Question

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

Answer 1

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

`2PbS + 3O_2 rarr 2SO_2+2PbO`

Explanation:

`S^(2-) + 2H_2O rarr SO_2 + 4H^(+) + 6e^(-)` `(i)`

`O_2 +4e^(-) rarr 2O^(2-)` `(ii)`

So take `2xx(i)+3xx(ii)`

`3O_2 +2S^(2-) + 4H_2O rarr 2SO_2+6O^(2-)+8H^+`

But we can simplify the right hand side slightly: `4O^(2-) +8H^(+)-=4H_2O`

`3O_2 +2S^(2-) + cancel(4H_2O) rarr 2SO_2+2O^(2-)+cancel(4H_2O)`

`2S^(2-) + 3O_2 rarr 2SO_2+2O^(2-)`

In fact lead ion remains redox inert in this reaction, and remains as `Pb^(2+)`, and so we can add `2xxPb^(2+)` to BOTH SIDES of the reaction...

`2PbS + 3O_2 rarr 2SO_2+2PbO`

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.......