What is the answer to the NEET 2003 aluminium oxide electrolysis question?
The minimum required voltage is 2.14 V, option A. The key to this redox reactions and electrochemistry NEET 2003 question is matching the Gibbs energy and electron count to the same reaction amount: one mole of oxygen.
This Chemistry PYQ gives aluminium oxide formation as:
Formation releases 827 kJ of standard Gibbs energy per mole of oxygen consumed. Find the minimum external voltage needed to drive the reverse process.
The question bank classifies this question as hard and suggests a 90-second solving target. These are the question bank’s labels, not an official difficulty designation or exam allowance.
How do you reverse the reaction and change the Gibbs energy sign?
For electrolysis, reverse the formation reaction first. Its standard Gibbs energy changes from negative to positive because decomposition requires energy rather than releasing it. Keep the fractional coefficients unchanged: they preserve the supplied energy basis of one mole of oxygen.
For this reversed reaction as written:
This amount corresponds to producing one mole of oxygen, not decomposing one mole of aluminium oxide. The electrical input must supply this positive Gibbs energy of decomposition.
Keep the fractions for this calculation. Removing them is allowed, but the Gibbs energy and electron count must then be multiplied by the same factor.
Why are four electrons involved for one mole of oxygen?
One oxygen molecule contains two oxygen atoms, and each atom gains two electrons when its oxidation state changes from zero to negative two. The official solution therefore counts four electrons per oxygen molecule, or four moles of electrons per mole of oxygen.
Its electron-counting half-reaction is:
During electrolysis, this half-reaction runs in reverse:
Reversing it changes whether electrons are gained or lost. It does not change the magnitude of electron transfer.
The aluminium side confirms the count. Each aluminium ion needs three electrons, but the reaction produces four-thirds of a mole of aluminium:
Thus, for the supplied energy basis:
This count matches the energy quoted for one mole of oxygen. Choosing three from aluminium’s oxidation state alone ignores its coefficient.
How do you calculate the minimum required emf?
Answer: 2.14 V, option A. Divide the positive Gibbs energy of decomposition by the charge transferred. The official required-input method gives a positive voltage magnitude. Convert kilojoules to joules before substituting, because the Faraday constant is expressed in coulombs per mole.
The energy conversion is:
Substitute the matched energy and electron count:
The dimensional check is:
Do not confuse this required external voltage with signed cell emf. The standard cell convention is:
For decomposition, positive Gibbs energy therefore gives a negative cell emf:
The question asks for the positive magnitude of the applied voltage required. 2.14 V is the ideal thermodynamic minimum; practical electrolysis needs additional voltage to overcome losses and overpotentials.
Why does counting two electrons give option B, 4.28 V?
Counting two electrons uses the change for one oxygen atom, while retaining 827 kJ uses the energy for one mole of whole oxygen molecules. Those reaction amounts do not match. This error halves the denominator without changing the numerator, so the calculated voltage doubles.
The incorrect substitution is:
This is the doubled result represented by option B, listed as 4.28 V. Two electrons can be used consistently if the entire reaction is halved:
Rescaling changes the energy and charge, not the voltage. Before substitution, write which reaction amount both the Gibbs energy and electron count represent.
Which electrochemistry practice questions check this method?
Use these three checks: rescale the reaction, calculate the total charge, and reverse the emf sign for formation. These are original practice questions based on the same data, not additional verified NEET questions.
What voltage is needed to reverse the tripled formation reaction?
The minimum remains 2.14 V. Suppose formation is written as:
For its reversal, both the required energy and electron count triple:
Voltage does not change when the whole reaction is rescaled. The energy-to-charge ratio stays the same.
What theoretical charge completely electrolyses one mole of aluminium oxide?
The required charge is 579000 C. One mole of aluminium oxide contains two moles of aluminium ions:
Six moles of electrons therefore give:
This asks for charge, not voltage. Use the amount of aluminium oxide to count the total electrons transferred.
What is the signed standard cell emf for the original formation direction?
It is positive 2.14 V. Formation has negative standard Gibbs energy, so applying the signed relation gives:
Next, apply the same electron-counting method to oxide reduction and metal extraction: Metallurgy Practice Questions NEET: 5 Worked MCQs.
Frequently asked questions
What is the answer to the NEET 2003 aluminium oxide electrolysis question?
The minimum required voltage is 2.14 V, option A. For decomposition producing one mole of oxygen, use ΔG = +827000 J and n = 4, giving E = 827000/(4 × 96500). This is the ideal thermodynamic minimum; practical electrolysis requires additional voltage.
Why is n equal to 4 in the aluminium oxide electrolysis question?
The supplied Gibbs energy corresponds to one mole of oxygen molecules, which contains two moles of oxygen atoms. Each oxygen atom changes oxidation state by two units, so four moles of electrons are transferred per mole of oxygen. The aluminium count agrees: (4/3) × 3 = 4.
Why is 4.28 V the wrong answer for this electrolysis question?
Using n = 2 with 827 kJ mixes the electron count for one oxygen atom with the energy basis for one whole oxygen molecule. This halves the denominator and doubles the calculated voltage. If the entire reaction is halved consistently, ΔG becomes 413.5 kJ and n becomes 2, still giving 2.14 V.
Why is the required electrolysis voltage positive if the cell emf is negative?
Decomposition has positive Gibbs energy, so ΔG° = −nFE° gives a signed cell emf of −2.14 V. The question asks for the positive magnitude of the external voltage needed to drive decomposition, which is 2.14 V at the ideal thermodynamic limit.