Question

Number of moles of $K_{2}C{r}_2{O}_7$ reduced by one mole of $S{n}^{2+}$ is

• A. $1/3$
• B. $3$
• C. $1/6$
• D. $6$

Answer 1

The correct option is A $1/3$

$K_{2}C{r}_2{O}_7\to 2K^{+}+C{r}_2{O}_7^2$-

$C{r}_2{O}_7^{2-}+14H6++6e^{-}\to 2C{r}^{3+}+7H_{2}O$

$S{n}^{2+}\to S{n}^{4+}+2e^{-}$

Each $C{r}_2{O}_7^{2-}$ ion gains 6 electrons and $Sn6+2$ loses two electrons each to form $S{n}^{4+}$

Therefore you need 3 moles of $S{n}^{+2}$ to reduce 1 mole of $K_{2}C{r}_2{O}_7$

Answer is a 1/3 mole of $K_{2}C{r}_2{O}_7$ is reduced by 1 mole of $S{n}^{+2}$

Full equation looks like this

$Cr_2O_7^{2-} + 3Sn^{2+} +14H{+} \rightarrow 3Sn^{4+} + 2Cr^{3+} + 7H_2O$

Hence option A is correct.