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

enter image source here

1 Answer
Aug 24, 2017

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.......