JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

In acidic medium, K_2Cr_2O_7 acts as a strong oxidizing agent where Cr_2O_7^ 2- is reduced to Cr^ 3+ . The reduction half-reaction is: Cr_2O_7^ 2- + 14H^+ + 6e^- 2Cr^ 3+ + 7H_2O. For the oxidation of I^- to I_2: The balanced oxidation half-reaction is 2I^- I_2 + 2e^-. To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3: 6I^- 3I_2 + 6e^-. Thus, the number of electrons involved in the balanced redox reaction for I^- oxidation is X = 6. For the oxidation of S^ 2- to S: The balanced oxidation half-reaction is S^ 2- S + 2e^-. To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3: 3S^ 2- 3S + 6e^-. Thus, the number of electrons involved in the balanced redox reaction for S^ 2- oxidation is Y = 6. The value of X + Y = 6 + 6 = 12.