166.2 = -702.6\,\text{kJ mol}^{-1}

So,

|\Delta G^\circ| = 702.6\,\text{kJ mol}^{-1}

Use given enthalpy of combustion

\Delta H^\circ = -726\,\text{kJ mol}^{-1}

Hence,

\eta = \frac{702.6}{726}

\eta = 0.9678 \approx 96.8\%

Choose the nearest option

\eta \approx 97\%

So the correct option is C.

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