X and Y are the number of electrons involved, respectively, during the oxidation of to and to by acidified . The value of is ____ . [2026]

Question

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 ____ . [2026]

Smart Achievers · Accepted Answer

(12)

${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 {Cr}^{3 +} + 3 I_{2} + 7 H_{2} O$

$no. of moles e^{-} involved = x = 6$

${Cr}_{2} O_{7}^{2 -} + 3 S^{2 -} + 14 H^{+} \to S + 2 {Cr}^{3 +} + 7 H_{2} O$

$No. of moles e^{-} involved = y = 6$

$Sum of x + y = 6 + 6 = 12$

Solution

Q.
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 ____ . [2026]

Ans.
(12)

${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 {Cr}^{3 +} + 3 I_{2} + 7 H_{2} O$

$no. of moles e^{-} involved = x = 6$

${Cr}_{2} O_{7}^{2 -} + 3 S^{2 -} + 14 H^{+} \to S + 2 {Cr}^{3 +} + 7 H_{2} O$

$No. of moles e^{-} involved = y = 6$

$Sum of x + y = 6 + 6 = 12$

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Solution

Q. X and Y are the number of electrons involved, respectively, during the oxidation of I- to I2 and S2- to S by acidified K2Cr2O7. The value of X+Y is ____ . [2026]

Ans. (12) Cr2O72-+14H++6I-→2Cr3++3I2+7H2O no. of moles e- involved=x=6 Cr2O72-+3S2-+14H+→S+2Cr3++7H2O No. of moles e- involved=y=6 Sum of x+y=6+6=12

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Q.
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 ____ . [2026]

Ans.
(12)

${Cr}_{2} O_{7}^{2 -} + 14 H^{+} + 6 I^{-} \to 2 {Cr}^{3 +} + 3 I_{2} + 7 H_{2} O$

$no. of moles e^{-} involved = x = 6$

${Cr}_{2} O_{7}^{2 -} + 3 S^{2 -} + 14 H^{+} \to S + 2 {Cr}^{3 +} + 7 H_{2} O$

$No. of moles e^{-} involved = y = 6$

$Sum of x + y = 6 + 6 = 12$