In this previous question the mechanism for the reaction of Tollens' reagent was outlined. As I understand it the oxidising agent in Tollens' is the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ but the mechanism only showed the silver ion.

What is the purpose of adding ammonia to the silver nitrate to form the diamminesilver(I) complex: why not just use silver nitrate? Is it purely to provide a source of hydroxide ions or is the structure of the complex ion actually important?

  • $\begingroup$ Complexation reduces redox potential. $\endgroup$
    – Mithoron
    Apr 28, 2015 at 23:31
  • $\begingroup$ @Mithoron Can you expand upon this a bit. I had a feeling this might be the case but can you give some more data and possibly explain why $\endgroup$
    – bon
    Apr 29, 2015 at 18:45
  • $\begingroup$ Maybe I'll find something :) $\endgroup$
    – Mithoron
    Apr 29, 2015 at 23:15

1 Answer 1


The formation of the diamminesilver(I) complex $\ce{[Ag(NH3)2]+}$ due to addition of ammonia has mainly three functions.

The Tollens test takes place under alkaline aqueous conditions. The complexation of $\ce{Ag+}$ with $\ce{NH3}$ prevents precipitation of brown $\ce{Ag2O}$:

$$\ce{2Ag+ + 2OH- <=> 2 AgOH <=> Ag2O + H2O}$$

The Tollens test shall be selective for aldehydes. $\ce{[Ag(NH3)2]+}$ is a milder oxidising agent than $\ce{Ag+}$.

$$\begin{alignat}{2} \ce{Ag+ + e- \;&<=> Ag}\quad &&E^\circ = +0.799\ \mathrm{V}\\ \ce{[Ag(NH3)2]+ + e- \;&<=> Ag + 2 NH3}\quad &&E^\circ = +0.373\ \mathrm{V} \end{alignat}$$

The difference in redox potential can be explained using the stability constant of $\ce{[Ag(NH3)2]+}$:

$$K_\text{B} = \frac{\left[\ce{[Ag(NH3)2]+}\right]}{\left[\ce{Ag+}\right]\left[\ce{NH3}\right]^2}$$

$$\begin{aligned} E&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\left[\ce{Ag+}\right]\\ &=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{\left[\ce{[Ag(NH3)2]+}\right]}{K_\text{B}\cdot\left[\ce{NH3}\right]^2} \\ E_{\ce{[Ag(NH3)2]+}}^\circ&=E_{\ce{Ag+}}^\circ+\frac{RT}{F}\cdot\ln\frac{1}{K_\text{B}} \\ &\approx E_{\ce{Ag+}}^\circ-0.059\ \mathrm{V}\cdot\log K_\text{B} \end{aligned} $$

The complex formation equilibrium slows down the overall reaction. A slow, controlled reaction is important for creating the desired silver mirror. If $\ce{Ag+}$ is reduced too quickly, colloidal silver metal would appear, which would create a black cloudy liquid.

  • $\begingroup$ How have you got from the third to the fourth line of the redox potential explanation with $K_B$ $\endgroup$
    – bon
    May 9, 2015 at 17:39
  • $\begingroup$ @bon For the standard electrode potential $E_{\ce{[Ag(NH3)2]+}}^\circ$, I assume that the activity $a$ of each solute (except $\ce{Ag+}$) is 1 (i.e. $c\approx 1\ \mathrm{mol/l}$); thus $\left[\ce{[Ag(NH3)2]+}\right]=1$ and $\left[\ce{NH3}\right]=1$. $\endgroup$
    – user7951
    May 9, 2015 at 18:00
  • $\begingroup$ Ah ok that makes sense $\endgroup$
    – bon
    May 9, 2015 at 18:02

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