Electrochemistry JEE 2025: Butane Fuel-Cell Solution

What is the answer to the electrochemistry JEE 2025 butane fuel-cell question?

The answer to this electrochemistry JEE 2025 numerical is 105.50: the reaction energy and the 26-electron count must both refer to one mole of butane. This is a JEE Advanced 2025, Paper 2, Chemistry numerical-answer question, not a JEE Main question.

A cell obtains electrical energy from butane combustion at 1 bar and 298 K. Its standard cell potential is given below, where F is the Faraday constant:

E^{\circ} = \frac{X}{F} \times 10^{3} V

Find X using these supplied standard Gibbs energies of formation at 298 K:

Carbon dioxide, C O_{2} : & - 394 k J m o l^{- 1} Water: & - 237 k J m o l^{- 1} Butane: & - 18 k J m o l^{- 1}

The question bank tags this as medium and estimates 180 seconds. These are question-bank labels, not an official exam difficulty classification or time allowance.

How do you balance butane combustion and calculate its Gibbs energy change?

The standard reaction Gibbs energy is negative 2743 kJ for one mole of butane consumed. Calculate it from the balanced equation and the supplied formation energies, using products minus reactants. Keep this one-mole basis throughout the solution.

C_{4} H_{10} + \frac{13}{2} O_{2} \to 4 C O_{2} + 5 H_{2} O

Four carbon atoms require four carbon dioxide molecules. Ten hydrogen atoms require five water molecules, giving thirteen oxygen atoms on the product side. Since oxygen enters as diatomic molecules, its coefficient is thirteen divided by two.

Multiply each formation energy by its coefficient in the balanced equation. Then subtract the reactant sum from the product sum:

Δ G_{r e a c t i o n}^{\circ} = \sum_{p r o d u c t s} ν Δ_{f} G^{\circ} - \sum_{r e a c t a n t s} ν Δ_{f} G^{\circ}

Oxygen is an element in its standard state, so its standard Gibbs energy of formation is zero. Its contribution to the reactant sum therefore vanishes:

Δ_{f} G_{O_{2}}^{\circ} = 0

Δ G_{r e a c t i o n}^{\circ} = [4 (- 394) + 5 (- 237)] - [(- 18) + \frac{13}{2} (0)]

4 (- 394) = - 1576, 5 (- 237) = - 1185

Δ G_{r e a c t i o n}^{\circ} = - 1576 - 1185 + 18 = - 2743 k J

The final positive 18 comes from subtracting a negative reactant formation energy. Do not add all three supplied negative values: apply the coefficients and the products-minus-reactants rule.

Why does one butane molecule transfer 26 electrons?

One butane molecule loses 26 electrons on complete oxidation. Count the total change in carbon oxidation numbers, not the number of atoms or a coefficient in the equation.

In butane, hydrogen contributes a total of positive ten. Because the molecule is neutral, the four carbon atoms have a combined oxidation number of negative ten:

\sum O N (H) = + 10, \sum O N (C) = - 10

In four carbon dioxide molecules, each carbon has oxidation number positive four. The increase in the combined carbon oxidation number gives the electrons lost:

\sum O N (C) = 4 (+ 4) = + 16

$+ 16 - (- 10) = 26$ Therefore, one mole of butane transfers 26 moles of electrons for the reaction as written. Cross-check with oxygen: each oxygen molecule gains four electrons. $n = 26$

\frac{13}{2} \times 4 = 26

The electron-transfer count is neither a reactant coefficient nor the number of oxygen atoms.

How do you calculate the cell potential and obtain X?

Δ G^{\circ} = - n F E^{\circ} \Rightarrow E^{\circ} = - \frac{Δ G^{\circ}}{n F}

Substituting the reaction energy and electron count gives the requested entry, 105.50. First convert kilojoules to joules because the Faraday constant is used in coulombs per mole of electrons.

