Match List I with List II. List I (Redox Transformation) List II (Charge required in Faraday) A 1 mol of Cr ₂ O ₇²⁻ to Cr ³⁺ I 1 F B 1 mol of NO ₃⁻ to N ₂ O II 2 F C 2 mol of Ag ⁺ to Ag III 4 F D 1 mol of S ₂ O ₃²⁻ to S ₄ O ₆²⁻ IV 6 F Choose the correct answer from the options given below:

Step-by-step solution

For A: The reduction half-reaction is Cr ₂ O ₇²⁻ + 14 H ^+ + 6 e ^- 2 Cr ³⁺ + 7 H ₂ O . Thus, 1 mol of Cr ₂ O ₇²⁻ requires 6 moles of electrons, which corresponds to 6 F . (A IV) For B: The reduction half-reaction is 2 NO ₃^- + 10 H ^+ + 8 e ^- N ₂ O + 5 H ₂ O . Here, 2 moles of NO ₃^- require 8 moles of electrons. Therefore, 1 mol of NO ₃^- requires 4 moles of electrons, corresponding to 4 F . (B III) For C: The reduction half-reaction is Ag ^+ + e ^- Ag . Thus, 2 moles of Ag ^+ require 2 moles of electrons, corre