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JEE Main202624 January 2026Morning ShiftChemistryRedox ReactionsActual

X and Y are the number of electrons involved, respectively during the oxidation of I ⁻ to I ₂ and S ²⁻ to S by acidified K ₂ Cr ₂ O ₇ . The value of X + Y is _ _ _ _ .

Correct answer

0

Step-by-step solution

In acidic medium, K₂Cr₂O₇ acts as a strong oxidizing agent where Cr₂O₇²⁻ is reduced to Cr³⁺ . The reduction half-reaction is: Cr₂O₇²⁻ + 14H^+ + 6e^- 2Cr³⁺ + 7H₂O . For the oxidation of I^- to I₂ : The balanced oxidation half-reaction is 2I^- I₂ + 2e^- . To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3: 6I^- 3I₂ + 6e^- . Thus, the number of electrons involved in the balanced redox reaction for I^- oxidation is X = 6 . For the oxidation of S²⁻ to S

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