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  4. How many moles of acidified K2Cr2O7 is required to liberate 6 moles of I2 from an aqueous solution of I- ?

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Q. How many moles of acidified $K_2Cr_2O_7$ is required to liberate $6$ moles of $I_2$ from an aqueous solution of $I^-$ ?

KCETKCET 2020The d-and f-Block Elements

Solution:

The balanced redox reaction will be:

$K _{2} Cr _{2} O _{7}+6 KI +7 H _{2} SO _{4} \rightarrow $
$4 K _{2} SO _{4}+3 I_{2}+ Cr _{2}\left( SO _{4}\right)_{3}+7 H _{2} O$

$3$ moles of $I_{2}$ requires $1$ mole of $K _{2} Cr _{2} O _{7}$

Hence, $6$ moles of $I_{2}$ will require $2$ moles of $K _{2} Cr _{2} O _{7}$