Zinc Oxide reduction

**mheslep** · Nov8-09, 09:59 PM

The two commercial methods of reducing ZnO to Zn appear to be
Thermal, in the presence of C:
2 ZnO + C → 2 Zn + CO₂
and electrowinning:
ZnO + H₂SO₄ → ZnSO₄ + H₂O (followed by electrolysis on the ZnSO₄)

But what is the effect of straight electrolysis on ZnO in the presence of some electrolyte to begin with? That is, doesn't electrolysis directly reduce ZnO (as in a charging a depleted Zn-Air battery), and if that's the case why isn't ZnO electrolysis an economic way of producing Zn from ore?

**chemisttree** · Nov9-09, 12:07 AM

zinc oxide is nearly insoluble in water or non-acidic electrolyte.

I'm not sure about the economics of the two processes you reference. I would imagine the electrowinning process produces very pure material and the carbon reduction would produce less pure product.

**lightarrow** · Nov9-09, 03:05 PM

Probably is possible to get Zn from electrolysis of melted ZnO at high temperatures using a flux, as the way Al is produced from Al₂O₃ using Na₃AlF₆ as flux. In the case of Al there is quite no other way (you don't get Al from the oxide reducing it with C and you don't get Al from electrolysis of acqueous solutions of its salts); in the case of Zn the two processes you described are certainly much more simple and cheap, so I presume it's for this reason they are used.

**mheslep** · Nov9-09, 07:19 PM

Probably my OP question is ill phrased. Let me try again.

Given a Zn anode, who's surface is largely oxidized to ZnO, and the ZnO is in contact with some alkaline electrolyte, should not the application of an electric potential to the electrode force reduction of the ZnO? Assume anything you like for the cathode, i.e. air via some porous material:

ZnO + 2e + H₂O → Zn(OH)₂ → 2OH^- + Zn

This is basically a Zinc Air battery being charged. The Zn need never enter solution, and no high temperatures required. It should be inexpensive?

**mheslep** · Nov9-09, 08:15 PM

Oh, I see that's a non trivial amount of charge per kg of Zn:

Zn @ 65g/mole
1 mole e = 96485 C, 3600 C / amp-hour
1 mole e = 26 amp-hours, or
52 amp-hours per mole of Zn,
800 amp-hours per kg of Zn @ 2V = 1.6 kWh per kg of Zn.

That's about $0.12 per kg Zn, with see Zinc @ $2.1/kg, about 6% of the cost. Seems reasonable.