Redox Reactions and Electrochemistry JEE 2013: Full Solution

What was asked in the Redox Reactions and Electrochemistry JEE 2013 question?

The redox reactions and electrochemistry JEE 2013 match-list question has option D as its correct answer: P–3, Q–4, R–1, S–2. It appeared in JEE Advanced 2013 Paper 2, Chemistry, under Redox Reactions and Electrochemistry. The question bank assigns it the hard difficulty tier and an expected solving time of 240 seconds.

The four unknown quantities were:

P: standard potential for

E^{\circ} (F e^{3 +} / F e)

Q: standard potential for

4 H_{2} O ⇌ 4 H^{+} + 4 O H^{-}

R: standard potential for

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

S: standard potential for

E^{\circ} (C r^{3 +} / C r^{2 +})

The supplied standard reduction potentials were:

E^{\circ} (F e^{3 +} / F e^{2 +}) = + 0.77 V

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

E^{\circ} (C u^{2 +} / C u^{+}) = + 0.34 V

E^{\circ} (C u^{+} / C u) = + 0.52 V

E^{\circ} (O_{2} + 4 H^{+} + 4 e^{-} \to 2 H_{2} O) = + 1.23 V

E^{\circ} (O_{2} + 2 H_{2} O + 4 e^{-} \to 4 O H^{-}) = + 0.40 V

E^{\circ} (C r^{3 +} / C r) = - 0.74 V

E^{\circ} (C r^{2 +} / C r) = - 0.91 V

List II contained:

The four code choices were:

What rule should you use to combine electrode potentials?

Combine reactions through standard Gibbs energy, not by directly adding electrode potentials. In the redox reactions and electrochemistry JEE 2013 problem, this is essential for iron and chromium because their successive reductions involve different electron counts. Standard Gibbs energy is extensive and additive, while electrode potential is intensive. $Δ G^{\circ} = - n F E^{\circ}$

An intensive quantity cannot generally be added or averaged. Convert each potential into Gibbs energy, add the Gibbs energies, and convert the total back into a potential using the total number of transferred electrons.

Reversing a half-reaction reverses the sign of its standard Gibbs energy. Its electrode-potential contribution also changes sign. Use the Faraday constant symbolically because it cancels, so no numerical value is required.

This differs from the valid cell relation:

E_{c e l l}^{\circ} = E_{r e d u c t i o n}^{\circ} + E_{o x i d a t i o n}^{\circ} = E_{c a t h o d e}^{\circ} - E_{a n o d e}^{\circ}

That relation combines a reduction and an oxidation after balancing the electrons. It does not permit direct addition of successive reduction potentials with different electron counts.

How are the iron and water potentials calculated?

For P, convert the one-electron and two-electron iron reductions into Gibbs energies before combining them. For Q, retain the alkaline oxygen reduction and reverse the acidic oxygen reduction. Both oxygen half-reactions involve four electrons, so their reduction and oxidation contributions can be subtracted directly after reversal.

How do you calculate P from the two iron reduction steps?

P equals approximately −0.04 V because the two iron steps must be weighted by their electron counts. The first reduction transfers one electron, while the second transfers two. Adding or averaging their potentials without this weighting gives the wrong result.

The two steps are:

F e^{3 +} + e^{-} \to F e^{2 +} E^{\circ} = + 0.77 V

F e^{2 +} + 2 e^{-} \to F e E^{\circ} = - 0.44 V

Convert both potentials into Gibbs energies:

Δ G_{t o t a l}^{\circ} = - F (0.77) - 2 F (- 0.44)

Δ G_{t o t a l}^{\circ} = - 0.77 F + 0.88 F = 0.11 F

The overall reduction transfers three electrons:

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

Therefore:

E^{\circ} = - \frac{0.11 F}{3 F} = - 0.0367 V \approx - 0.04 V

Thus, P maps to (3), not to the unweighted average of +0.77 V and −0.44 V.

