In the sepia "loning" of a black-and-white photograph, the image is converted to a rich brownish violet by placing the finished photograph in a solution of gold (III) ions, in which metallic gold replaces the metallic silver. Use Appendix D to explain the chemistry of this process.

The given information is when gold (III) ions are mixed with silver metal, the gold replaces the silver - that is, the silver metals are oxidized and turn into silver (I) ions

A coordination complex consists of a central atom or ion, which is typically metallic and is termed the coordination centre, and a surrounding array of bound molecules or ions, that are successively referred to as ligands or complexing agents.

When gold (III) ions are mixed with silver metal, the gold replaces the silver - that is, the silver metals are oxidized and turn into silver (I) ions, while the gold metals are oxidized, and turn from $A{u}^{3+}$ions to solid gold metal.

This happens because the $E_{\text{half}-\text{cell}}^o$of $A{u}^{3+}$ is more positive than that of $A{g}^{+}$.

Value $E_{\text{half - oell}}^o=1.50V$ for Au (III) to Au and value $E_{\text{half - eell}}^o=0.80V$ for Ag (l) to Ag gives :

$E_{\text{eell}}^o=1.50-0.80=0.70\text{V}$

Since the potential is less than that of , it will be oxidized and the reaction will be reversed. In the photograph, Ag will be the anode and produce ${\text{Ag}}^{+}$while Au(III) will be reduced in the cathode, and replace the silver metal in the photograph.

This occurs readily because the $E_{\text{cell}}^a$ of the reaction is positive, indicating that it occurs spontaneously.