Match the column and choose correct option:
`{:("(A) MnO"_(4)^(-)rarrMn^(+2)" (1 mol)","P. Required 1 F"),("(B) "CuSO_(4)rarr"Cu (1 mol)","Q. Required 5 F"),("(C) "Al_(2)O_(3)rarrAl" (1 mol)","R. Required 3 F"),("(D) "NaCl rarr Na" (1 mol)","S. Required 2 F"):}`

Options

A. `A-Q, B-P, C-S, D-R`
B. `A-P, B-Q, C-S, D-R`
C. `A-Q, B-S, C-P, D-R`
D. `A-Q, B-S, C-R, D-P`

Answer

To solve the problem of matching the columns based on the required Faraday's (F) for each reaction, we will analyze each reaction step-by-step.

### Step 1: Analyze the reaction of MnO₄⁻ to Mn²⁺
- The half-reaction is: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
- Here, manganese (Mn) is reduced from +7 oxidation state in MnO₄⁻ to +2 in Mn²⁺.
- This involves the transfer of 5 electrons.
- Since 1 mole of electrons corresponds to 1 Faraday, the total Faraday's required = 5 F.

### Step 2: Analyze the reaction of CuSO₄ to Cu
- The half-reaction is: Cu²⁺ + 2e⁻ → Cu
- Copper (Cu) is reduced from +2 oxidation state to 0 in elemental copper.
- This involves the transfer of 2 electrons.
- Therefore, the total Faraday's required = 2 F.

### Step 3: Analyze the reaction of Al₂O₃ to Al
- The half-reaction is: Al₂O₃ + 6e⁻ → 2Al
- Aluminum (Al) is reduced from +3 oxidation state in Al₂O₃ to 0 in elemental aluminum.
- This involves the transfer of 6 electrons.
- Therefore, the total Faraday's required = 6 F.

### Step 4: Analyze the reaction of NaCl to Na
- The half-reaction is: Na⁺ + e⁻ → Na
- Sodium (Na) is reduced from +1 oxidation state to 0 in elemental sodium.
- This involves the transfer of 1 electron.
- Therefore, the total Faraday's required = 1 F.

### Summary of Results
- (A) MnO₄⁻ → Mn²⁺ requires 5 F (matches with Q)
- (B) CuSO₄ → Cu requires 2 F (matches with S)
- (C) Al₂O₃ → Al requires 6 F (matches with R)
- (D) NaCl → Na requires 1 F (matches with P)

### Final Matching
- A → Q (5 F)
- B → S (2 F)
- C → R (6 F)
- D → P (1 F)

### Correct Option
The correct option that matches the columns is **D**.