How do you calculate Q from acidic and alkaline oxygen reductions?

Q equals −0.83 V. Keep the alkaline oxygen reduction as written and reverse the acidic oxygen reduction. Because both involve four electrons, the oxygen and electrons cancel, leaving the required water equilibrium.

Use the alkaline reduction:

O_{2} + 2 H_{2} O + 4 e^{-} \to 4 O H^{-} E^{\circ} = + 0.40 V

Reverse the acidic oxygen reduction, whose forward reduction potential is +1.23 V:

2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-} E_{o x i d a t i o n}^{\circ} = - 1.23 V

Adding the half-reactions cancels oxygen and electrons: $4 H_{2} O \to 4 H^{+} + 4 O H^{-}$

Hence:

E^{\circ} = 0.40 - 1.23 = - 0.83 V

Thus, Q maps to (4).

How are copper and chromium matched to the final code?

R equals −0.18 V and S equals −0.40 V. Copper is a reduction-plus-oxidation combination with equal electron counts, so direct subtraction is valid. Chromium involves one-electron and two-electron stages, so it must be solved by adding Gibbs energies.

How do you calculate R for the copper comproportionation reaction?

R equals −0.18 V. Use the reduction of copper(II) to copper(I), then reverse the reduction of copper(I) to copper. The reduction and oxidation each transfer one electron, so their potential contributions can be added directly.

Use:

C u^{2 +} + e^{-} \to C u^{+} E^{\circ} = + 0.34 V

Reverse the second reduction:

C u \to C u^{+} + e^{-} E_{o x i d a t i o n}^{\circ} = - 0.52 V

Adding them gives:

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

Since the electron counts are equal:

E^{\circ} = 0.34 - 0.52 = - 0.18 V

Thus, R maps to (1).

How do you calculate S from the two chromium potentials?

S equals −0.40 V. Treat the reduction of chromium(III) to chromium as the sum of a one-electron reduction to chromium(II) and a two-electron reduction from chromium(II) to chromium. Combine their Gibbs energies, not their potentials.

Let:

E^{\circ} (C r^{3 +} / C r^{2 +}) = x

The unknown half-reaction is:

C r^{3 +} + e^{-} \to C r^{2 +}

For the complete three-electron reduction:

C r^{3 +} + 3 e^{-} \to C r

Δ G^{\circ} = - 3 F (- 0.74) = 2.22 F

For the two-electron reduction:

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

Δ G^{\circ} = - 2 F (- 0.91) = 1.82 F

Additive Gibbs energies give: $2.22 F = - F x + 1.82 F$

Therefore: $- F x = 0.40 F$ $x = - 0.40 V$

Thus, S maps to (2).

The final audit is:

P = −0.04 V → 3
Q = −0.83 V → 4
R = −0.18 V → 1
S = −0.40 V → 2

The final code is P–3, Q–4, R–1, S–2, so the correct choice is option D.

Why does row-wise matching give the wrong option C?

Option C is P–1, Q–2, R–3, S–4. It merely pairs each List I entry with the value printed on the same row of List II. A match-list layout is not pre-aligned. Every quantity must be calculated first and then converted into its List II label.

P alone contradicts option C:

E^{\circ} (F e^{3 +} / F e) \approx - 0.04 V

The value −0.04 V has label 3. Therefore, P must map to label 3 rather than label 1.

Use a two-column audit after completing the calculations:

First column: calculated numerical potential.
Second column: corresponding List II label.

This prevents a correct numerical calculation from becoming an incorrect match-list code.

Which related electrochemistry questions test the same method?

These questions test electron weighting, reversal signs and the connection between cell potential and Gibbs energy. For each one, decide first whether you are combining successive reductions through Gibbs energy or combining a balanced reduction and oxidation through the cell-potential relation.

How do you combine two successive reductions of X?

Given the following potentials, find the standard potential for the complete reduction of the trivalent ion to the element. Use the electron-weighted Gibbs-energy method because the two stages transfer different numbers of electrons.

