General Principle

To find the standard electrode potential ($E^o$) for a redox reaction that is a combination of other half-reactions, we must use the relationship between Gibbs free energy and potential, $\Delta G^o = -nFE^o$. Gibbs free energies are additive, whereas electrode potentials are not (unless the number of electrons transferred, n, is the same in all combined reactions). The procedure is as follows:

1. Write the target half-reaction or full reaction.
2. Express the target reaction as an algebraic sum of the given reactions.
3. Calculate $\Delta G^o$ for each given reaction.
4. Calculate $\Delta G^o$ for the target reaction by summing the $\Delta G^o$ values from step 3.
5. Use $\Delta G^o_{target} = -n_{target}FE^o_{target}$ to find the required $E^o$.

P. $E^o(Fe^{3+} , Fe)$

1. The target half-reaction is: $Fe^{3+} + 3e^- \to Fe$, for which $n=3$.
2. We are given: (i) $Fe^{3+} + e^- \to Fe^{2+}$; $E^o_1 = +0.77V$, $n_1=1$ (ii) $Fe^{2+} + 2e^- \to Fe$; $E^o_2 = -0.44V$, $n_2=2$
3. The target reaction is the sum of (i) and (ii).
4. Calculate Gibbs free energies: $\Delta G^o_1 = -n_1 F E^o_1 = -1 \times F \times (+0.77) = -0.77F$ $\Delta G^o_2 = -n_2 F E^o_2 = -2 \times F \times (-0.44) = +0.88F$
5. Sum the Gibbs free energies: $\Delta G^o_{target} = \Delta G^o_1 + \Delta G^o_2 = -0.77F + 0.88F = +0.11F$
6. Calculate $E^o_{target}$: $\Delta G^o_{target} = -n_{target} F E^o_{target}$ $+0.11F = -3 \times F \times E^o_{target}$ $E^o_{target} = -\frac{0.11}{3} \approx -0.0367V \approx -0.04V$ Therefore, P matches with 3 (-0.04 V).

Q. $E^o(4H_2O \rightleftharpoons 4H^+ + 4OH^-)$

1. The target reaction is: $4H_2O \rightleftharpoons 4H^+ + 4OH^-$. This is not a redox reaction, but its potential can be determined by combining two given redox processes.
2. We are given: (i) $O_2(g) + 2H_2O + 4e^- \to 4OH^-$; $E^o_1 = +0.40V$ (ii) $O_2(g) + 4H^+ + 4e^- \to 2H_2O$; $E^o_2 = +1.23V$
3. The target reaction can be obtained by subtracting reaction (ii) from reaction (i): (i) $O_2 + 2H_2O + 4e^- \to 4OH^-$ -(ii) $2H_2O \to O_2 + 4H^+ + 4e^-$ Sum: $4H_2O \to 4H^+ + 4OH^-$
4. The potential of the overall cell reaction is $E^o_{cell} = E^o_{cathode} - E^o_{anode}$. Here, reaction (i) acts as the cathode and reaction (ii) as the anode. $E^o_{target} = E^o_1 - E^o_2 = 0.40V - 1.23V = -0.83V$ Therefore, Q matches with 4 (-0.83 V).

S. $E^o(Cr^{3+}, Cr^{2+})$

1. The target half-reaction is: $Cr^{3+} + e^- \to Cr^{2+}$, for which $n=1$.
2. We are given: (i) $Cr^{3+} + 3e^- \to Cr$; $E^o_1 = -0.74V$, $n_1=3$ (ii) $Cr^{2+} + 2e^- \to Cr$; $E^o_2 = -0.91V$, $n_2=2$
3. The target reaction is reaction (i) minus reaction (ii).
4. Calculate Gibbs free energies: $\Delta G^o_1 = -n_1 F E^o_1 = -3 \times F \times (-0.74) = +2.22F$ $\Delta G^o_2 = -n_2 F E^o_2 = -2 \times F \times (-0.91) = +1.82F$
5. Calculate the target Gibbs free energy: $\Delta G^o_{target} = \Delta G^o_1 - \Delta G^o_2 = 2.22F - 1.82F = +0.40F$
6. Calculate $E^o_{target}$: $\Delta G^o_{target} = -n_{target} F E^o_{target}$ $+0.40F = -1 \times F \times E^o_{target}$ $E^o_{target} = -0.40V$ Therefore, S matches with 2 (-0.4 V).

R. $E^o(Cu^{2+} + Cu \to 2Cu^+)$

1. The target reaction is: $Cu^{2+} + Cu \to 2Cu^+$.
2. We are given: (i) $Cu^{2+} + 2e^- \to Cu$; $E^o_1 = +0.34V$, $n_1=2$ (ii) $Cu^+ + e^- \to Cu$; $E^o_2 = +0.52V$, $n_2=1$
3. To obtain the target reaction, we can use reaction (i) and the reverse of reaction (ii) multiplied by 2. (i) $Cu^{2+} + 2e^- \to Cu$ (rev ii) $\times 2$: $2Cu \to 2Cu^+ + 2e^-$ Summing these gives: $Cu^{2+} + 2Cu + 2e^- \to Cu + 2Cu^+ + 2e^-$, which simplifies to $Cu^{2+} + Cu \to 2Cu^+$.
4. Calculate Gibbs free energies: $\Delta G^o_1 = -n_1 F E^o_1 = -2 \times F \times (+0.34) = -0.68F$ For reaction (ii), $\Delta G^o_2 = -1 \times F \times (+0.52) = -0.52F$. For $2Cu \to 2Cu^+ + 2e^-$, the Gibbs energy is $-2 \times \Delta G^o_2 = -2(-0.52F) = +1.04F$.
5. Sum the Gibbs free energies: $\Delta G^o_{target} = \Delta G^o_1 + (-2\Delta G^o_2) = -0.68F + 1.04F = +0.36F$
6. Calculate $E^o_{target}$. For the reaction as constructed ($Cu^{2+} + 2e^- \to Cu$ and $2Cu \to 2Cu^+ + 2e^-$), the number of electrons transferred is $n=2$. $\Delta G^o_{target} = -n_{target} F E^o_{target}$ $+0.36F = -2 \times F \times E^o_{target}$ $E^o_{target} = -\frac{0.36}{2} = -0.18V$ Therefore, R matches with 1 (-0.18 V).

Conclusion

Based on the calculations:

• P matches with 3.
• Q matches with 4.
• R matches with 1.
• S matches with 2.

The correct matching is P-3, Q-4, R-1, S-2. This corresponds to option D.