Question

The number of moles of $K_{2}C{r}_2{O}_7$ reduced by one moles of $S{n}^{2+}$ ions is:

• A. 1/3
• B. 1/6
• C. 2/3
• D. 3/4

Answer 1

Answer: (A)

1/3

Ionic form of reaction is:

$C{r}_2{O}_7^{2-}+S{n}^{+2}+H^{+}\to S{n}^{+4}+C{r}^{+3}+H_{2}O$

Step 1: Assign the oxidation state using oxidation number of $O=-2$.

we get:

$\stackrel{+6}{C{r}_2}{O}_7^{2-}+\stackrel{+2}{S{n}^{2+}}+H^{+}\to \stackrel{+4}{S{n}^{4+}}+\stackrel{+3}{C{r}^{3+}}+H_{2}O$

Reduction half-reaction:

$\stackrel{+6}{C{r}_2}{O}_7^{2-}\to 2\stackrel{+3}{C{r}^{3+}}$: Gain of 6 electrons

Oxidation half reaction:

$\stackrel{+2}{S{n}^{2+}}\to \stackrel{+4}{S{n}^{4+}}$: Loss of 2 electrons

Step 2: Equalise the number of electrons as:

Oxidation half reaction:

$3\stackrel{+2}{S{n}^{2+}}\to 3\stackrel{+4}{S{n}^{4+}}$

Step 3: balance O atoms by adding $H_{2}O$ and then H by $H^{+}$

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

Step 4: overall reaction:

$\stackrel{+6}{C{r}_2}{O}_7^{2-}+14H^{+}+3\stackrel{+2}{S{n}^{2+}}\to 2\stackrel{+3}{C{r}^{3+}}+3\stackrel{+4}{S{n}^{4+}}+7H_{2}O$

Thus 3 mole of $S{n}^{2+}$will reduce 1 moles of $K_{2}C{r}_2{O}_7$.

Therefore, 1 mole of $S{n}^{2+}$will reduce $\frac{1}{3}$ moles of $K_{2}C{r}_2{O}_7$.