# How is pyrophosphate anion, P_2O_7^(4-) reduced to elemental phosphorus, with oxidation of hydrogen sulfide to sulphur?

Sep 22, 2016

The overal redox equation is:

$2 {P}_{2} {O}_{7}^{4 -} + 8 {H}^{+} + 10 {H}_{2} S \rightarrow {P}_{4} + 10 S + 14 {H}_{2} O$

Whew!! Arithmetic.

#### Explanation:

I take it you mean $\text{pyrophosphate anion}$, ${P}_{2} {O}_{7}^{4 -}$. This is $P \left(V +\right)$, which is reduced to elemental phosphorus, ${P}^{0}$.

${P}_{2} {O}_{7}^{4 -} + 14 {H}^{+} + 10 {e}^{-} \rightarrow \frac{1}{2} {P}_{4} + 7 {H}_{2} O$ $\left(i\right)$

Charge and mass are balanced here. Why is that important?

Sulfide anion is oxidized to elemental sulfur:

$8 {H}_{2} S \rightarrow {S}_{8} + 16 {H}^{+} + 16 {e}^{-}$ $\left(i i\right)$

Charge and mass are balanced again!

So we cross multiply, and $16 \times \left(i\right) + 10 \times \left(i i\right) =$

$16 {P}_{2} {O}_{7}^{4 -} + 64 {H}^{+} + 80 {H}_{2} S \rightarrow 8 {P}_{4} + 10 {S}_{8} + 112 {H}_{2} O$

We could make this a little simpler by dividing thru by $8$, i.e. we treat ${S}_{8}$ simply as $S$:

$2 {P}_{2} {O}_{7}^{4 -} + 8 {H}^{+} + 10 {H}_{2} S \rightarrow {P}_{4} + 10 S + 14 {H}_{2} O$

Here again, charge and mass are balanced, as required.