Question

In the given reaction, $K_{2}C{r}_2{O}_7+X{H}_{2}S{O}_4+YS{O}_2\to K_{2}S{O}_4+C{r}_2(S{O}_4{)}_3+Z{H}_{2}O$. Find X, Y and Z.

Answer 1

$S{O}_2+2H_{2}O⟶{\left(S{O}_4\right)}^{2-}+4H^{+}+2e^{-}\dots \dots \left(1\right)$

$(C{r}_2{O}_7{)}^{2-}+14H^{+}+6e^{-}⟶2G^{3+}+7H_{2}O\dots (2)$

Multiplying $\left(1\right)$ by $3$ and adding $\left(2\right)$,

$3S{O}_2+{\left(C{r}_2{O}_7\right)}^{2-}+2H^{+}⟶3{\left(S{O}_4\right)}^{2-}+2C{r}^{3+}+H_{2}O$

Completing the reaction,

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

Here, $X=1,Y=3,Z=1$