Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 29

Q: Why do we use $n=2$ for the cell reaction?

In the reaction $H_2 \rightarrow 2H^+ + 2e^-$, one mole of hydrogen gas releases two moles of electrons. Thus, the number of moles of electrons transferred per mole of $H_2$ is 2.

Q: Why is the heat exchange $q$ taken as zero?

The problem explicitly states that the gas is compressed in a "thermally insulated container." An insulated container does not allow heat to flow in or out, making the process adiabatic ($q=0$).

Q: Why do we multiply by $10^{-3}$ when calculating the work?

The standard Gibbs free energy ($\Delta G^\circ$) we calculated is for 1 mole of $H_2$. Since only $1.0 \times 10^{-3}$ moles of $H_2$ are consumed, we must scale the work proportionally.

Q: What if the gas was diatomic instead of monoatomic?

For a diatomic gas, the molar heat capacity at constant volume ($C_v$) is $\frac{5}{2}R$ instead of $\frac{3}{2}R$. This would result in a smaller temperature change for the same amount of work done.

Q: Does the 70% efficiency apply to the voltage or the total energy?

The efficiency applies to the total work output. The maximum possible work is given by $\Delta G^\circ$, so the actual useful work derived is 70% of this maximum theoretical work.

Question

Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is
$H_2 (g) + \frac{1}{2} O_2(g) ightarrow H_2O(l)$.
The work derived from the cell on the consumption of $1.0 	imes 10^{-3}	ext{ mol}$ of $H_2(g)$ is used to compress 1.00 mol of a monoatomic ideal gas in a thermally insulted container. What is the change in the temperature (in K) of the ideal gas ?
The standard reduction potentials for the two half-cells are given below.
$O_2(g) + 4 H^+ (aq.) + 4e^- ightarrow 2H_2O (l) , E^\circ = 1.23 	ext{ V}$,
$2H^+ (aq.) + 2e^- ightarrow H_2(g), E^\circ = 0.00	ext{V}$.
Use $F = 96500 	ext{ C mol}^{-1}, R = 8.314 	ext{ J mol}^{-1}	ext{K}^{-1}$

Sigma Percentile · Accepted Answer

The beauty of JEE Advanced lies in its ability to seamlessly weave multiple concepts into a single, elegant narrative. In this problem, we are witnessing a beautiful handshake between **Electrochemistry** and **Thermodynamics**. We have a hydrogen-oxygen fuel cell generating electrical work, which is then entirely utilized to compress a monoatomic ideal gas.

Let's break this down step-by-step and uncover the physics behind the math.

### The Fuel Cell: Generating Power

Our journey begins inside the fuel cell. The core reaction powering this cell is the combustion of hydrogen:

$H_2(g) + \frac{1}{2}O_2(g) ightarrow H_2O(l)$

To find out how much punch this cell packs, we first need its standard cell potential, $E^\circ_{	ext{cell}}$. The problem provides the standard reduction potentials for both half-reactions. Oxygen is reduced at the cathode, and hydrogen is oxidized at the anode.

$E^\circ_{	ext{cell}} = E^\circ_{	ext{cathode}} - E^\circ_{	ext{anode}}$

Substituting the given values:

$E^\circ_{	ext{cell}} = 1.23 	ext{ V} - 0.00 	ext{ V} = 1.23 	ext{ V}$

### Calculating the Available Work

Now that we have the voltage, we can determine the maximum theoretical work this cell can perform per mole of hydrogen consumed. This is given by the standard Gibbs free energy change, $\Delta G^\circ$.

$W_{	ext{max}} = |\Delta G^\circ| = nFE^\circ_{	ext{cell}}$

For the oxidation of one mole of $H_2$, exactly 2 moles of electrons are transferred, meaning $n = 2$.

