How do you balance this redox equation? Au(s) + CN−(aq) + O2(g) → Au(CN)4−(aq)?
I'm having trouble figuring this one out..
Au(s) + CN^−(aq) + O2(g) → Au(CN)4^−(aq). It's occurring in a basic aqueous medium so you'd add either H20(l) or OH^-(aq).
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O2(g) + 2 H2O + 4 e⁻ ⇋ 4 OH−(aq) E° = +0.40V
Au3+ + 3 e⁻ ⇋ Au(s) E° = +1.52 V (one of the highest known; Au is "noble")
http://en.wikipedia.org/wiki/Table_of_standard_ele...
The reason why the rxn goes is the binding by the strong CN^- ligand stabilizes Au(III) as the soluble [Au(CN)4]^- cmplx anion.
So we need to × the top rxn by 3 and the one underneath by 4 so we don't have any ¼ e⁻ wandering around. 4 Au so 16 CN^- ligands:
4Au(s) + 16CN^−(aq) + 3O2(g) + 6H2O → 4Au(CN)4^−(aq) + 12OH^-
By Gad Sir - I think you've got it. Elementary dear Watson, elementary.
Actually in the extraction of Au it is the Au(I) cmplx [Au(CN)2]^- that is employed.