Standard Electrode Potential NEET 2026: Iron Reduction

What is the answer to the standard electrode potential NEET 2026 question?

Option B, +0.76 V, is correct for this standard electrode potential NEET 2026 question. The calculation requires Gibbs-energy changes, not direct subtraction of the given potentials: $Δ G^{\circ} = - n F E^{\circ}$

At 298 K, the standard reduction potentials for iron(III) to iron metal and iron(II) to iron metal are −0.04 V and −0.44 V, respectively. Find the standard potential for reducing iron(III) to iron(II).

The supplied options are:

The supplied question bank classifies this as medium and gives an expected solving time of 90 seconds. These are the bank’s estimates, not an official difficulty rating or a measured student average.

How do you write the half-reactions and count electrons?

The required reduction transfers one electron. Adding it to the iron(II)-to-metal reduction gives the iron(III)-to-metal reduction, establishing the reaction relationship before any arithmetic.

Reaction 1:

F e^{3 +} + 3 e^{-} \to F e

E_{1}^{\circ} = - 0.04 V, n_{1} = 3

Reaction 2:

F e^{2 +} + 2 e^{-} \to F e

E_{2}^{\circ} = - 0.44 V, n_{2} = 2

Required reaction:

F e^{3 +} + e^{-} \to F e^{2 +}, n = 1

Iron changes from oxidation state +3 to +2. It gains one electron, not three or two.

Add the required reaction and reaction 2:

F e^{3 +} + e^{-} + F e^{2 +} + 2 e^{-} \to F e^{2 +} + F e

Cancel the iron(II) ion appearing on both sides. Combine the one electron and the two electrons on the left:

F e^{3 +} + 3 e^{-} \to F e

The result is reaction 1. Use this balanced relationship to decide which energy changes to subtract.

How do you convert the given potentials into Gibbs-energy changes?

Convert each potential using its own electron count, then subtract the energies. Gibbs-energy changes add when reactions add; electrode potentials do not. Potential is an intensive quantity: it does not scale with the amount of reaction.

Use the relation: $Δ G^{\circ} = - n F E^{\circ}$

The symbols mean: $n = stoichiometric electron count$ $F = Faraday constant, the charge per mole of electrons$

Apply the equation separately to the two reductions:

Δ G_{1}^{\circ} = - 3 F (- 0.04) = + 0.12 F

Δ G_{2}^{\circ} = - 2 F (- 0.44) = + 0.88 F

These expressions represent energies, not potentials. The voltage factors are implicit in this shorthand: multiplying the Faraday constant by volts gives energy per mole of reaction.

Because reaction 1 equals the required reaction plus reaction 2:

Δ G_{1}^{\circ} = Δ G_{r e q u i r e d}^{\circ} + Δ G_{2}^{\circ}

Subtract reaction 2’s energy from reaction 1’s energy:

Δ G_{r e q u i r e d}^{\circ} & = Δ G_{1}^{\circ} - Δ G_{2}^{\circ} & = 0.12 F - 0.88 F & = - 0.76 F

Keep the Faraday constant symbolic because it cancels when converting back to potential. Substituting its numerical value adds unnecessary arithmetic.

How do you recover the potential and confirm option B?

Use the target reaction’s one-electron count to convert its energy back into a potential. The negative energy change gives a positive reduction potential, confirming option B.

For the required reduction:

F e^{3 +} + e^{-} \to F e^{2 +}, n = 1

Substitute the calculated energy: $- 1 \times F \times E^{\circ} = - 0.76 F$

Divide both sides by the negative Faraday constant: $E^{\circ} = + 0.76 V$

A compact check follows directly from the energy calculation:

E_{r e q u i r e d}^{\circ} & = \frac{3 (- 0.04) - 2 (- 0.44)}{1} V & = (- 0.12 + 0.88) V & = + 0.76 V

The factors three and two come from converting potentials into energies. They do not multiply the physical electrode potentials. The negative calculated Gibbs-energy change and positive potential are consistent with: $Δ G^{\circ} = - n F E^{\circ}$

Why does direct subtraction give the wrong option A?

