Understanding the Problem

We are given standard reduction potentials for several half-cells and need to find the standard reduction potentials for four specific redox pairs: (P) Fe³⁺/Fe, (Q) 4H₂O ⇌ 4H⁺ + 4OH^–, (R) Cu²⁺ + Cu → 2Cu⁺, and (S) Cr³⁺/Cr²⁺. We must match them to their correct values from List II.

The key concept is that the standard reduction potential for a reaction that is a combination of other half-reactions can be found by relating the Gibbs free energy changes, since $\mathrm{\Delta G°}=–\mathrm{nFE°}$. The free energy is an additive property.

Step-by-Step Solutions

Step 1: Find E° for (P) Fe³⁺/Fe

The reduction ${\mathrm{Fe}}^{3+}+3e^{–}\to \mathrm{Fe}$ can be obtained by adding two known half-reactions:

1. ${\mathrm{Fe}}^{3+}+e^{–}\to {\mathrm{Fe}}^{2+}$;    $\mathrm{\Delta G}1°=–1\times F\times \left(0.77\right)$

2. ${\mathrm{Fe}}^{2+}+2e^{–}\to \mathrm{Fe}$;    $\mathrm{\Delta G}2°=–2\times F\times (–0.44)$

Adding these gives the desired reaction: ${\mathrm{Fe}}^{3+}+3e^{–}\to \mathrm{Fe}$;    $\mathrm{\Delta G}°=\mathrm{\Delta G}1°+\mathrm{\Delta G}2°$

Therefore, $–3F{E}_P°=[–1F(0.77\left)\right]+[–2F(–0.44\left)\right]$

Solving for E_P°: $–3E_P°=–0.77+0.88=0.11$

$E_P°=\frac{–0.11}{3}=–0.0367V\approx –0.04V$

Match: (P) with (3) –0.04 V

Step 2: Find E° for (Q) 4H₂O ⇌ 4H⁺ + 4OH^–

This represents the autoionization of water. The reaction is the reverse of the reduction reaction given for acidic and basic media:

Given: $O_2\left(g\right)+4H^{+}+4e^{–}\to 2H_{2}O$;    E° = +1.23 V

Given: $O_2\left(g\right)+2H_{2}O+4e^{–}\to 4{\mathrm{OH}}^{–}$;    E° = +0.40 V

The difference between these two reactions is precisely the autoionization reaction (Q) multiplied by 4. Subtracting the second reaction from the first:

$(O_2+4H^{+}+4e^{–})–(O_2+2H_{2}O+4e^{–})=4H^{+}+2H_{2}O–4{\mathrm{OH}}^{–}$

Simplifying: $4H^{+}+4{\mathrm{OH}}^{–}\rightleftharpoons 4H_{2}O$ (This is the reverse of reaction Q)

The E° for the reaction as written in List I (4H₂O ⇌ 4H⁺ + 4OH^–) is found from the relationship between the Gibbs free energies:

$\mathrm{\Delta G}°\left(Q\right)=\mathrm{\Delta G}°\left(acidic\right)–\mathrm{\Delta G}°\left(basic\right)$

$–4F{E}_Q°=[–4F(1.23\left)\right]–[–4F(0.40\left)\right]$

$–4F{E}_Q°=–4F(1.23–0.40)=–4F\left(0.83\right)$

Therefore, $E_Q°=–0.83V$

Match: (Q) with (4) –0.83 V

Step 3: Find E° for (R) Cu²⁺ + Cu → 2Cu⁺