Redox Reactions and Electrochemistry NEET 2008: Fuel Cell EMF

What is the answer to the Redox Reactions and Electrochemistry NEET 2008 fuel-cell question?

Option C, 1.0968 V, is correct. The Redox Reactions and Electrochemistry NEET 2008 pentane–oxygen question requires one consistency check: the Gibbs energy and electron count must refer to the same balanced reaction.

Calculate the standard EMF of a cell using pentane and oxygen, given these standard Gibbs energies of formation at 298 K:

Liquid water:

H_{2} O (l) : - 237.2 k J m o l^{- 1}

Carbon dioxide gas:

C O_{2} (g) : - 394.4 k J m o l^{- 1}

Pentane gas:

C_{5} H_{12} (g) : - 8.2 k J m o l^{- 1}

The answer choices are:

How do you balance the pentane–oxygen reaction?

Use one mole of pentane as the reaction basis. Its five carbon atoms require five carbon dioxide molecules, and its twelve hydrogen atoms require six water molecules. The products contain sixteen oxygen atoms, requiring eight oxygen molecules.

C_{5} H_{12} (g) + 8 O_{2} (g) \to 5 C O_{2} (g) + 6 H_{2} O (l)

Keep water liquid and pentane gaseous. The supplied formation energies apply to those states, not to steam or liquid pentane.

Every Gibbs-energy change and electron count in the main calculation refers to this reaction as written, consuming one mole of pentane. Using the same basis guards against a factor-of-two error.

How do you calculate the standard Gibbs energy without a sign error?

The standard Gibbs-energy change is negative 3387.0 kJ per mole of reaction as written. Calculate products minus reactants, including each balancing coefficient. Subtracting pentane’s negative formation energy contributes a positive term.

Δ G^{\circ} = \sum_{products} ν Δ G_{f}^{\circ} - \sum_{reactants} ν Δ G_{f}^{\circ}

Oxygen gas has zero standard Gibbs energy of formation because it is oxygen in its standard elemental state. This does not mean oxygen can generally be ignored in energy calculations.

Δ G^{\circ} = [5 (- 394.4) + 6 (- 237.2)] - [(- 8.2) + 8 (0)]

The product contributions are calculated separately. Subtracting the negative pentane term then changes its sign: $5 (- 394.4) = - 1972.0$ $6 (- 237.2) = - 1423.2$ $- (- 8.2) = + 8.2$

Δ G^{\circ} = - 1972.0 - 1423.2 + 8.2 = - 3387.0 k J m o l^{- 1}

The negative result indicates that the forward reaction is thermodynamically spontaneous under standard conditions. Keep that sign when applying the EMF relation.

Why are 32 electrons transferred in pentane combustion?

One pentane molecule releases 32 electrons on complete oxidation. Count the total change in carbon oxidation numbers rather than treating every carbon atom as identical.

In neutral pentane, the twelve hydrogens contribute a total of positive twelve. The five carbon oxidation numbers must therefore sum to negative twelve:

\sum O N (C) = - 12

In the five carbon dioxide molecules, each carbon has oxidation number positive four. The increase in the total gives the electrons released:

\sum O N (C) = 5 (+ 4) = + 20

20 - (- 12) = 32 \Rightarrow n = 32

Cross-check using oxygen: eight oxygen molecules contain sixteen atoms. Each changes from zero to negative two, accepting two electrons: $16 \times 2 = 32$

The average carbon oxidation number in pentane is: $- \frac{12}{5}$

That average does not mean all five carbon atoms have identical oxidation numbers. Do not add electrons lost to electrons gained: both counts describe the same transferred electrons.

How do you obtain 1.0968 V from the Gibbs energy?

Convert kilojoules to joules, then divide the negative of the Gibbs-energy change by the electron count and Faraday constant. Using 32 electrons and the supplied Faraday constant gives a positive EMF slightly greater than one volt.

Δ G^{\circ} = - n F E_{c e l l}^{\circ}

E_{c e l l}^{\circ} = - \frac{Δ G^{\circ}}{n F}

Δ G^{\circ} = - 3387.0 k J m o l^{- 1} = - 3387 \times 10^{3} J m o l^{- 1}

F = 96500 C m o l^{- 1}, n = 32

E_{c e l l}^{\circ} = \frac{3387000}{32 \times 96500} = \frac{3387000}{3088000} \approx 1.0968 V

Before selecting option C, make three checks:

Units: joules per coulomb give volts.

