Question #aaf68

1 Answer
Sep 12, 2017

Answer:

We gots sulfur #(VI+)#

Explanation:

And so we have #Cu^(2+)# and #SO_4^(2-)#. Now for sulfate, the SUM of the individual elemental oxidation numbers is equal to the charge on the ion.....

And thus #S_"oxidation number"+4xxO_"oxidation number"=-2#

Now #O_"oxidation number"# is usually #-II#, and it is here, and clearly, #S_"oxidation number"=+VI#.

For #"thiosulfate anion"#, #S_2O_3^(2-)#, the given approach would yield #S(+II)#. Here, however, I like to think that sulfur has replaced ONE of the oxygens in SULFATE, #SO_4^(2-)#, and assumed its oxidation state, i.e. #S(-II)#. The other sulfur is of course remains #S(VI+)# by this formalism. The average oxidation state is still #S(+II)#, but of course both ideas are necessarily formalisms.

For #"sulfite anion"#, #SO_3^(2-)# we gots #S(IV+)#, and of course for #"sulfide dianion"#, #S^(2-)#, we got #S(-II)#.

So here is a question.....we can add elemental sulfur (possibly #S_8#) to a sulfide solution as shown.....

#S_8(s)+S^(2-) rarr S_10^(2-)#

The sulfur goes up to give polysulfide chains of various length. What is the oxidation state of the sulfur here?