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This Bulletin describes the simultaneous determination of gold and copper by potentiometric titration using an Fe(II) solution as titrant. Fe(II) reduces Au(III) directly to the free metal, whereas Cu(II) does not react. By the addition of fluoride ions the Fe(III) is complexed and a shift of the redox potential is effected. Afterwards, potassium iodide is added, thus reducing the Cu(II) to Cu(I), and the free iodine is again titrated with the Fe(II) solution using a Pt Titrode.

Chemical reactions:

Au(III) + 3 Fe(II)  → Au + 3 Fe(III)

2 Cu(II) + 2 I-   → 2 Cu(I) + I2

I2 + 2 Fe(II) → 2 I- + 2 Fe(III)


Metrohm AG

9100 Herisau