cuso4+water+agcn --- copper red type of presifitate what chemical missed here?

Answer 1

NH4OH, Ammonium Hydroxide

Answer 2

Lancenigo di Villorba (TV), Italy

YOUR STUFFS
As you know, copper sulphate penta-hydrate (e.g. CuSO4.5H2O) is the most common copper salt in lab's stores.
Silver nitrate (e.g. AgNO3) is the most common silver salt in lab's stores.
Sodium cyanide (e.g. NaCN) is one among the most common available cyanide salts.

WHAT YOU MADE
As you know, when silver and cyanide ions are present in an aqueous solutions, they react toward most stable ion, as follows :
Ag+(aq) + 2 CN-(aq) ---> [Ag(CN)2]-(aq)
where you retrieve [Ag(CN)2]- as dicyanoargentate(I) anion ; the latter is known as silver/cyanide's reaction of LIEBIG.
Now, you can find the case of a white tarnished solution or the case of a clear and uncoloured one, this depending on silver/cyanide ion's amount Ratio.
If the Ratio is LOWER THAN 0.5, cyanide's amount exceeds silver ion's one thus you can write :
Ag+NO3-(aq) + 2 Na+CN-(aq) --->
---> Na+[Ag(CN)2]-(aq) + Na+NO3-(aq)
thus you have the uncoloured liquid.
On the contrary, if the Ratio is GREATER THAN 0.5, you write :
2 Ag+NO3-(aq) + 2 Na+CN-(aq) --->
---> Ag+[Ag(CN)2]-(s) + 2 Na+NO3-(aq)
thus you have a white solid precipitate Ag[Ag(CN)2] as silver dicyanoargentate(I). The referee where you retrieve the strange writing as AgCN simplify the CORRECT EXPRESSION
like Ag[Ag(CN)2] it is.
Since I assume your aqueous media present any solid body of silver compounds, Silver/Cyanide Ion's amount Ratio is lower than 0.5 and they exist "free" cyanide ions able to react with copper ones. The latters follow the reaction :
2 Cu++(aq) + 6 CN-(aq) ---> 2 [Cu(CN)2]-(aq) + (CN)2(g)
where you retrieve [Cu(CN)2]--- as dicyanocuprate(I) anion and (CN)2 as the cyanogen gas. You may see the disappearing of blue-greenish coloration of aqueous media while some gas bubbles lift upward.

WHAT YOU CAN DO
Now, you must evaluate the reversible electrical potential interesting [Cu(CN)2]- anion in the following half-reaction :
[Cu(CN)2]-(aq) + e ---> Cu(s) + 2 CN-(aq)
You can start from another half-reaction more known than the previous one, I refere to :
Cu+(aq) + e ---> Cu(s)
which reversible electrical potential is evaluated in 0.52 V.
By applying the well-known Nernst's relations you calculate the corrective factor, it is connecting the second equilibrium to the former one. The calculus involved is :
R * T * (1 / F) * ln (Kinst * |[Cu(CN)2]-| / |CN-|^2) =
= 8.31 * 298.16 * (1 / 96500) * ln (1E-24 * 1 / 1^2) = - 0.62 V
If I refer the calculus to standard conditions, in aqueous media I must assume dicyanocuprate(I) and cyanide solution as 1 M concentrated. Thus, the corrective factor is evaluated as - 0.62 V and the electrical potential becomes - 0.10 V for the
half-reaction :
[Cu(CN)2]-(aq) + e ---> Cu(s) + 2 CN-(aq)

Finally, your conclusion involve a chemical reducing agent having an reversible electrical potential LOWER THAN - 0.10 V.
You understand this is NOT the case of common reducing stuffs as sulphurous acid.
YOU MUST ADD POWDER OF METALIC ZINC
(e.g. [Zn(CN)4]-(aq) + 2 e ---> Zn(s) + 4 CN-(aq) - 1.26 V).

I hope this helps you.