The standard reduction potential data at `25^(@)C` is given below
`E^(@) (Fe^(3+), Fe^(2+)) = +0.77V`,
`E^(@) (Fe^(2+), Fe) = -0.44V`,
`E^(@) (Cu^(2+),Cu) = +0.34V`,
`E^(@)(Cu^(+),Cu) = +0.52 V`,
`E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V`
`E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V`,
`E^(@) (Cr^(3+), Cr) =- 0.74V`,
`E^(@) (Cr^(2+),Cr) = - 0.91V`,
Match `E^(@)` of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists:
`{:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):}`
Codes:

Options

A. `{:(P,Q,R,S),(4,1,2,3):}`
B. `{:(P,Q,R,S),(2,3,4,1):}`
C. `{:(P,Q,R,S),(1,2,3,4):}`
D. `{:(P,Q,R,S),(3,4,1,2):}`

Explanation

To solve the problem, we need to find the standard reduction potentials (E°) for the redox pairs listed in List-I and match them with the values in List-II. Let's go through each redox pair step by step.

### Step 1: Calculate E° for (P) Fe³⁺ + e⁻ ⇌ Fe²⁺
Given:
- E°(Fe³⁺/Fe²⁺) = +0.77 V

This is already provided in the question.

### Step 2: Calculate E° for (Q) 4H₂O ⇌ 4H⁺ + 4e⁻
We can use the provided half-reaction:
- O₂(g) + 4H⁺ + 4e⁻ ⇌ 2H₂O, E° = +1.23 V
- O₂(g) + 2H₂O + 4e⁻ ⇌ 4OH⁻, E° = +0.40 V

To find the potential for the reaction 4H₂O ⇌ 4H⁺ + 4OH⁻, we can use the following relationship:
1. From the first equation, we can derive:
- 2H₂O ⇌ O₂ + 4H⁺ + 4e⁻ (reverse the first reaction)
- E° = -1.23 V

2. Adding the two reactions:
- 2H₂O ⇌ O₂ + 4H⁺ + 4e⁻ (E° = -1.23 V)
- O₂ + 2H₂O + 4e⁻ ⇌ 4OH⁻ (E° = +0.40 V)

The overall reaction becomes:
- 4H₂O ⇌ 4H⁺ + 4OH⁻

Calculating the E°:
- E° = +0.40 V - 1.23 V = -0.83 V

### Step 3: Calculate E° for (R) Cu²⁺ + e⁻ ⇌ Cu⁺
Given:
- E°(Cu²⁺/Cu) = +0.34 V
- E°(Cu⁺/Cu) = +0.52 V

To find E°(Cu²⁺/Cu⁺), we can use the following relationship:
- E°(Cu²⁺/Cu) = E°(Cu²⁺/Cu⁺) + E°(Cu⁺/Cu)

Rearranging gives us:
- E°(Cu²⁺/Cu⁺) = E°(Cu²⁺/Cu) - E°(Cu⁺/Cu)
- E°(Cu²⁺/Cu⁺) = +0.34 V - (+0.52 V) = -0.18 V

### Step 4: Calculate E° for (S) Cr³⁺ + e⁻ ⇌ Cr²⁺
Given:
- E°(Cr³⁺/Cr) = -0.74 V
- E°(Cr²⁺/Cr) = -0.91 V

To find E°(Cr³⁺/Cr²⁺), we can use the relationship:
- E°(Cr³⁺/Cr) = E°(Cr³⁺/Cr²⁺) + E°(Cr²⁺/Cr)

Rearranging gives us:
- E°(Cr³⁺/Cr²⁺) = E°(Cr³⁺/Cr) - E°(Cr²⁺/Cr)
- E°(Cr³⁺/Cr²⁺) = -0.74 V - (-0.91 V) = +0.17 V

### Summary of Results
- (P) E°(Fe³⁺/Fe²⁺) = +0.77 V
- (Q) E°(4H₂O ⇌ 4H⁺ + 4OH⁻) = -0.83 V
- (R) E°(Cu²⁺/Cu⁺) = -0.18 V
- (S) E°(Cr³⁺/Cr²⁺) = +0.17 V

### Matching with List-II
- (P) E°(Fe³⁺/Fe²⁺) matches with (1) -0.18 V
- (Q) E°(4H₂O ⇌ 4H⁺ + 4OH⁻) matches with (4) -0.83 V
- (R) E°(Cu²⁺/Cu⁺) matches with (3) -0.04 V
- (S) E°(Cr³⁺/Cr²⁺) matches with (2) -0.40 V

### Final Answer
The correct matching is:
- (P) - (1)
- (Q) - (4)
- (R) - (3)
- (S) - (2)