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NVAEasyJEE 2026Oxidation Number & Redox Reactions

JEE Chemistry 2026 Question with Solution

X and Y are the number of electrons involved, respectively during the oxidation of I−\mathrm{I^-} to I2\mathrm{I_2} and S2−\mathrm{S^{2-}} to S\mathrm{S} by acidified K2Cr2O7\mathrm{K_2Cr_2O_7}. The value of X+YX + Y is _____.

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: We need the number of electrons involved in the oxidation of I−\mathrm{I^-} to I2\mathrm{I_2} and S2−\mathrm{S^{2-}} to S\mathrm{S}.

Find: The value of X+YX + Y.

Step 1: Oxidation of iodide ion

2I−→I2+2e−2\mathrm{I^-} \rightarrow \mathrm{I_2} + 2e^-

So, X=2X = 2.

Step 2: Oxidation of sulfide ion

S2−→S+2e−\mathrm{S^{2-}} \rightarrow \mathrm{S} + 2e^-

So, Y=2Y = 2.

Step 3: Total electrons transferred

X+Y=2+2=4X + Y = 2 + 2 = 4

Therefore, the total number of electrons involved in both oxidations is 44.

Common mistakes

  • A common mistake is counting electrons for one iodine atom instead of the balanced reaction. Since iodine forms I2\mathrm{I_2}, the correct half-reaction is 2I−→I2+2e−2\mathrm{I^-} \rightarrow \mathrm{I_2} + 2e^-, so use the balanced equation before counting electrons.

  • Another mistake is assuming sulfur loses only one electron. In S2−\mathrm{S^{2-}}, sulfur goes from oxidation state −2-2 to 00 in elemental sulfur, so it loses 22 electrons, not 11.

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