X and Y are the number of electrons involved, respectively during the oxidation of $$\ I^{-}\text{ to }I_{2}\text{ and }S^{2-}$$ to S by acidified $$K_{2}Cr_{2}O_{7}$$. The value of X + Y is __ .

We need to find X + Y where X is the number of electrons involved in the oxidation of $$I^-$$ to $$I_2$$ and Y is for $$S^{2-}$$ to S, by acidified $$K_2Cr_2O_7$$.

The half-reaction for oxidation of $$I^-$$ to $$I_2$$ is $$2I^- \rightarrow I_2 + 2e^-$$, and the reduction half-reaction of dichromate under acidic conditions is $$Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$. Multiplying the oxidation half-reaction by 3 balances the electrons exchanged, giving $$6I^- \rightarrow 3I_2 + 6e^-$$. Thus, X = 6 electrons.

For the oxidation of $$S^{2-}$$ to S, the half-reaction is $$S^{2-} \rightarrow S + 2e^-$$, while the reduction half-reaction remains the same. To balance the electrons, the oxidation half-reaction is multiplied by 3, resulting in $$3S^{2-} \rightarrow 3S + 6e^-$$. Hence, Y = 6 electrons.

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

The answer is 12.