$|\Delta G^\circ| = 2 	imes 96500 	ext{ C mol}^{-1} 	imes 1.23 	ext{ V} = 237390 	ext{ J mol}^{-1}$

However, the real world is rarely perfect. Our fuel cell operates at a **70% efficiency**, and we are only consuming a tiny fraction of hydrogen—specifically, $1.0 	imes 10^{-3}$ moles. The actual electrical work derived is:

$W_{	ext{derived}} = 0.70 	imes 237390 	imes 10^{-3} 	ext{ J}$

$W_{	ext{derived}} = 166.173 	ext{ J}$

### The Thermodynamic Compression

This $166.173 	ext{ J}$ of electrical work doesn't just vanish; it is channeled into compressing 1 mole of a monoatomic ideal gas. The container is thermally insulated, which is a crucial keyword. It means no heat can enter or escape the system, making the process **adiabatic** ($q = 0$).

According to the First Law of Thermodynamics:

$\Delta U = q + W_{	ext{on\_gas}}$

Since $q = 0$, all the work done on the gas goes directly into increasing its internal energy:

$W_{	ext{on\_gas}} = \Delta U = n_{	ext{gas}} C_{v,m} \Delta T$

### The Final Temperature Surge

We know the gas is monoatomic, which means its molar heat capacity at constant volume, $C_{v,m}$, is $\frac{3}{2}R$. We can now equate the work derived from the fuel cell to the internal energy increase of the gas:

$166.173 = 1 	imes \left(\frac{3}{2} 	imes 8.314ight) 	imes \Delta T$

$166.173 = 12.471 	imes \Delta T$

Solving for the temperature change:

**$\Delta T = 13.32 	ext{ K}$**

And there we have it! The electrical energy from a tiny amount of hydrogen was enough to raise the temperature of a whole mole of gas by over 13 degrees. This problem is a masterclass in energy conservation across different domains of physical chemistry.

Enter Numerical Value:

Visualized Solution

\text{System Overview}

\text{Standard Cell Potential}

\text{Calculating } E^\circ_{\text{cell}}

\text{Maximum Work Available}

\text{Calculating } \Delta G^\circ

\text{Actual Work Derived}

\text{Calculating } W_{\text{derived}}

\text{Thermodynamic Compression}

\text{Equating Work and Energy}

\text{Solving for } \Delta T

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Electrochemical Cells

The Fuel Cell

Calculating the Available Work

The Thermodynamic Compression

The Final Temperature Surge

Similar Questions

For the following cell with hydrogen electrodes at two different pressures $p_{1}$ and $p_{2}$ $P t (H_{2}) ∣ H^{+} (a q) ∣ P t (H_{2})$ $p_{1}$ $1 M$ $p_{2}$ emf is given by

What would be the electrode potential for the given half-cell reaction at $pH = 5$ ? …… . $2 H_{2} O \to O_{2} + 4 H^{\oplus} + 4 e^{-}$ ; $E_{red}^{\circ} = 1.23 V$ ( $R = 8.314 J mol^{- 1} K^{- 1}$ ; Temp $= 298 K$ ; oxygen under std. atm. pressure of $1 bar$ )

The standard Gibbs energy for the given cell reaction in $kJ mol^{- 1}$ at $298 K$ is $Zn (s) + Cu^{2 +} (a q) ⟶ Zn^{2 +} (a q) + Cu (s), E^{\circ} = 2 V$ at $298 K$ (Faraday's constant, $F = 96000 C mol^{- 1}$ )

The Gibbs energy change (in J) for the given reaction at $[C u^{2 +}] = [S n^{2 +}] = 1 M$ and $298 K$ is $C u (s) + S n^{2 +} (a q) ⟶ C u^{2 +} (a q) + S n (s)$ ; $(E_{S n^{2 +} / S n}^{\circ} = - 0.16 V, E_{C u^{2 +} / C u}^{\circ} = 0.34 V$ , Take, $F = 96500 C mol^{- 1})$

In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to

The standard emf of a cell, involving one electron change is found to be $0.591 V$ at $2 5^{\circ} C$ . The equilibrium constant of the reaction is ( $F = 96500 C mol^{- 1}$ )

For the redox reaction $Z n (s) + C u^{2 +} (0.1 M) ⟶ Z n^{2 +} (1 M) + C u (s)$ taking place in a cell, $E_{ce l l}^{\circ}$ is $1.10 V$ . $E_{ce l l}$ for the cell will be $(2.303 \frac{R T}{F} = 0.0591)$