# What is the reaction between sulfur dioxide, "SO"_2, and potassium dichromate, "K"_2 "Cr"_2"O"_7 ?

Apr 1, 2017

In presence of ${\text{H"_2"SO}}_{4}$,potassium dicromate reacts with ${\text{SO}}_{2}$ as following.

#### Explanation:

$3 {\text{SO}}_{2}$ + ${\text{K"_2"Cr"_2"O}}_{7}$ + ${\text{H"_2"SO}}_{4}$ $\to$ ${\text{K"_2"SO}}_{4}$ + "Cr"_2("SO"_4)_3 + $\text{H"_2"O}$

Apr 1, 2017

Well, we would assume that $\text{sulfur dioxide}$ is oxidized to $\text{sulfur trioxide}$.

#### Explanation:

I think this is done industrially by ${V}_{2} {O}_{5}$.

But sulfur is oxidized to from $S \left(+ I V\right)$ to $S \left(+ V I\right) :$

$S {O}_{2} \left(g\right) + {H}_{2} O \left(l\right) \rightarrow S {O}_{3} \left(g\right) + 2 {H}^{+} + 2 {e}^{-}$ $\left(i\right)$

And dichromate is reduced from $C r \left(+ V I\right)$ to $C r \left(+ I I I\right) :$

$C {r}_{2} {O}_{7}^{2 -} + 14 {H}^{+} + 6 {e}^{-} \rightarrow 2 C {r}^{3 +} + 7 {H}_{2} O \left(l\right)$ $\left(i i\right)$

We add $3 \times \left(i\right) + \left(i i\right)$ to eliminate the electrons:

$C {r}_{2} {O}_{7}^{2 -} + 8 {H}^{+} + 3 S {O}_{2} \left(g\right) \rightarrow 2 C {r}^{3 +} + 4 {H}_{2} O \left(l\right) + 3 S {O}_{3} \left(g\right)$

Are mass and charge balanced? This is your problem not mine.