# How to Balance this Equation by Oxidation number method? H_2SO_4+HI->H_2O+H_2S+I_2

Jun 17, 2017

Well, $\text{sulfur(VI+)}$ is REDUCED to $S \text{(-II)}$......

#### Explanation:

Well, $\text{sulfur(VI+)}$ is REDUCED to $S \text{(-II)}$......a formal 8 electron transfer.......

${\text{SO}}_{4}^{2 -} + 8 {H}^{+} + 8 {e}^{-} \rightarrow {S}^{2 -} + 4 {H}_{2} O \left(l\right)$ $\left(i\right)$

And of course sulfide anion has the same oxidation state as the charge on the ion, i.e. $S \left(- I I\right)$; a formal eight electron reduction.

And for every reduction there is a corresponding oxidation......

${I}^{-} \rightarrow \frac{1}{2} {\stackrel{0}{I}}_{2} \left(s\right) + {e}^{-}$ $\left(i i\right)$

And we takes $\left(i\right) + 8 \times \left(i i\right)$:

$8 {I}^{-} + S {O}_{4}^{2 -} + 8 {H}^{+} \rightarrow 4 {\stackrel{0}{I}}_{2} \left(s\right) + {S}^{2 -} + 4 {H}_{2} O \left(l\right)$

The which, I think, is balanced with respect to mass and charge; as indeed it must be if it is to represent physical reality.