In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is CH3O

Q: Why is the efficiency of a fuel cell defined as $\Delta G / \Delta H$ and not work output over heat input like a heat engine?

A fuel cell is not a heat engine; it doesn't operate on a temperature difference cycle (like the Carnot cycle). It converts chemical energy directly into electrical energy. The maximum electrical work it can do at constant temperature and pressure is given by the change in Gibbs free energy ($\Delta G$). The total energy released by the reaction is the enthalpy change ($\Delta H$). Thus, its theoretical maximum efficiency is $\Delta G / \Delta H$.

Q: Why did we take the standard Gibbs free energy of formation of $\text{O}_2(g)$ as zero?

By thermodynamic convention, the standard enthalpy of formation and standard Gibbs free energy of formation of any element in its most stable reference state at standard conditions (1 bar, 298 K) are defined as exactly zero. Since $\text{O}_2$ is the most stable form of oxygen under these conditions, its $\Delta G_f^\circ$ is 0.

Q: What does the negative sign in $\Delta G$ and $\Delta H$ indicate?

The negative sign indicates that the process is exergonic and exothermic, respectively. A negative $\Delta G$ means the reaction is spontaneous and can do useful work on the surroundings. A negative $\Delta H$ means the reaction releases heat to the surroundings. When calculating efficiency, we take the ratio of their magnitudes.

Q: Can the efficiency of a fuel cell be greater than 100%?

Theoretically, yes! If a reaction has a positive entropy change ($\Delta S > 0$), then $\Delta G = \Delta H - T\Delta S$ means that $|\Delta G|$ can be greater than $|\Delta H|$ (since $\Delta H$ is negative). In such a case, the fuel cell would absorb heat from the surroundings to convert into electrical work, leading to an efficiency $> 100\%$. However, for the methanol fuel cell, $\Delta S$ is negative, so efficiency is $< 100\%$.

Q: Why did we multiply the $\Delta G_f$ of water by 2?

The balanced chemical equation for the combustion of 1 mole of methanol produces 2 moles of water. Since the given $\Delta G_f$ values are per mole of the substance formed, we must multiply the value for water by its stoichiometric coefficient (2) to find the total free energy change for the reaction as written.

Question

In a fuel cell, methanol is used as fuel and oxygen gas is used as an oxidiser. The reaction is
$	ext{CH}_3	ext{OH}(l) + \frac{3}{2}	ext{O}_2(g) \longrightarrow 	ext{CO}_2(g) + 2	ext{H}_2	ext{O}(l)$
At 298 K standard Gibb's energies of formation for $	ext{CH}_3	ext{OH}(l)$, $	ext{H}_2	ext{O}(l)$ and $	ext{CO}_2(g)$ are $-166.2$, $-237.2$ and $-394.4 	ext{ kJ mol}^{-1}$, respectively. If standard enthalpy of combustion of methanol is $-726 	ext{ kJ mol}^{-1}$, efficiency of the fuel cell will be

Options:

(A) 80%

(B) 87%

(C) 90%

(D) 97%

Sigma Percentile · Accepted Answer

Imagine a world where we don't have to burn coal to boil water, to spin a turbine, to turn a generator, just to get electricity. Every step in that traditional process loses massive amounts of energy as waste heat. What if we could bypass all those clumsy mechanical steps and convert chemical energy directly into electrical energy? Welcome to the elegant world of **Fuel Cells**.

In this problem, we are analyzing a methanol fuel cell. Methanol ($	ext{CH}_3	ext{OH}$) is a fantastic liquid fuel, and when it reacts with oxygen, it releases a lot of energy. But how much of that energy can we actually harness as useful electricity?

### The Thermodynamic Limit: Gibbs vs. Enthalpy

To understand efficiency, we need to consult the two heavyweights of chemical thermodynamics: **Enthalpy ($\Delta H$)** and **Gibbs Free Energy ($\Delta G$)**.

Enthalpy ($\Delta H$) represents the total heat energy released during the combustion of methanol. Think of it as the total budget available. The problem generously gives us this value: $\Delta H = -726 	ext{ kJ mol}^{-1}$.

However, the universe demands a tax in the form of entropy. We cannot convert all that heat into useful work. The maximum amount of non-expansion work (like electrical work) we can extract from a reaction at constant temperature and pressure is given by the change in Gibbs Free Energy ($\Delta G$).

Therefore, the theoretical maximum efficiency ($\eta$) of our fuel cell is simply the ratio of the useful work we can extract to the total energy available:

$$ \eta = \frac{\Delta G}{\Delta H} 	imes 100 $$

### Crunching the Numbers: The Methanol Combustion

Our mission now is to find $\Delta G$ for the combustion reaction:

$$ 	ext{CH}_3	ext{OH}(l) + \frac{3}{2}	ext{O}_2(g) \longrightarrow 	ext{CO}_2(g) + 2	ext{H}_2	ext{O}(l) $$

We can calculate the standard Gibbs free energy of the reaction ($\Delta G_r^\circ$) using the standard Gibbs free energies of formation ($\Delta G_f^\circ$) of the reactants and products. The formula is beautifully simple: products minus reactants.

$$ \Delta G_r^\circ = \sum \Delta G_f^\circ(	ext{products}) - \sum \Delta G_f^\circ(	ext{reactants}) $$

Let's plug in the values. We have 1 mole of $	ext{CO}_2$ and 2 moles of $	ext{H}_2	ext{O}$ as products. For reactants, we have 1 mole of $	ext{CH}_3	ext{OH}$ and $\frac{3}{2}$ moles of $	ext{O}_2$.

**Crucial Concept:** The standard Gibbs free energy of formation for any element in its most stable standard state is exactly zero. So, for $	ext{O}_2(g)$, $\Delta G_f^\circ = 0$.

$$ \Delta G_r^\circ = [\Delta G_f^\circ(	ext{CO}_2) + 2\Delta G_f^\circ(	ext{H}_2	ext{O})] - [\Delta G_f^\circ(	ext{CH}_3	ext{OH}) + \frac{3}{2}\Delta G_f^\circ(	ext{O}_2)] $$

$$ \Delta G_r^\circ = [-394.4 + 2(-237.2)] - [-166.2 + 0] $$

$$ \Delta G_r^\circ = [-394.4 - 474.4] + 166.2 $$

$$ \Delta G_r^\circ = -868.8 + 166.2 = -702.6 	ext{ kJ mol}^{-1} $$

This means that out of the total $726 	ext{ kJ}$ of energy released per mole of methanol, a massive $702.6 	ext{ kJ}$ is available to do useful electrical work!

### The Final Verdict: How Efficient Are We?

Now, we just plug our $\Delta G$ and $\Delta H$ back into our efficiency formula. Since both values are negative (indicating energy leaving the system), the signs will cancel out.

$$ \eta = \frac{-702.6}{-726} 	imes 100 $$

$$ \eta = 0.96776 	imes 100 = 96.78\% $$

Rounding to the nearest integer, we get an astonishing **97%**.

Think about that for a second. A typical car engine has an efficiency of around 25-30%. Even the best power plants struggle to hit 50%. But this fuel cell, by bypassing the thermal cycle and converting chemical energy directly into electricity, achieves a theoretical efficiency of 97%. This is why fuel cells are considered a cornerstone of future clean energy technologies!

Select Answer:

Visualized Solution

$Efficiency of Fuel Cell$

$Reaction and Δ G_{r}$

$Calculating Δ G_{r}$

$Calculating Efficiency$

The Sigma Insight: Entropy and Free Energy

The Thermodynamic Limit

Crunching the Numbers

The Final Verdict

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