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

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1 Answer
Aug 24, 2017

Answer:

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.......