Δ G^{\circ} = - 2743 \times 10^{3} J

Dividing energy by transferred charge gives joules per coulomb, which is volts. Use the reaction energy and electron count for the same one-mole butane basis:

1 J C^{- 1} = 1 V

E^{\circ} = - \frac{- 2743 \times 10^{3}}{26 F} = \frac{2743 \times 10^{3}}{26 F} V

Compare directly with the question’s expression. The Faraday constant and the power-of-ten factor cancel, so the numerical value of F is unnecessary:

E^{\circ} = \frac{X}{F} \times 10^{3} V

X = \frac{2743}{26} = 105.50

Final numerical answer: 105.50. This is X, not a cell potential of 105.50 volts. The sign checks: negative Gibbs energy for the forward reaction gives a positive standard cell potential.

Why would counting 13 electrons give a wrong answer?

Counting thirteen oxygen atoms as thirteen transferred electrons gives 211.00, a derived wrong numerical answer. This is a numerical-answer question with no supplied answer options. The value below comes from a hypothetical error, not an official distractor.

The hypothetical calculation keeps the correct reaction energy but uses the wrong electron count. It treats each oxygen atom as accepting only one electron: $n_{w r o n g} = 13$

X_{w r o n g} = \frac{2743}{13} = 211.00

Each oxygen atom actually starts at zero in elemental oxygen and ends at negative two in the products. It therefore gains two electrons, not one: $0 ⟶ - 2$

Recover the correct count by multiplying the number of oxygen atoms by the change per atom. Then use that count with the original reaction energy: $n = 13 \times 2 = 26$

X = \frac{2743}{26} = 105.50

What three practice questions check whether you understand the method?

Changing the reaction scale leaves the potential unchanged; reversing it changes the potential’s sign. The three questions below are practice variations built from the supplied data, not additional verified PYQs. Attempt each before reading its worked answer.

If the combustion equation is doubled, what happens to Gibbs energy, electron count and potential?

Both the reaction Gibbs energy and electron count double. Their ratio stays unchanged, so the cell potential does too:

Δ G^{\circ} = 2 (- 2743) = - 5486 k J, n = 52

E^{\circ} = \frac{5486 \times 10^{3}}{52 F} = \frac{2743 \times 10^{3}}{26 F} V

$X = 105.50$ Doubling only the energy while retaining the original electron count mixes two reaction scales. Scale both quantities together.

For the formal reverse reaction, what are the Gibbs energy and cell potential?

The Gibbs energy changes sign, while the magnitude of the electron-transfer count remains 26. The cell potential therefore changes sign:

Δ G^{\circ} = + 2743 k J, n = 26

E^{\circ} = - \frac{105.50}{F} \times 10^{3} V

The reverse direction is non-spontaneous under the stated standard conditions. Reversal changes direction, not the magnitude of energy or electron transfer.

What maximum electrical work can one mole of butane deliver reversibly?

The obtainable electrical-work magnitude equals the negative of the Gibbs energy change. This positive magnitude describes work delivered by the cell, while the reaction Gibbs energy is negative:

w_{e l e c t r i c a l, o b t a i n a b l e, m a x} = - Δ G^{\circ} = 2743 k J m o l^{- 1} o f b u t a n e

Before substituting in your next fuel-cell numerical, write the fuel quantity beside both the reaction energy and electron count. Use them together only when those quantities match.

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Frequently asked questions

What is the answer to the JEE Advanced 2025 butane fuel-cell question?

The numerical answer is X = 105.50. For one mole of butane, the reaction Gibbs energy is −2743 kJ and the electron-transfer count is 26, giving X = 2743/26. This is the value of X in the supplied expression, not a cell potential of 105.50 volts.

Why does butane transfer 26 electrons during combustion?

The four carbon atoms in butane have a combined oxidation number of −10. In four carbon dioxide molecules, their combined oxidation number is +16, so complete oxidation loses 26 electrons. Equivalently, the 13 oxygen atoms each gain two electrons, giving the same count.

How is the Gibbs energy change for butane combustion calculated?

For one mole of butane, the balanced combustion equation produces four moles of carbon dioxide and five moles of water. Using products minus reactants gives ΔG° = [4(−394) + 5(−237)] − (−18) = −2743 kJ. Oxygen contributes zero because its standard Gibbs energy of formation is zero.

Does doubling the combustion equation change the cell potential?

No: doubling the equation doubles both the reaction Gibbs energy and the electron-transfer count. For butane, these become −5486 kJ and 52, so their ratio in E° = −ΔG°/(nF) remains unchanged. The cell potential is unchanged, and X remains 105.50.