Question

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

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

Answer 1

The correct option is A $1/3$

The balanced chemical reactions (ionic reactions)for reduction of $K_{2}C{r}_2{O}_7$ by $S{n}^{2+}$ are:

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

$(S{n}^{2+}\to S{n}^{4+}+2e^{-})\times 3$

Balanced Net reaction : $3S{n}^{2+}+{C{r}_2{O}_7}^{2-}+14H^{+}\to 3S{n}^{4+}+2C{r}^{3+}+7H_{2}O$

Thus according to the balanced reaction : 1 mole of ${C{r}_2{O}_7}^{2-}$ will be reduced by 3 moles of $S{n}^{2+}$

Thus 1 mole of $S{n}^{2+}$ will reduce $=\frac{1}{3}$ moles of${C{r}_2{O}_7}^{2-}$

Hence, the correct option is $A$