Hello Kevin; I sent this a couple of days ago, but somehow it got
sent to a chemical supplier rather than Orchid. How that happened
is beyond me. I apologize for being late.
It seems to me that you are saying in your post that the term
"oxidize" is not correct when talking about the formation of black
silver sulfide on silver metal because an oxide is not formed.
Maybe that is not what you intended to say? Chemically speaking the
term “oxidize” is correct under the conditions given.
Oxidation is defined as the loss of electrons. Oxygen or an oxide
is not necessarily involved. For silver metal to form the sulfide
it is necessary that the silver give up electrons. That can be
Ag(zero charge) = Ag(1 plus) + e(1 minus)
When something is oxidized something else has to be reduced.
Reduction is defined as the gaining of electrons. The species being
reduced in the case of sulfide patination of silver is a little more
complicated than, for example, the reaction of zinc with
hydrochloric acid. Looking at that reaction for a minute we have;
Zn(zero charge) + 2 H(1 plus) + 2 Cl(1 minus) = Zn(2 plus) + 2
Cl(1-minus) + H2
Here the zinc gave up two electrons to become the zinc cation. The
electrons were accepted by the hydrogen ions to give hydrogen gas.
The zinc was oxidized and the H(1 plus) was reduced. The oxidizing
agent (that which caused the oxidation of the zinc) was the hydrogen
ions and the reducing agent (that which caused the reduction of the
hydrogen ions) was the zinc. Getting back to the case of sulfide
patination, the oxidizing agent is the hydrosulfide anion no matter
what sulfur compound is used. This is because there is essentially
no sulfide anion in aqueous solutions of sulfide salts (Na2S for
example). The sulfide anion is an extremely strong base which
abstracts a hydrogen ion from water to give the hydrosulfide anion
(1) S(2 minus) + H2O = HS(1 minus) + OH(1 minus)
The H(1 plus) part of the HS(1 minus) accepts the electron given up by
the silver and is thereby reduced to H(zero). The reactions are:
(2) Ag(zero) = Ag(1 plus) + e(1 minus) Oxidation
(3) HS(1 minus) + e(1 minus) = H + S(2 minus) Reduction (gain
(4) 2 Ag(1 plus) + S(2 minus) = Ag2S (on the Ag
(5) H(zero) + H(zero) = H2 (very
Any sulfur anion from reaction (3) that is not close enough to the
silver ions on the silver surface to form Ag2S is instantly
converted to HS(1 minus) via reaction (1).
To summarize, any time a metal reacts to produce a metal compound
the metal has been “oxidized” even if oxygen was not involved. Be
aware that it is not necessary that the oxidized entity have a zero
charge as is the case with metals. For example, the half reaction;
2 Cl(1 minus) = Cl2 + 2 e(1 minus) is an oxidation. It is an
oxidation because the chloride anions lost an electron to become
zero charge chlorine atoms.
I hope the above is germane to the previous posts on the topic.
Confusion about “redox” reactions seems to be very common. If I have
not read the previous posts correctly, perhaps the here
will still be worthwhile. I just hope I have not created more
confusion than may already exist.
"Marlinespike Seamanship in Precious Metals"