E^{\circ} (X^{3 +} / X^{2 +}) = + 0.60 V

E^{\circ} (X^{2 +} / X) = - 0.30 V

The first step transfers one electron and the second transfers two:

Δ G^{\circ} = - F (0.60) - 2 F (- 0.30) = 0

The complete reduction transfers three electrons:

E^{\circ} (X^{3 +} / X) = - \frac{0}{3 F} = 0 V

The requested check answer of −0.10 V is inconsistent with the supplied potentials. The electron-weighted thermodynamic result is 0 V.

How do you find the potential for a Y comproportionation reaction?

Given the following reduction potentials, find the standard potential for the reaction between the divalent ion and elemental Y. Reverse the second half-reaction and combine its oxidation contribution with the first reduction.

E^{\circ} (Y^{2 +} / Y^{+}) = + 0.25 V

E^{\circ} (Y^{+} / Y) = + 0.55 V

Reverse the second half-reaction:

Y \to Y^{+} + e^{-} E_{o x i d a t i o n}^{\circ} = - 0.55 V

The reduction and oxidation each transfer one electron:

E^{\circ} = 0.25 - 0.55 = - 0.30 V

Therefore:

Y^{2 +} + Y \to 2 Y^{+}

has a standard potential of −0.30 V.

How do you calculate Gibbs energy for a two-electron cell?

For a two-electron cell with a standard cell potential of +0.40 V, calculate the standard Gibbs-energy change in terms of the Faraday constant. Then use its sign to decide whether the standard reaction is spontaneous. $n = 2$

E_{c e l l}^{\circ} = + 0.40 V

Therefore:

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

Δ G^{\circ} = - 2 F (0.40) = - 0.80 F

The negative Gibbs-energy change means the reaction is spontaneous under standard conditions. For further practice, the free past-paper archive lets you search every JEE Main paper from 2002 and every Advanced paper from 2007 by year, subject or chapter, each with a worked solution.

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Frequently asked questions

What is the correct answer to the JEE Advanced 2013 electrochemistry match-list question?

The correct answer is option D: P–3, Q–4, R–1 and S–2. The corresponding potentials are −0.04 V, −0.83 V, −0.18 V and −0.40 V.

Can standard electrode potentials be added directly?

Successive electrode potentials cannot generally be added directly because potential is an intensive quantity. Convert each value using ΔG° = −nFE°, add the Gibbs energies and divide by the total electron count.

How is the Fe3+/Fe standard potential calculated?

Weight the Fe3+/Fe2+ potential by one electron and the Fe2+/Fe potential by two electrons. This gives E°(Fe3+/Fe) = −0.0367 V, which rounds to −0.04 V.

Why is option C wrong in the JEE 2013 match-list question?

Option C simply matches entries occupying the same rows in the two lists. The calculated value of P is −0.04 V, which has label 3 rather than label 1, so the row-wise match fails immediately.

Redox Reactions and Electrochemistry JEE 2013: Full Solution

What was asked in the Redox Reactions and Electrochemistry JEE 2013 question?

What rule should you use to combine electrode potentials?

How are the iron and water potentials calculated?

How do you calculate P from the two iron reduction steps?

How do you calculate Q from acidic and alkaline oxygen reductions?

How are copper and chromium matched to the final code?

How do you calculate R for the copper comproportionation reaction?

How do you calculate S from the two chromium potentials?

Why does row-wise matching give the wrong option C?

Which related electrochemistry questions test the same method?

How do you combine two successive reductions of X?

How do you find the potential for a Y comproportionation reaction?

How do you calculate Gibbs energy for a two-electron cell?

Frequently asked questions

What is the correct answer to the JEE Advanced 2013 electrochemistry match-list question?

Can standard electrode potentials be added directly?

How is the Fe3+/Fe standard potential calculated?

Why is option C wrong in the JEE 2013 match-list question?

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