Question

How many moles of $K_{2}C{r}_2{O}_7$ can be reduced by $1$ mole of $S{n}^{2+}$?

• A. $1/3$
• B. $1/6$
• C. $2/3$
• D. $1$

Answer 1

The correct option is A $1/3$

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

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

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

From the above equation each $C{r}_2{O}_7^{2-}$ ion gains 6 electrons

But $S{n}^{+2}$ loses two electrons each to form $S{n}^{4+}$

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

so 1 mole of $S{n}^{+2}$ can reduce 1/3 mole of $K_{2}C{r}_2{O}_7$

Hence option A is correct.