Redox Reactions and Electrochemistry NEET 2009: Copper

What is the answer to the copper-potential question in NEET 2009?

Option A, 0.52 V, is correct for the redox reactions and electrochemistry NEET 2009 copper question. The worked calculation gives 0.521 V: add Gibbs free energy changes, not electrode potentials.

The question gives two standard reduction potentials. Copper(II) can reduce directly to copper metal, or accept one electron to form copper(I):

C u^{2 +} + 2 e^{-} \to C u, E_{o v e r a l l}^{\circ} = 0.337 V

C u^{2 +} + e^{-} \to C u^{+}, E_{1}^{\circ} = 0.153 V

Find the unknown standard reduction potential:

C u^{+} + e^{-} \to C u, E^{\circ} = ?

The supplied options are:

The direct reduction transfers two electrons. Each component step transfers one, so their potentials cannot be added directly.

Why must we add Gibbs energies instead of electrode potentials?

Gibbs free energy changes add because they depend only on the initial and final states, not the reaction path. Electrode potentials measure energy change per unit charge, so electron counts must be included before combining them.

The first step forms copper(I). The required step consumes it:

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

C u^{+} + e^{-} \to C u, n_{2} = 1, E_{2}^{\circ} = E^{\circ}

Add the equations and cancel copper(I), which appears on opposite sides. The two electrons remain on the reactant side:

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

$n_{o v e r a l l} = 2$ Connect each potential to its Gibbs energy change: $Δ G^{\circ} = - n F E^{\circ}$

Here, the electron count is for the half-reaction as written. The Faraday constant is the charge per mole of electrons:

n = electron count, F = Faraday constant

The count of two belongs only to the overall reaction, not either individual step. Multiplying each potential by its transferred charge converts it into an energy change that can be added.

How does the Gibbs-energy calculation give option A?

The unknown potential is 0.521 V, which rounds to option A, 0.52 V. Apply the Gibbs-energy relation separately to all three half-reactions, using two electrons for the overall reaction and one for each component step.

For the overall reduction:

Δ G_{o v e r a l l}^{\circ} = - 2 F \times 0.337

For the first step: $Δ G_{1}^{\circ} = - F \times 0.153$

For the required step: $Δ G_{2}^{\circ} = - F \times E^{\circ}$

The reaction sum requires:

Δ G_{o v e r a l l}^{\circ} = Δ G_{1}^{\circ} + Δ G_{2}^{\circ}

Substitute without changing any signs:

- 2 F \times 0.337 = - F \times 0.153 - F \times E^{\circ}

The minus signs come from the Gibbs-energy relation. Neither component reaction has been reversed, so no extra sign reversal is needed.

Cancel the common negative Faraday factor: $2 \times 0.337 = 0.153 + E^{\circ}$ $0.674 = 0.153 + E^{\circ}$

E^{\circ} = 0.674 - 0.153 = 0.521 V

E^{\circ} \approx 0.52 V Option A

No numerical value of the Faraday constant is required, because it cancels. Check the result against the given overall potential:

\frac{0.153 + 0.521}{2} = 0.337 V

This arithmetic average works because both component steps transfer one electron. It is a check based on equal electron counts, not permission to average arbitrary electrode potentials.

How can a reaction-scaling mistake lead to option C?

Copying the overall potential to the required reaction can lead to option C through a nearest-option choice. This is a possible error route, not a claim about how students actually answered.

The faulty method treats copper(I) reduction as a rescaled version of copper(II) reduction and copies 0.337 V. Among the listed choices, C is then closer than D:

| 0.337 - 0.30 | = 0.037 V

| 0.38 - 0.337 | = 0.043 V

0.30 V is not the correctly rounded value of 0.337 V. It is a nearest-option selection after a method error.

Halving the original reaction gives:

\frac{1}{2} C u^{2 +} + e^{-} \to \frac{1}{2} C u

It does not give: $C u^{+} + e^{-} \to C u$

Multiplying an unchanged reaction scales its Gibbs energy change and electron count together, leaving its potential unchanged. Changing the oxidation-state couple is different chemistry, not reaction scaling.

Check the species before comparing coefficients. Use NEET Silly Mistakes: Find the Cause and Fix the Check to separate a concept error from a rounding error.

What happens if the copper half-reaction is doubled?

The potential remains 0.337 V, while the Gibbs energy change doubles. This is an original practice variation, not another verified PYQ.

Given the original reduction below, find the standard potential and Gibbs energy change when its equation is doubled:

C u^{2 +} + 2 e^{-} \to C u, E^{\circ} = 0.337 V

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

Start with the original Gibbs energy change: $Δ G^{\circ} = - 2 F \times 0.337$

Doubling every coefficient doubles the Gibbs energy change and electron count:

Δ G^{\circ'} = 2 Δ G^{\circ} = - 4 F \times 0.337, n^{'} = 4

Recover the potential from the doubled equation:

E^{\circ'} = - \frac{Δ G^{\circ'}}{4 F} = - \frac{- 4 F \times 0.337}{4 F} = 0.337 V

The numerator and denominator increase by the same factor, so Gibbs energy doubles but electrode potential does not. No new electrode-potential data are needed because the reaction has only been scaled.

What changes when the copper(I) reduction is reversed?

The standard oxidation potential becomes negative 0.521 V, approximately negative 0.52 V. This original practice variation uses the reduction potential just derived.

Find the standard oxidation potential for copper forming copper(I), given:

C u^{+} + e^{-} \to C u, E_{r e d u c t i o n}^{\circ} = 0.521 V

Start with its Gibbs energy change:

Δ G_{r e d u c t i o n}^{\circ} = - F \times 0.521

Reverse the reaction and the Gibbs-energy sign: $C u \to C u^{+} + e^{-}$

Δ G_{o x i d a t i o n}^{\circ} = + F \times 0.521

The electron count remains positive and equal to one. Apply the same relation:

Δ G^{\circ} = - n F E^{\circ}, n = 1

E_{o x i d a t i o n}^{\circ} = - \frac{+ F \times 0.521}{F} = - 0.521 V \approx - 0.52 V

Before calculating your next answer, classify the operation:

Scaling an unchanged reaction: leave its potential unchanged.
Reversing a reaction: change the potential’s sign.
Combining different steps: add their Gibbs energy changes first.

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Frequently asked questions

What is the answer to the NEET 2009 copper electrode potential question?

Option A, 0.52 V, is correct for the reduction of Cu+ to Cu. Using ΔG° = −nFE°, the unknown potential is 2 × 0.337 − 0.153 = 0.521 V, which rounds to 0.52 V.

Why can't we add standard electrode potentials directly?

Electrode potentials represent energy change per unit charge, so the electron count matters when combining reaction steps. Convert each potential to a Gibbs energy change using ΔG° = −nFE°, add those energy changes, and then calculate the resulting potential.

Can we average the two copper reduction potentials?

The overall potential is the arithmetic average of the two component potentials here because each component step transfers one electron. The check is (0.153 + 0.521)/2 = 0.337 V. This does not justify averaging arbitrary electrode potentials.

Does doubling a half-reaction double its electrode potential?

No, doubling an unchanged half-reaction doubles both its Gibbs energy change and its electron count. Their ratio in E° = −ΔG°/(nF) remains unchanged, so the copper(II)-to-copper potential stays at 0.337 V.

What happens to electrode potential when a half-reaction is reversed?

Reversing a half-reaction changes the sign of its electrode potential. Since Cu+ reduction to Cu has a standard potential of +0.521 V, the reverse oxidation of Cu to Cu+ has a standard potential of −0.521 V.