\frac{J}{C} = V

Sign: negative Gibbs-energy change gives positive EMF for the forward reaction under standard conditions.
Scale: 3.387 million divided by 3.088 million must be slightly greater than one, not close to zero.

How can a division mistake produce option D, 0.0968 V?

Discarding the integer part of the quotient produces option D. This is an illustrative error route, not a documented account of student responses or a claim about how the examiner designed the distractor.

Split the numerator into one full denominator plus a remainder. The quotient therefore contains an integer part and a fractional part: $3387000 = 3088000 + 299000$

\frac{3387000}{3088000} = 1 + \frac{299000}{3088000} \approx 1 + 0.0968

Keeping only the remainder divided by the denominator gives 0.0968 V. That remainder supplies the fractional part, not the entire answer. The missing one volt is an arithmetic error, not an electrochemistry error.

Compare numerator and denominator before long division. A numerator larger than its positive denominator cannot produce a quotient below one. That check rejects D immediately after the correct substitution.

What two related questions test whether you understand the method?

Halving the reaction leaves EMF unchanged; reversing it changes the EMF’s sign. These are original related practice questions, not additional verified NEET PYQs. They test whether your Gibbs energy and electron count use the same reaction basis.

If every reaction coefficient is halved, what are the Gibbs energy, electron count and EMF?

The Gibbs-energy change and electron count both halve, but the standard EMF remains 1.0968 V. The new quantities refer to the halved reaction, which consumes half a mole of pentane:

Δ G^{\circ} = - 1693.5 k J m o l^{- 1}, n = 16

E_{c e l l}^{\circ} = \frac{1693500}{16 \times 96500} \approx 1.0968 V

Scaling the reaction changes the numerator and denominator by the same factor. EMF does not scale with the amount reacted, so halving the voltage would be incorrect.

If the original reaction is reversed, what are the Gibbs energy and EMF, and is it spontaneous?

The Gibbs-energy change becomes positive and the standard EMF becomes negative, so the reverse direction is non-spontaneous under standard conditions. The electron-count magnitude remains 32: reversing electron flow does not change how many electrons are transferred.

Δ G^{\circ} = + 3387.0 k J m o l^{- 1}, n = 32

E_{c e l l}^{\circ} = - \frac{3387000}{32 \times 96500} \approx - 1.0968 V

Cover both worked answers and reproduce the scaling and sign changes before doing any division. Check that the Gibbs energy and electron count refer to the same written reaction each time.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

For a worked example of the same idea, see Silvered Lens NEET 2000: Why the Answer Is 10 cm.

Frequently asked questions

What is the answer to the NEET 2008 pentane fuel-cell question?

Option C, 1.0968 V, is correct. For the balanced reaction consuming one mole of pentane, ΔG° = −3387.0 kJ mol⁻¹ and n = 32. Converting Gibbs energy to joules and using E° = −ΔG°/(nF), with F = 96500 C mol⁻¹, gives 1.0968 V.

Why is n equal to 32 in the pentane fuel cell?

The carbon oxidation numbers in one pentane molecule sum to −12, while those in the five CO₂ molecules sum to +20. The increase is 20 − (−12) = 32, so 32 electrons are transferred. Oxygen accepts those same 32 electrons; do not add electrons lost to electrons gained.

Why is 0.0968 V the wrong answer for the pentane fuel cell?

The correct division is 3387000/3088000, which equals approximately 1 + 0.0968. Keeping only the fractional part gives 0.0968 V and discards one full volt. Since the numerator exceeds the positive denominator, the answer must be greater than 1 V.

Does halving the balanced reaction halve the cell EMF?

No. Halving every reaction coefficient halves both the Gibbs-energy change and the electron count, leaving their ratio unchanged. For the pentane fuel cell, ΔG° becomes −1693.5 kJ mol⁻¹ and n becomes 16, but the standard EMF remains 1.0968 V.