Question

Gas (A) and Gas (B) both turns $K_{2}C{r}_2{O}_2/H^{+}$ green. Gas (A) also turns lead acetate paper black. When gas (A) is passed into gas (B) in aqueous solution, yellowish white turbidity appears.

If gas (A) is $H_{2}S$ enter 1, else enter 0.

Answer 1

Gas A is $H_{2}S$. It is oxidised to $S$ by $C{r}_2{O}_7^{2-}/H^{+}$.
${C{r}_2{O}_7}^{2-}$ $+3H_{2}S+8H^{+}\to 3S+2C{r}^{3+}+7H_{2}O$

Gas B is $S{O}_2$. It is oxidised to $S{O}_4^{2-}$ by $C{r}_2{O}_7^{2-}/H^{+}$.

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

The green colour is due to $C{r}^{3+}$.

$S{O}_2+2H_{2}O\to {S{O}_4}^{2-}+4H^{+}+2e^{-}$.........(ii)
(i) + 3 $\times$ (ii)

${C{r}_2{O}_7}^{2-}+3S{O}_2+2H^{+}\to 2C{r}^{3+}+3{S{O}_4}^{2-}+H_{2}O$

$K_{2}C{r}_2{O}_7+H_{2}S{O}_4+3S{O}_2\to K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+H_{2}O$

When gas A is passed into gas B in aqueous solution, yellowish turbidity appears.

$2H_{2}S+S{O}_2\to \underset{yellowishwhite}{3S}+2H_{2}O$
Ans-1