Redox Reactions and Electrochemistry JEE 2020: 13.32 K

Redox Reactions and Electrochemistry JEE 2020: 13.32 K Solution

JEE Advanced 2020 Chemistry Redox Reactions and Electrochemistry Fuel cell, Gibbs free energy and adiabatic compression

By Founder, JEEnius - IIT Kanpur Alumni · Sep 11, 2026 · 3 min read

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Consider a $70 %$ efficient hydrogen-oxygen fuel cell working under standard conditions at $1$ bar and $298$ K. Its cell reaction is

$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)$

The work derived from the cell on the consumption of $1.0 \times 10^{- 3}$ mol of $H_{2} (g)$ is used to compress $1.00$ mol of a monoatomic ideal gas in a thermally insulated 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.

$O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l)$ , $E^{\circ} = 1.23$ V

$2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$ , $E^{\circ} = 0.00$ V

Use $F = 96500$ C mol $^{- 1}$ , $R = 8.314$ J mol $^{- 1}$ K $^{- 1}$ .

Show answerAnswer

13.32

Explanation

For the hydrogen-oxygen fuel cell, the cathode reduction is:

$\frac{1}{2} O_{2} (g) + 2 H^{+} (a q) + 2 e^{-} \to H_{2} O (l)$

$E_{cathode}^{\circ} = 1.23$ V

The anode oxidation is:

$H_{2} (g) \to 2 H^{+} (a q) + 2 e^{-}$

$E_{anode}^{\circ} = 0.00$ V

So, the standard cell potential is:

$E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ}$

$E_{cell}^{\circ} = 1.23 - 0.00$

$E_{cell}^{\circ} = 1.23$ V

For the consumption of $1$ mol of $H_{2}$ , the number of electrons transferred is $2$ mol. For $1.0 \times 10^{- 3}$ mol of $H_{2}$ , maximum electrical work is obtained from Gibbs free energy change:

$Δ G^{\circ} = - n F E_{cell}^{\circ}$

$Δ G^{\circ} = - 2 \times 96500 \times 1.23 \times 1.0 \times 10^{- 3}$ J

$Δ G^{\circ} = - 237.39$ J

The fuel cell is only $70 %$ efficient, so useful work obtained is:

$W = 0.70 \times 237.39$

$W = 166.173$ J

This work is used to compress $1.00$ mol of a monoatomic ideal gas in a thermally insulated container. Since the container is thermally insulated:

$q = 0$

The work done on the gas increases its internal energy:

$W_{on} = Δ U$

For a monoatomic ideal gas:

$C_{V} = \frac{3}{2} R$

Therefore:

$Δ U = n C_{V} Δ T$

$166.173 = 1.00 \times \frac{3}{2} \times 8.314 \times Δ T$

$Δ T = \frac{166.173 \times 2}{3 \times 8.314}$

$Δ T = 13.32$ K

Therefore, the change in temperature of the ideal gas is $13.32$ K.

What was the 2020 JEE Advanced numerical on 70% efficient hydrogen-oxygen fuel cell driving adiabatic compression?

The temperature change is 13.32 K. A 70% efficient hydrogen-oxygen fuel cell operates at 1 bar and 298 K with cell reaction H₂(g) + ½O₂(g) → H₂O(l). It consumes exactly 1.0 × 10^{-3} mol H₂. The electrical work obtained is entirely used to compress 1.00 mol monoatomic ideal gas inside a thermally insulated container. Standard reduction potentials supplied are 1.23 V for O₂/H₂O and 0.00 V for H⁺/H₂, with F = 96500 C mol^{-1} and R = 8.314 J mol^{-1} K^{-1}.

Hydrogen-oxygen fuel cell with H2 inlet on anode side and O2 inlet on cathode side producing electrical current that drives a piston to compress 1 mol monoatomic ideal gas inside a thermally insulated cylinder labelled q=0, arrows for work transfer W from cell to gas and final

How does the official JEE solution calculate the temperature change step by step?

The official method yields exactly 13.32 K. Cathode half-cell: ½O₂ + 2H⁺ + 2e– → H₂O, E° = 1.23 V. Anode: H₂ → 2H⁺ + 2e–, E° = 0.00 V. Thus E°_cell = 1.23 V and n = 2 electrons per mole H₂. $Δ G^{\circ} = - n F E_{cell}^{\circ}$

For 1.0 × 10^{-3} mol H₂ the value is –2 × 96500 × 1.23 × 10^{-3} = –237.39 J. Useful work after 70% efficiency equals 0.70 × 237.39 = 166.173 J.