Direct subtraction gives option A because it treats two potentials as additive reaction energies. It ignores that one supplied reduction transfers three electrons and the other transfers two.

The incorrect calculation is:

- 0.04 - (- 0.44) = + 0.40 V (incorrect here)

The familiar cell equation is still valid:

E_{c e l l}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}

That equation gives the voltage of a complete cell using its cathode and anode reduction potentials. This question instead asks you to derive the potential of a specified one-electron half-reaction. Subtracting the supplied values does not perform that derivation.

Use this repair sequence rather than memorising a special iron formula:

Write balanced half-reactions.
Record each electron count.
Calculate each Gibbs-energy change.
Combine the energies according to the reaction relationship.
Convert back using the target reaction’s electron count.

Can you solve three related electrode-potential questions?

Use the same energy method for reaction reversal, reaction scaling and rebuilding the original reduction. All three checks below are original practice questions based on the supplied iron half-reactions, not additional verified PYQs.

Try each before reading its worked answer. Keep the electron count beside the reaction throughout.

What is the potential when iron(II) is oxidised to iron(III)?

Answer: −0.76 V. Reversing the reaction reverses the Gibbs-energy change and the sign of the potential:

F e^{2 +} \to F e^{3 +} + e^{-}

Δ G^{\circ} = + 0.76 F, E^{\circ} = - \frac{0.76 F}{F} = - 0.76 V

The electron-count magnitude remains one. This is the oxidation potential for the written reaction, not the reduction potential for its reverse.

What happens if the target reduction is doubled?

Answer: +0.76 V, unchanged. Doubling the reaction doubles both the energy change and the electron count:

2 F e^{3 +} + 2 e^{-} \to 2 F e^{2 +}

Δ G^{\circ} = - 1.52 F, n = 2

E^{\circ} = - \frac{- 1.52 F}{2 F} = + 0.76 V

Can you recover the iron(III)-to-metal reduction potential?

Answer: −0.04 V. Add the energy changes of the two successive reductions, using these given potentials:

E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.76 V

E^{\circ} (F e^{2 +} / F e) = - 0.44 V

The reactions add to the three-electron reduction:

Δ G^{\circ} = - F (0.76) - 2 F (- 0.44) = + 0.12 F

E^{\circ} = - \frac{0.12 F}{3 F} = - 0.04 V

Recovering the supplied value checks both the electron count and the signs. Next, find and verify chapterwise NEET PYQs and write the energy calculation before checking each solution.

Frequently asked questions

How do you calculate the standard potential for Fe3+ to Fe2+?

Use ΔG° = −nFE° to combine reaction energies, not electrode potentials. With the supplied Fe3+/Fe and Fe2+/Fe reduction potentials of −0.04 V and −0.44 V, the required potential is [3(−0.04) − 2(−0.44)]/1 = +0.76 V. This gives option B.

Why is +0.40 V wrong in this electrode potential question?

Direct subtraction ignores that the supplied reductions transfer three and two electrons, respectively. Gibbs-energy changes add when reactions add, but electrode potentials do not. Convert each potential into ΔG°, subtract the energies, then convert back using the target reaction's one-electron count.

How many electrons are transferred when Fe3+ becomes Fe2+?

One electron is transferred: Fe3+ + e− → Fe2+. The oxidation state decreases from +3 to +2, so use n = 1 when converting the target reaction's Gibbs-energy change into its potential.

Does doubling a half-reaction double its electrode potential?

No, electrode potential is an intensive quantity and does not change when a reaction is doubled. Both ΔG° and the electron count n double, leaving E° = −ΔG°/(nF) unchanged. For the iron(III)-to-iron(II) reduction calculated here, the potential remains +0.76 V.