Can someone help me please? This is the last two questions I need help with for tonight. Thank you and have a great night!

Aug 24, 2017

As regards $16$ sulfide is oxidized, and oxygen is reduced......

$2 P b S + 3 {O}_{2} \rightarrow 2 S {O}_{2} + 2 P b O$

Explanation:

${S}^{2 -} + 2 {H}_{2} O \rightarrow S {O}_{2} + 4 {H}^{+} + 6 {e}^{-}$ $\left(i\right)$

${O}_{2} + 4 {e}^{-} \rightarrow 2 {O}^{2 -}$ $\left(i i\right)$

So take $2 \times \left(i\right) + 3 \times \left(i i\right)$

$3 {O}_{2} + 2 {S}^{2 -} + 4 {H}_{2} O \rightarrow 2 S {O}_{2} + 6 {O}^{2 -} + 8 {H}^{+}$

But we can simplify the right hand side slightly: $4 {O}^{2 -} + 8 {H}^{+} \equiv 4 {H}_{2} O$

$3 {O}_{2} + 2 {S}^{2 -} + \cancel{4 {H}_{2} O} \rightarrow 2 S {O}_{2} + 2 {O}^{2 -} + \cancel{4 {H}_{2} O}$

$2 {S}^{2 -} + 3 {O}_{2} \rightarrow 2 S {O}_{2} + 2 {O}^{2 -}$

In fact lead ion remains redox inert in this reaction, and remains as $P {b}^{2 +}$, and so we can add $2 \times P {b}^{2 +}$ to BOTH SIDES of the reaction...

$2 P b S + 3 {O}_{2} \rightarrow 2 S {O}_{2} + 2 P b O$

And this is balanced with respect to mass and charge as is absolutely required. Sulfide anion is OXIDIZED to sulfur dioxide, and dioxygen is reduced to oxide.......