Redox Reactions and Electrochemistry NEET 2000: 0.38 V

What is the answer to the Redox Reactions and Electrochemistry NEET 2000 copper question?

Option C, 0.38 V, is correct for the supplied Redox Reactions and Electrochemistry NEET 2000 question. The task is to calculate the standard potential when copper(I) ions disproportionate, forming copper(II) ions and copper metal:

2 {C u}^{+} \to {C u}^{2 +} + C u

The supplied standard reduction potentials are:

E_{{C u}^{2 +} / C u}^{\circ} = 0.34 V, E_{{C u}^{2 +} / {C u}^{+}}^{\circ} = 0.15 V

The options are:

This Chemistry PYQ is tagged hard by the question bank. Its 120-second expected solving time is the bank’s estimate, not an exam rule. First find the missing copper(I)-to-copper reduction potential.

Which copper(I) ion is oxidised, and which is reduced?

One copper(I) ion loses an electron; another gains that electron. Copper(I) therefore acts as both the reducing agent and the oxidising agent. This is disproportionation: the same initial oxidation state produces a higher and a lower oxidation state.

For oxidation, copper changes from plus one to plus two:

{C u}^{+} \to {C u}^{2 +} + e^{-}

For reduction, copper changes from plus one to zero:

{C u}^{+} + e^{-} \to C u

Adding these half-reactions cancels one electron:

2 {C u}^{+} \to {C u}^{2 +} + C u

The electron count is one, not two. Two copper(I) ions appear, but only one electron passes between them.

Both supplied values are reduction potentials. Neither directly gives the copper(I)-to-copper reduction potential needed for the second half-reaction.

How do you find the missing reduction potential using Gibbs energy?

Convert both supplied potentials into standard Gibbs energies, then subtract the reactions. Gibbs energies are additive and scale with reaction amounts. Electrode potentials are intensive quantities: they do not scale with coefficients and cannot simply be added to construct the missing half-reaction.

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

The electron count is: $n = number of electrons transferred$

The symbol F represents the Faraday constant. Its numerical value is unnecessary because it cancels later.

For the first supplied reduction:

{C u}^{2 +} + 2 e^{-} \to C u, n = 2

Δ G_{1}^{\circ} = - 2 F (0.34) = - 0.68 F

For the second supplied reduction:

{C u}^{2 +} + e^{-} \to {C u}^{+}, n = 1

Δ G_{2}^{\circ} = - F (0.15) = - 0.15 F

Subtracting the second reaction means adding its reverse. Reversal changes the sign of its Gibbs energy:

{C u}^{+} \to {C u}^{2 +} + e^{-}, Δ G_{reverse 2}^{\circ} = + 0.15 F

Adding this to the first reaction gives:

{C u}^{2 +} + 2 e^{-} + {C u}^{+} \to C u + {C u}^{2 +} + e^{-}

Cancel one copper(II) ion and one electron from each side. The remaining reaction is the missing reduction:

{C u}^{+} + e^{-} \to C u

Its Gibbs energy is:

Δ G_{3}^{\circ} = - 0.68 F - (- 0.15 F) = - 0.53 F

The remaining half-reaction transfers one electron. Convert its Gibbs energy back to a potential using that count:

n = 1, E_{{C u}^{+} / C u}^{\circ} = - \frac{Δ G_{3}^{\circ}}{F} = 0.53 V

How do you calculate the disproportionation potential from 0.53 V?

Add the newly found reduction potential to the oxidation potential for copper(I) forming copper(II). The supplied copper(II)-to-copper(I) potential is a reduction potential, so reverse its sign once when reversing that half-reaction.

The reduction half-reaction is:

{C u}^{+} + e^{-} \to C u, E_{r e d}^{\circ} = + 0.53 V

The oxidation half-reaction is:

{C u}^{+} \to {C u}^{2 +} + e^{-}, E_{o x}^{\circ} = - 0.15 V

Therefore:

E_{c e l l}^{\circ} = E_{r e d}^{\circ} + E_{o x}^{\circ} = 0.53 + (- 0.15) = + 0.38 V

Correct option: C, 0.38 V. The equivalent convention uses both values as reduction potentials:

E_{c e l l}^{\circ} = E_{c a t h o d e, r e d}^{\circ} - E_{a n o d e, r e d}^{\circ} = 0.53 - 0.15

Do not subtract an already reversed oxidation potential. That would reverse its sign twice.

