X and Y are the number of electrons involved, respectively during...

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 $X + Y = 6 + 6 = 12$.