For the thermally insulated container q = 0, therefore W_on gas = ΔU. Cv = (3/2)R for monoatomic gas so ΔU = 1 × (3/2) × 8.314 × ΔT.

Δ T = \frac{166.173 \times 2}{3 \times 8.314} = 13.32 K .

Why does ignoring the 70 percent efficiency factor produce an incorrect 19.03 K answer?

Treating the entire |ΔG| of 237.39 J as the actual work delivered yields ΔT ≈ 19.03 K. The arithmetic simply replaces 166.173 with 237.39 in the final expression, producing (237.39 × 2) / (3 × 8.314) = 19.03. This is a procedural flaw.

The mistake arises from forgetting that real fuel cells deliver only a fraction of the maximum Gibbs free energy as useful electrical work. The official solution explicitly scales by 0.70 before equating to ΔU.

How do Gibbs free energy and the first law connect in this fuel cell problem?

ΔG equals the maximum non-expansion work obtainable from the cell at constant T and P. Only 70% of that maximum is realised as electrical work in the given fuel cell. For the insulated container q = 0, so any work done on the gas appears entirely as increase in internal energy.

Monoatomic ideal gas internal energy depends only on temperature: ΔU = nCvΔT with Cv = 3/2 R. The scaled |ΔG| becomes the magnitude of W_on gas, which equals ΔU, which fixes ΔT through the Cv expression.

Search the past-paper archive to find every JEE Main paper from 2002 and every Advanced paper from 2007 by year, subject or chapter, each with a worked solution (free).

What similar problems combine Gibbs energy with thermodynamics?

Calculate ΔG and maximum work for the same fuel cell when 0.05 mol H2 is consumed at 298 K with 80% efficiency (numerical).

A Daniell cell supplies electrical energy to drive an isothermal reversible expansion of 2 mol ideal gas. Find final volume if initial is 10 L and E_cell = 1.1 V (link nFE to work).

Determine temperature rise when work from a 60% efficient methanol–oxygen fuel cell is used for adiabatic compression of 0.5 mol diatomic gas (compare Cv = 5/2 R).

What checklist prevents mistakes in these fuel cell numericals?

Write half-cell reactions and confirm n from the balanced cell equation before calculating ΔG.
Multiply |ΔG| by the efficiency decimal before calling it useful work.
State q = 0 explicitly when the container is thermally insulated and set W = ΔU.
Use Cv = 3/2 R only for monoatomic gas and carry at least three decimal places until final rounding to match 13.32.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What is the temperature change in the 2020 JEE Advanced hydrogen oxygen fuel cell problem?

The temperature change is exactly 13.32 K. The 70% efficient fuel cell produces useful electrical work of 166.173 J from 1.0 × 10^{-3} mol H₂. This work is fully converted to ΔU of 1 mol monoatomic gas in the q=0 container, so ΔT = (2 × 166.173) / (3 × 8.314) = 13.32 K.

How does the official JEE solution calculate the temperature change in the fuel cell problem?

Calculate ΔG° = –nFE°_cell with n=2 and E°_cell=1.23 V for the given 10^{-3} mol H₂ to get –237.39 J. Multiply by 0.70 efficiency to obtain 166.173 J useful work. For the adiabatic container q=0 so W = ΔU = (3/2)R ΔT. Solve for ΔT = 13.32 K.

Why is the answer 13.32 K and not 19.03 K in the JEE 2020 electrochemistry numerical?

19.03 K appears if the full |ΔG| of 237.39 J is taken as work delivered. The fuel cell is only 70% efficient, so useful electrical work is 0.70 × 237.39 J = 166.173 J. This scaled value equals ΔU for the monoatomic gas, producing ΔT = 13.32 K.

How are Gibbs free energy and the first law connected in the JEE Advanced 2020 fuel cell question?

ΔG equals maximum non-expansion work at constant T and P. Only 70% of |ΔG| is realised as electrical work. This work is done on the thermally insulated gas (q=0), so first law gives ΔU = W. For monoatomic ideal gas ΔU = n(3/2)RΔT, fixing ΔT at 13.32 K.

What efficiency factor must be applied in the redox reactions and electrochemistry JEE 2020 fuel cell numerical?

The given 70% efficiency must be multiplied to the magnitude of ΔG before equating to ΔU. Real fuel cells deliver only a fraction of the maximum possible Gibbs free energy change as useful electrical work. Ignoring this produces the wrong 19.03 K instead of the official 13.32 K.

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