Question

Number of moles of $K_{2}C{r}_2{O}_7$ reduced by one mole of $S{n}^{2+}$ ions (in acidic medium) is:

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

Answer 1

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

Reaction that occurs between $C{r}_2{O}_7^{2-}$ and $S{n}^{2+}$ ion is:
$C{r}_2{O}_7^{2-}+S{n}^{2+}\stackrel{H^{+}}{\to }C{r}^{3+}+S{n}^{4+}+H_{2}O$

Reduction half: ${\stackrel{+6}{Cr}}_2{O}_7^{2-}\to 2C{r}^{3+};\text{n-factor}=3\times 2=6$
Oxidation half: $S{n}^{2+}\to S{n}^{4+};\text{n-factor}=2$

Since 6 mol of $S{n}^{2+}$ are used by 2 mol of $C{r}_2{O}_7^{2-}$. Hence 1 mol of $S{n}^{2+}$ will be used by $\frac{1}{3}$ mol of $C{r}_2{O}_7^{2-}$.