Question

The number of moles of $K_2{Cr}_2{O}_7$ reduced by $1$ mol of ${Sn}^{2+}$ ions is :

• A. $\frac{1}{3}$
• B. $3$
• C. $\frac{1}{6}$
• D. $6$

Answer 1

The correct option is A $\frac{1}{3}$

$6e^{-}+{Cr}_2{O}_7^{2-}⟶2{Cr}^{3+}$
${Sn}^{2+}⟶{Sn}^{4+}+2e^{-}$
Equivalent of ${Cr}_2{O}_7^{2-}=$ Equivalent of ${Sn}^{2+}$
$n=6$ $n=2$
$1$ equivalent ${Cr}_2{O}_7^{2-}$ = $1$ equivalent of ${Sn}^{2+}$
or, $\frac{1}{6}mol=\frac{1}{2}mol$
Therefore, $\frac{1}{3}$ mol of ${Cr}_2{O}_7^{2-}=1$ mol of ${Sn}^{2+}$