JEE Main 2026 — Chemistry Physical Chemistry

2026integermedium

X and Y are the number of electrons involved, respectively during the oxidation of $\mathrm{I}^{-}$ to $\mathrm{I}_{2}$ and $\mathrm{S}^{2-}$ to S by acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ . The value of $\mathrm{X}+\mathrm{Y}$ is $\_\_\_\_$ .

Official previous-year question

Held on 24 Jan 2026 · Verified 6 Jul 2026.

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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: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ .

For the oxidation of $I^-$ to $I_2$ :

The balanced oxidation half-reaction is $2I^- \rightarrow I_2 + 2e^-$ .

To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3:

$6I^- \rightarrow 3I_2 + 6e^-$ .

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 $S^{2-} \rightarrow S + 2e^-$ .

To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3:

$3S^{2-} \rightarrow 3S + 6e^-$ .

Thus, the number of electrons involved in the balanced redox reaction for $S^{2-}$ oxidation is $Y = 6$ .

The value of $X + Y = 6 + 6 = 12$ .

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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: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ .

For the oxidation of $I^-$ to $I_2$ :

The balanced oxidation half-reaction is $2I^- \rightarrow I_2 + 2e^-$ .

To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3:

$6I^- \rightarrow 3I_2 + 6e^-$ .

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 $S^{2-} \rightarrow S + 2e^-$ .

To balance the electrons with the reduction half-reaction (6 electrons), we multiply the oxidation half-reaction by 3:

$3S^{2-} \rightarrow 3S + 6e^-$ .

Thus, the number of electrons involved in the balanced redox reaction for $S^{2-}$ oxidation is $Y = 6$ .

The value of $X + Y = 6 + 6 = 12$ .

JEE Main 2026 — Chemistry Physical Chemistry

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JEE Main 2026 — Chemistry Physical Chemistry

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Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

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