X and Y are the number of electrons involved, | vedclass

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(C) The balanced chemical equation for the oxidation of $I^{-}$ to $I_{2}$ by acidified $K_{2}Cr_{2}O_{7}$ is: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6I^{-} \r

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) The balanced chemical equation for the oxidation of $I^{-}$ to $I_{2}$ by acidified $K_{2}Cr_{2}O_{7}$ is: $Cr_{2}O_{7}^{2-} + 14H^{+} + 6I^{-} \rightarrow 2Cr^{3+} + 3I_{2} + 7H_{2}O$. Here, $6$ electrons are involved, so $X = 6$. The balanced chemical equation for the oxidation of $S^{2-}$ to $S$ by acidified $K_{2}Cr_{2}O_{7}$ is: $Cr_{2}O_{7}^{2-} + 3S^{2-} + 14H^{+} \rightarrow 3S + 2Cr^{3+} + 7H_{2}O$. Here, $6$ electrons are involved, so $Y = 6$. Therefore, the sum $X + Y = 6 + 6 = 12$.

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

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