Part of the preparation of Tollens' reagent occurs according to the formula:

$$\ce{Ag2O (s) + 4 NH3 + 2 NaNO3 + H2O → 2 [Ag(NH3)2]NO3 + 2 NaOH}$$

I wonder how exactly the oxygen gets "unstuck" from the silver under the influence of ammonia. I've read about coordination complexes, so I understand what $\ce{[Ag(NH3)2]NO3}$ is, but I'm puzzled at what exactly happens to pull that oxygen apart from $\ce{Ag2O}$

I've done very little exploration of the issue, so this may be too basic a question - in which case just please give a hint on what to read up.

I found this (now deleted) comment:

$\ce{NH3}$ dissolves silver oxide because the nitrogen has a pair of electrons that it can share with the silver forming a diammine complex. It has nothing to do with pH adjustment. – A.K. Jan 20 '16 at 17:17

Chemguide states this:

This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate.


3 Answers 3


It's quite elementary (perhaps). Silver (and gold) are known to show two main coordination numbers in its compounds, namely, $2$ and $4$. To start, we can consider $\ce{[Ag(OH)2]-}$ to be our initial complex as in aqueous solutions it can be thought of to be existing in an equilibrium with $\ce{Ag2O}$. See footnote [1].

A quick glance at the order for the strength of ligands tells us that $\ce{NH3}$ is a stronger ligand than $\ce{OH-}$. What I now intend to do is to compare the thermodynamic stability of $\ce{[Ag(OH)2]-}$ with that of the diammine complex $\ce{[Ag(NH3)2]+}$.

The strength of ligands is directly proportional to the thermal stability of a complex so yes, the diammine complex is more stable.

A similar case is seen with copper(I). Cyanide complexes of copper(I) are known to be extremely stable and hence it prefers forming a complex with cyanide over other ligands.

[1] Nevertheless, saying this may be highly controversial because Ellingham diagrams tell us that $\ce{Ag(OH)}$ being unstable gets converted into $\ce{Ag2O}$ which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist.

  • 1
    $\begingroup$ How do you quantify the "strength" of a ligand? Different ligands bind to different metals with different affinities, and the usual way to measure this affinity is by the formation constant of the complex $$\ce{M + n L <=> ML_n}; \qquad K_\mathrm d = \frac{[\ce{ML_n}]}{[\ce{M}][\ce{L}]^n}.$$ Your argument then becomes tautological, because it is essentially now "it is more thermodynamically stable because it is more thermodynamically stable". $\endgroup$ Commented Dec 23, 2018 at 21:09

Some thoughts on the issue received via email from Jim Clark, the author of the Chemguide website. I hope these thoughts will prove helpful to readers, thus I post them here (with the author's permission).

I don’t really see the problem here. The silver oxide is formed in the reaction by adding hydroxide ions to silver(I) ions. Instead of precipitating silver(I) hydroxide, it dehydrates (either totally or partially - I don’t know) to silver(I) oxide. But as with any other insoluble compound, there will be an equilibrium between the solid compound and its ions in solution - albeit strongly to the solid side of the equilibrium.

When you add ammonia solution, it complexes with silver(I) ions to give diamminesilver(I) ions which are soluble. The equilibrium position of that reaction is well to the complex side.

So you can think of this in two ways.

The ammonia reacts with silver(I) ions in the solid silver(I) oxide and goes into solution. The oxide ions react with water to produce hydroxide ions. Or...

The ammonia reacts with the tiny amount of silver(I) ions already in solution, and the precipitation equilibrium shifts to replace them. The solid therefore goes back into solution producing the hydroxide ions that were added originally, but with the silver ions now mainly complexed.

The first way is not only simpler, but also, of course, applies to any solid silver(I) oxide you might have - for example a silver(I) oxide tarnish on some item of silver which can be removed by wiping it with ammonia solution (a source of both ammonia to complex the silver ions and water to react with the oxide ions.


Actually both comments perhaps, to quote:

"because Ellingham Diagrams tell us that Ag(OH) being unstable gets converted into Ag2O which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist"

"NH3 is a stronger ligand than OH−"

Assuming a starting Ag2O presence from the reaction, some Ag+ ions from limited solubility, and noting, "a ppt of AgOH appears and dissolves in excess ammonia", see https://books.google.com/books?id=uwJbSFRIqyAC&pg=PA37&lpg=PA37&dq=radical+attack+on+silver+Ag+%2B+%E2%80%A2OH++--%3E+AgOH&source=bl&ots=jG9-2NG9x3&sig=V9nVefopAXXKP3lFnxQYnJriG14&hl=en&sa=X&ved=2ahUKEwjasar_g7ffAhWlhOAKHQEDA_4Q6AEwGHoECAcQAQ#v=onepage&q&f=false :

2 AgOH = Ag2O + H2O

both comments may be right.

Further, in the presence of visible light leading to the photocatalyst Ag2O/Ag, which per research out of P.R. China (published 2011 in the ‘Chemistry, a European Journal’), which has been cited as a powerful stable visible light photocatalyst capable of generating electrons (e-) and holes (h+) (see https://onlinelibrary.wiley.com/doi/abs/10.1002/chem.201101032 ):

Ag2O/Ag + hv --> e− + h+

With electrons:

Ag+ + e- --> Ag

And, in the presence of active electron holes, any OH- created via:

XOH = X+ + OH- (where X = H or NH3 or Ag)

could be converted into •OH radical:

OH- (aq) + h+ --> •OH (aq)

which may attack Silver metal directly (see https://pubs.acs.org/doi/abs/10.1021/acs.jpca.7b08081?src=recsys&journalCode=jpcafh and superacidity related redox couples at http://beta.chem.uw.edu.pl/people/WGrochala/Ag2+_mechanism.pdf, and more generally radical attacks on metals in their lower valence states at https://srd.nist.gov/NSRDS/NSRDS-NBS-65.pdf , where one could postulate the half cell reactions: Ag = Ag+ + e- and •OH + e- = OH- with the net reaction below):

Ag + •OH (aq) --> AgOH (aq)

2 AgOH = Ag2O + H2O

Or: AgOH (aq) = Ag+ (aq) + OH- (aq) (based on some limited solubility of Ag2O)

producing the oxygen free Silver ion to complex with NH3 (or OH-, depending on the respect ion concentration) in a photo cyclic reaction. Note: the hydroxyl radical can be largely scavenged at high ammonia concentration (creating •NH2), so this path is pH sensitive.


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