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

JEE Main · 2009 · 12 Apr 2009 · Offline · Redox Reactions and Electrochemistry

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

CH_{3} OH (ℓ) + \frac{3}{2} O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (ℓ)

At 298 K standard Gibb's energies of formation for

CH_{3} OH (ℓ), H_{2} O (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

Answer: (D)

The efficiency of a fuel cell is the fraction of the combustion enthalpy that can appear as useful electrical work,

η = \frac{∣Δ G ∣}{∣Δ H ∣}

. The reaction free energy is

Δ G^{\circ} = [- 394.4 + 2 (- 237.2)] - (- 166.2) = - 702.6 kJ mo l^{- 1}

, and with the combustion enthalpy

Δ H = - 726 kJ mo l^{- 1}

the efficiency is

\frac{702.6}{726} \times 100

, which is about 97 per cent.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option