JEE Challenger
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ChemHardAdv 2013 • Paper 2

Standard Reduction Potential Values and Redox Pair Matching

The standard reduction potential values measured at 25 ∘C25\ ^\circ\text{C} are provided below: E∘(Fe3+,Fe2+)=+0.77 VE^\circ(Fe^{3+}, Fe^{2+}) = +0.77\text{ V} E∘(Fe2+,Fe)=−0.44 VE^\circ(Fe^{2+}, Fe) = -0.44\text{ V} E∘(Cu2+,Cu)=+0.34 VE^\circ(Cu^{2+}, Cu) = +0.34\text{ V} E∘(Cu+,Cu)=+0.52 VE^\circ(Cu^+, Cu) = +0.52\text{ V} E∘[O2(g)+4H++4e−→2H2O]=+1.23 VE^\circ[O_2(g) + 4H^+ + 4e^- \rightarrow 2H_2O] = +1.23\text{ V} E∘[O2(g)+2H2O+4e−→4OH−]=+0.40 VE^\circ[O_2(g) + 2H_2O + 4e^- \rightarrow 4OH^-] = +0.40\text{ V} E∘(Cr3+,Cr)=−0.74 VE^\circ(Cr^{3+}, Cr) = -0.74\text{ V} E∘(Cr2+,Cr)=−0.91 VE^\circ(Cr^{2+}, Cr) = -0.91\text{ V}

Match E∘E^\circ of the given redox pair in List I with the values listed in List II and select the correct option using the provided code:

List IList IIP. E∘(Fe3+,Fe)1. −0.18 VQ. E∘(4H2O⇌4H++4OH−)2. −0.4 VR. E∘(Cu2++Cu→2Cu+)3. −0.04 VS. E∘(Cr3+,Cr2+)4. −0.83 V\begin{array}{ll} \text{List I} & \text{List II} \\ \text{P. } E^\circ(Fe^{3+}, Fe) & 1.\ -0.18\text{ V} \\ \text{Q. } E^\circ(4H_2O \rightleftharpoons 4H^+ + 4OH^-) & 2.\ -0.4\text{ V} \\ \text{R. } E^\circ(Cu^{2+} + Cu \rightarrow 2Cu^+) & 3.\ -0.04\text{ V} \\ \text{S. } E^\circ(Cr^{3+}, Cr^{2+}) & 4.\ -0.83\text{ V} \end{array}

Options

A

P →\rightarrow 4, Q →\rightarrow 1, R →\rightarrow 2, S →\rightarrow 3

B

P →\rightarrow 2, Q →\rightarrow 3, R →\rightarrow 4, S →\rightarrow 1

C

P →\rightarrow 1, Q →\rightarrow 2, R →\rightarrow 3, S →\rightarrow 4

D

P →\rightarrow 3, Q →\rightarrow 4, R →\rightarrow 1, S →\rightarrow 2

Correct

Topics & Concepts

ElectrochemistryElectrochemical CellNernst EquationOxidation State

Step-by-Step Solution

To match the standard reduction potentials in List I with the corresponding values in List II, we evaluate each item step-by-step using standard thermodynamic relations (ΔG∘=−nFE∘\Delta G^\circ = -nFE^\circ).

P. E∘(Fe3+,Fe)E^\circ(\text{Fe}^{3+}, \text{Fe})

The half-reactions and their Gibbs free energy changes are:

  1. Fe3++e−→Fe2+\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}, ΔG1∘=−1⋅F⋅(+0.77 V)=−0.77F\quad \Delta G_1^\circ = -1 \cdot F \cdot (+0.77\text{ V}) = -0.77F
  2. Fe2++2e−→Fe\text{Fe}^{2+} + 2e^- \rightarrow \text{Fe}, ΔG2∘=−2⋅F⋅(−0.44 V)=+0.88F\quad \Delta G_2^\circ = -2 \cdot F \cdot (-0.44\text{ V}) = +0.88F

Adding both half-reactions gives: Fe3++3e−→Fe\text{Fe}^{3+} + 3e^- \rightarrow \text{Fe}

ΔG3∘=ΔG1∘+ΔG2∘=−0.77F+0.88F=+0.11F\Delta G_3^\circ = \Delta G_1^\circ + \Delta G_2^\circ = -0.77F + 0.88F = +0.11F

Since ΔG3∘=−3FE∘(Fe3+,Fe)\Delta G_3^\circ = -3 F E^\circ(\text{Fe}^{3+}, \text{Fe}): −3FE∘(Fe3+,Fe)=+0.11F  ⟹  E∘(Fe3+,Fe)=−0.113 V≈−0.04 V-3 F E^\circ(\text{Fe}^{3+}, \text{Fe}) = +0.11F \implies E^\circ(\text{Fe}^{3+}, \text{Fe}) = -\frac{0.11}{3}\text{ V} \approx -0.04\text{ V}

Thus, P →\rightarrow 3.

