Uploaded October 2016 | Updated September 2026, 1 week ago
A copper coin in placed in a silver nitrate solution. After some time (video is 6x faster) a dark layer forms on the surface. The Silver is reduced to elemental form, but doesn't stick to the surface like a Mercury coating would do as no alloy forms. This is just like the Iron nails in the copper solution. The nails appeared to be coated by copper, but it can be easily peeled of.
A copper coin in placed in a silver nitrate solution. After some time (video is 6x faster) a dark layer forms on the surface. The Silver is reduced to elemental form, but doesn't stick to the surface like a Mercury coating would do as no alloy forms. This is just like the Iron nails in the copper solution. The nails appeared to be coated by copper, but it can be easily peeled of.
. If this is acidified, the vanadates can much like the Chromate and Dichromate to Polychromate condense by giving off water and form corner-sharing tetrahedra in long chains. At pH 8 to 13 those shorter Vanadates and the in this experiment used Metavanadates (VO3(-))n form. At pH 2 to 6 the red/orange Decavanadates form and below pH 2 which I am not sure whether we really reached it as we only added some drops to it Vanadyl-cations form [VO2](+) or probably better if you saw our Ammine-copper/nickel video a [VO2(H2O)4](+). This can now be reduced with nascent Hydrogen or Hg:Zn and probably other reducing agents as well to Vanadium(IV) which is blue and looks like this [VO(H2O)5](2+). What is happening is the VO2(+) changes to a VO(2+) and as you see we loose an oxygen in this reaction.
The next step forms the green Vanadium(III) [V(H2O)6](3+) and the purple Vanadium(II) [V(H2O)6](2+). The color green appears in between but seems to be a mixture of the yellow Vanadium(V) and the blue Vanadium(IV).
There is an additional note on the last page in the book which says:
For V(III) its not [V(H2O)6]Cl3 but [VCl2(H2O)4]Cl as trans-Isomer.
For V(II) its not [V(H2O)6]Cl2 but again trans-[VCl2(H2O)4].
V(II) will react with water to form Hydrogen and V(III) again all of them will react with Oxygen to form V(V). Vanadium Chemistry: V(V) reduction to V(IV), V(III) and V(II) using nasc. Hydrogen](https://i.ytimg.com/vi/wbsujp2dEMw/mqdefault.jpg)
 is deprotonated to form [Ni(H2O)4(OH)2] which is often just called Ni(OH)2. If more Ammonia is added the [Ni(NH3)6](2+) complex forms.
Here the color of the Ammine and the Aqua complex differ without any change in geometry or spin which makes this a good example to show how Ligands can cause the color of complexes. If you saw our Vanadium and Manganese reduction you saw how different oxidation states on the central atom change the color as well.
As there are no pi-bonds in water or ammonia there wont be any backbonding-effect so both can be easy compared. Ammonia is a way stronger ligand than water is, which means that the electron density of the electrons donated to the central atom will be closer to the ligand in water than in ammonia which results in a bigger ligand field splitting in the Ammine complex. Light of shorter wavelength must be absorbed now and the complex gets blue shifted from green to purple-blue. As both ligands have only Hydrogen we think the difference comes from the electronegativity. As Oxygen is more electronegative than Nitrogen it will pull the electron density closer to itself causing a red shift in comparison to the Ammonia. Nickel Chemistry: Hexamminenickelchloride](https://i.ytimg.com/vi/wcHXHVu0_TM/mqdefault.jpg)





![Copper Chemistry: Cu(II)/As(III) chameleon [no sound, 2x speed]
In this video, we show another way to make a beautiful spectrum of colors using Copper chemistry. The reaction is the following: Arsenic(III) is oxidized to Arsenic(V), while Copper(II) is reduced to Copper(I), all happening at a high pH value. But of course, we cannot just add cations and so different precipitates form and dissolve in this process creating many different colors.
We start off with a CuSO4 solution, which is light blue. To this, we add some NaAsO2. We already showed you Copperarsenite in another video. The blue-green precipitate also leaves a greenish aqueous layer on top. When we add KOH the remaining CuSO4 forms the dark blue Cu(OH)2. You can see the surface of the KOH turn dark blue. There is also an equilibrium between Arsenite and
Hydroxide and so the overall color turns more blueish in this process. Now the actual reaction also requires some heat. The KOH dissolving in so little water already causes it to become quite hot. This is why you suddenly see red-brown spots appearing everywhere. When we finally heat it all the Cu(II) is reduced to Cu(I) forming Cu2O.
Much like with our silver video, this is another case where some redox-reactions are just much easier accessible when high pH values are used.
For Arsenic(III to V) the potentials are 0,56 and -0,71 for pH=0 and 14 and for Copper(II to I) the potentials are 0,159 and -0,08.
To reduce Copper(II) to Copper(I) the redox potential of the reducing agent, here Arsenic(III) needs to be lower, which is only fulfilled at higher pH values. Copper Chemistry: Cu(II)/As(III) chameleon [no sound, 2x speed]](https://i.ytimg.com/vi/xHic2pew9bI/mqdefault.jpg)
![Sulfur Chemistry: [Fe(CN)5NOS] from Sodium Polysulfide
There is an interesting reagent called Sodium Nitroprusside. It is sometimes used to identify several ions and is capable of distinguishing between Sulfite and Sulfide for example as well as forming many colored complexes. As we showed in our Cadmium Sulfide video, Sodium Polysulfide, made from elemental Sulfur and Sodium Hydroxide seems to be an alternative for the rather unstable Sodium Sulfide. So we tried to make the famous Sulfide test using Nitroprusside and a selfmade Polysulfide solution.
Here is the result. Note, as we didnt have any other source of H2S or Sulfide we could not compare this to the real test. So we assume the compound formed should have the composition as given in the title. Sulfur Chemistry: [Fe(CN)5NOS] from Sodium Polysulfide](https://i.ytimg.com/vi/xm7IYOrwQfA/mqdefault.jpg)

