It was generally believed at one time that the precipitation of gold and silver by zinc was effected by simple displacement, according to the equation:

2KAuCy2 + Zn = K2ZnCy4 + 2Au

Some considerations have been adduced, however, which throw doubt on this view. It is well known that a considerable excess of free cyanide must be present to enable the precipitation to take place quickly and efficiently, and if the solution contains less than 0.05 per cent. KCy the solution must be in contact with the zinc for an hour. Pure sheet zinc in particular is quite unable to precipitate gold in the absence of free potassium cyanide, although the filiform zinc slowly extracts the gold from the same solution, perhaps owing to the presence of lead in the zinc. Some suggest that the cause of this is that hydrogen is set free by the first action of the aurocyanide of potassium on the zinc, thus:

2KAuCy2 + 2Zn + 2H2O = 2KOH + H2 + 2ZnCy2 +2Au

This equation practically demands the replacement of the K ions in the aurocyanide by zinc—thus:

2(K + AuCy2)  + Zn = (Zn + 2AuCy2) + 2K

followed by a re-arrangement of the molecule of zinc-gold cyanide and the replacement of the gold by another atom of zinc. This mechanism of change seems the more probable when it is remembered that the K ions in KCy itself are displaced by zinc in the presence of water.

Assuming, then, that hydrogen is set free as above, it would form an imperceptible layer of hydrogen on the surface of the sheet zinc, which would thereby become “polarised” and protected from further action, so that the reaction would soon cease. Hydrogen would also be set free on the surface of the thread zinc, but in this case the ragged edges of the shavings would assist the gas to form into bubbles and become detached, so that less “polarisation” would occur. Hydrogen is actually given off after some time has elapsed, but there are other theories to account for its formation, as noted below. It does not appear, moreover, how the presence of free KCy would prevent polarisation by hydrogen, and the prevention of the action may be due to the formation of an invisible protective layer of insoluble ZnCy2 on the surface of the zinc, which would be at once dissolved if a sufficient amount of potassium cyanide were present. In the absence of KCy, the reaction proceeds:

4ZnCy2 + 4KOH = 2ZnCy2 + ZnCy2. 2KCy + Zn(KO)2 + 2H2O

This reaction is, of course, incomplete, but unless the amount of potash in the solution were large, it is clear that some insoluble zinc cyanide would remain. The reaction is completed by the solution of ZnCy2 in free potassium cyanide.

ZnCy2 + 2KCy = K2ZnCy4

According to Christy’s formula, 1 part of zinc should precipitate about 3 parts of gold, but in practice 1 part of zinc precipitates only from 0.06 to 0.2 part of gold, the zinc being wasted by direct dissolution in potassium cyanide and by the reactions described below.

Pure zinc has a slower action on solutions of potassium cyanide than ordinary commercial zinc containing about 1 per cent, of’ lead, and the action of the gold-zinc couple, formed by the black deposit of gold (which may be really a compound of gold and zinc), and the unaltered zinc, is still more vigorous. This gold- zinc couple probably develops enough electromotive force to decompose water thus:

Zn + 2H2O = Zn(OH)2 + H2

The positive element, zinc, is thus oxidised, and subsequently dissolved by the cyanide solution. As a matter of fact, Messrs. Butters & Clennel observed a vigorous evolution of small bubbles, which proved to be mainly hydrogen, in the zinc boxes at the Robinson Mine. The hydrogen thus evolved doubtless carries off mechanically some hydrocyanic acid, and so the odour of the latter, noticeable above the zinc boxes, may be accounted for. The nascent hydrogen also unites directly with hydrocyanic acid, forming methylamine, thus:

HCN + 2H2 = CH2.NH2

The presence of the methylamine may in part account for the ammoniacal odour sometimes occurring above the zinc boxes, which may, however, be also caused by the hydrolytic decomposition. The hydrate of zinc, formed as shown above, dissolves at once in excess of the cyanide:

Zn(OH)2 + 4KCy = K2ZnCy4 + 2KOH,

and the increase in alkalinity of the solution is thus explained.
Caustic potash, however, acts directly on the zinc to some extent forming potassium zincate, thus:

Zn + 2KOH = Zn(OK)2 + H2

but the zincate is, according to N. Anderson, at once dissolved by potassium cyanide, as follows:

Zn(OK2) + 4KCy + 2H2O = ZnCy2. 2KCy + 4KOH

In this way the solution continually tends to become more alkaline, but is neutralised by carbonic acid absorbed from the atmosphere.

By the reactions in the zinc boxes, not only is the potassium cyanide solution weakened by decomposition, as above, but a large amount of zinc is dissolved. Goyder has shown that the presence of iron in contact with the zinc is injurious, checking the rate of precipitation of the gold, and increasing the waste of zinc and the tendency for the formation of ZnCy2 on the zinc. Iron screws and sieves should, therefore, not be used in the precipitation boxes. The double cyanide of zinc and potassium is not available for the solution of the precious metals, and, speaking generally, as long as an excess of zinc is present, no gold will be dissolved by a solution of potassium cyanide flowing through the boxes.

An exception may be cited, however, in cases where solutions are aerated as they pass to the boxes. Caldecott stirred zinc-copper shavings with a solution of cyanide through which air was passed, with the result that gold previously deposited on the metal was rapidly dissolved.

The presence of large quantities of the double cyanide of zinc and potassium in the solutions is not prejudicial to the solvent action of the simple cyanides. At the Mercur Mine the stock solution was apparently as efficacious after nine months’ use as at the start, although it must have contained large quantities of zinc cyanide. Feldtmann showed that gold in ores can be dissolved by zinc-potassium cyanide, but J. S. C. Wells points out that the double cyanide remains undecomposed by gold so long as any simple cyanide is present. An accumulation of base heavy metals in the solution can be got rid of by the addition of soluble sulphides.

Ehrmann has shown by experiments on solutions of various strength, that about as much gold is precipitated by zinc in twenty-four hours at 20° C. as in two hours at 80° C. In the latter case, however, there is, according to Mac Arthur, “an enormous waste of cyanide by the formation of urea, which manifests itself by its strong unpleasant odour.”