Q. E∘(4H2O⇌4H++4OH−)E^\circ(4\text{H}_2\text{O} \rightleftharpoons 4\text{H}^+ + 4\text{OH}^-)

Consider the following standard electrode reactions:

  1. O2(g)+2H2O+4e−→4OH−\text{O}_2(g) + 2\text{H}_2\text{O} + 4e^- \rightarrow 4\text{OH}^-, E1∘=+0.40 V\quad E^\circ_1 = +0.40\text{ V}
  2. O2(g)+4H++4e−→2H2O\text{O}_2(g) + 4\text{H}^+ + 4e^- \rightarrow 2\text{H}_2\text{O}, E2∘=+1.23 V\quad E^\circ_2 = +1.23\text{ V}

Subtracting reaction (2) from reaction (1): 4H2O⇌4H++4OH−4\text{H}_2\text{O} \rightleftharpoons 4\text{H}^+ + 4\text{OH}^-

The standard potential for this combined process is given by: E∘=E1∘−E2∘=0.40 V−1.23 V=−0.83 VE^\circ = E^\circ_1 - E^\circ_2 = 0.40\text{ V} - 1.23\text{ V} = -0.83\text{ V}

Thus, Q →\rightarrow 4.

R. E∘(Cu2++Cu→2Cu+)E^\circ(\text{Cu}^{2+} + \text{Cu} \rightarrow 2\text{Cu}^+)

First, determine E∘(Cu2+,Cu+)E^\circ(\text{Cu}^{2+}, \text{Cu}^+):

  1. Cu2++2e−→Cu\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}, ΔG1∘=−2⋅F⋅(0.34 V)=−0.68F\quad \Delta G_1^\circ = -2 \cdot F \cdot (0.34\text{ V}) = -0.68F
  2. Cu++e−→Cu\text{Cu}^+ + e^- \rightarrow \text{Cu}, ΔG2∘=−1⋅F⋅(0.52 V)=−0.52F\quad \Delta G_2^\circ = -1 \cdot F \cdot (0.52\text{ V}) = -0.52F

For Cu2++e−→Cu+\text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+: ΔG3∘=ΔG1∘−ΔG2∘=−0.68F−(−0.52F)=−0.16F\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ = -0.68F - (-0.52F) = -0.16F

Now, for the overall reaction Cu2++Cu→2Cu+\text{Cu}^{2+} + \text{Cu} \rightarrow 2\text{Cu}^+: ΔG∘=ΔG∘(Cu2+→Cu+)+ΔG∘(Cu→Cu+)\Delta G^\circ = \Delta G^\circ(\text{Cu}^{2+} \rightarrow \text{Cu}^+) + \Delta G^\circ(\text{Cu} \rightarrow \text{Cu}^+) ΔG∘=−0.16F+0.52F=+0.36F\Delta G^\circ = -0.16F + 0.52F = +0.36F

For this reaction involving two equivalents of charge transferred per mole of reaction (n=2n=2): ΔG∘=−nFE∘  ⟹  +0.36F=−2FE∘  ⟹  E∘=−0.18 V\Delta G^\circ = -nFE^\circ \implies +0.36F = -2FE^\circ \implies E^\circ = -0.18\text{ V}

Thus, R →\rightarrow 1.

S. E∘(Cr3+,Cr2+)E^\circ(\text{Cr}^{3+}, \text{Cr}^{2+})

The half-reactions and their Gibbs free energy changes are:

  1. Cr3++3e−→Cr\text{Cr}^{3+} + 3e^- \rightarrow \text{Cr}, ΔG1∘=−3⋅F⋅(−0.74 V)=+2.22F\quad \Delta G_1^\circ = -3 \cdot F \cdot (-0.74\text{ V}) = +2.22F
  2. Cr2++2e−→Cr\text{Cr}^{2+} + 2e^- \rightarrow \text{Cr}, ΔG2∘=−2⋅F⋅(−0.91 V)=+1.82F\quad \Delta G_2^\circ = -2 \cdot F \cdot (-0.91\text{ V}) = +1.82F

Subtracting reaction (2) from reaction (1) gives: Cr3++e−→Cr2+\text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+}

ΔG3∘=ΔG1∘−ΔG2∘=+2.22F−(+1.82F)=+0.40F\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ = +2.22F - (+1.82F) = +0.40F

Since ΔG3∘=−1FE∘(Cr3+,Cr2+)\Delta G_3^\circ = -1 F E^\circ(\text{Cr}^{3+}, \text{Cr}^{2+}): −1FE∘(Cr3+,Cr2+)=+0.40F  ⟹  E∘(Cr3+,Cr2+)=−0.40 V=−0.4 V-1 F E^\circ(\text{Cr}^{3+}, \text{Cr}^{2+}) = +0.40F \implies E^\circ(\text{Cr}^{3+}, \text{Cr}^{2+}) = -0.40\text{ V} = -0.4\text{ V}

Thus, S →\rightarrow 2.

Conclusion:

The correct matching sequence is: P→3,Q→4,R→1,S→2\text{P} \rightarrow 3, \quad \text{Q} \rightarrow 4, \quad \text{R} \rightarrow 1, \quad \text{S} \rightarrow 2

This corresponds to Option D.

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