The thermodynamic check uses one transferred electron:

n = 1, Δ G^{\circ} = - F (0.38) < 0

The negative Gibbs-energy change means the forward disproportionation is thermodynamically favourable under standard conditions. This is a check on the reaction’s direction, not its speed.

Why does adding the supplied potentials give the wrong option A?

The supplied values describe two reductions starting from copper(II), not the reduction and oxidation half-reactions of the requested disproportionation. Adding them produces option A, but the arithmetic has no valid reaction basis:

Incorrect: 0.34 + 0.15 = 0.49 V

The deeper mistake is treating potential as an additive reaction quantity. It also ignores that the first supplied reduction transfers two electrons while the second transfers one. Those electron counts belong in the Gibbs-energy conversion.

Use this sequence instead:

Convert each potential to Gibbs energy using its electron count.
Subtract reactions to isolate the missing reduction.
Recover its potential, 0.53 V.
Calculate the target potential:

0.53 - 0.15 = 0.38 V

Before any arithmetic, write the half-reactions and count their electrons.

What changes if you reverse or double the copper reaction?

Reversing the reaction reverses the potential’s sign. Doubling the equation leaves the potential unchanged but doubles the electron count and Gibbs-energy change. These are original practice questions based on the supplied data, not additional verified PYQs.

What is the potential for copper(II) and copper forming copper(I)?

The standard potential is −0.38 V because this reverses the original reaction. Copper(II) now undergoes reduction, while copper metal undergoes oxidation.

Practice question 1: Using the same supplied potentials, calculate the standard potential for:

{C u}^{2 +} + C u \to 2 {C u}^{+}

Subtract the copper(I)-to-copper reduction potential from the copper(II)-to-copper(I) reduction potential:

E^{\circ} = 0.15 - 0.53 = - 0.38 V

One electron is still transferred:

n = 1, Δ G^{\circ} = - F (- 0.38) = + 0.38 F

The reversed reaction is not thermodynamically favourable under standard conditions. This does not apply to every possible concentration: non-standard conditions can change the reaction’s Gibbs energy.

What happens when all coefficients in the reaction are doubled?

The standard potential stays +0.38 V; the electron count and Gibbs-energy change double. Both scale by the same factor, leaving their ratio unchanged.

Practice question 2: Determine the standard potential, electron count and standard Gibbs-energy change for:

4 {C u}^{+} \to 2 {C u}^{2 +} + 2 C u

This doubles the original reaction without reversing it. Two electrons now pass between the oxidised and reduced copper ions:

E^{\circ} = + 0.38 V, n = 2

Δ G^{\circ} = - 2 F (0.38) = - 0.76 F

Before recalculating, mark the operation. Reversing a reaction reverses the potential’s sign; multiplying its coefficients does not multiply the potential.

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Frequently asked questions

What is the answer to the NEET 2000 copper disproportionation question?

Option C, 0.38 V, is correct for the supplied question. The missing Cu+/Cu reduction potential is 0.53 V, so the standard disproportionation potential is 0.53 − 0.15 = 0.38 V.

How do I calculate the missing Cu+/Cu reduction potential?

Use ΔG° = −nFE° to convert the supplied reduction potentials into Gibbs energies: −0.68F for Cu2+/Cu and −0.15F for Cu2+/Cu+. Subtract the second half-reaction from the first to obtain Cu+ + e− → Cu, with ΔG° = −0.53F. Since this half-reaction transfers one electron, its standard reduction potential is 0.53 V.

Why can't I add 0.34 V and 0.15 V to get the answer?

Both supplied potentials describe reductions starting from Cu2+, not the oxidation and reduction half-reactions required for disproportionation. Electrode potentials are not additive reaction quantities; combine Gibbs energies with the correct electron counts to find the missing potential first.

Why is the electron count one in copper(I) disproportionation?

In 2Cu+ → Cu2+ + Cu, one Cu+ ion loses one electron and the other gains that same electron. Only one electron is transferred, so n = 1 even though two Cu+ ions appear in the equation.

Does doubling the copper reaction double its electrode potential?

No: doubling the reaction leaves the standard potential at +0.38 V. The electron count doubles to two and ΔG° doubles to −0.76F, leaving the ratio −ΔG°/(nF) unchanged.

Redox Reactions and Electrochemistry NEET 2000: Copper Potential