Question

Metallic gold is collected from below the anode when a mixture of copper and gold metals is refined by electrolysis. Explain this behavior.

Short answer

During the electrolytic refining of a copper and gold mixture, gold is deposited below the anode due to its lower reactivity and higher standard electrode potential (\(+1.50\, V\)) compared to copper (\(+0.34\, V\)). Both copper and gold are oxidized at the anode and form ions, but gold ions are more likely to be reduced back to their metallic form and collect below the anode. Meanwhile, copper ions migrate toward the cathode, where they are reduced to form metallic copper. This selective reduction process separates and collects metallic gold during electrolysis.

Step-by-step solution

Understand the electrolysis process

Electrolysis is a technique used to drive non-spontaneous redox reactions by applying an external electric current. In an electrolytic cell, the anode is the electrode where oxidation occurs, and the cathode is the electrode where reduction takes place. During electrolysis, ions present in the solution will move towards the electrodes, with positive ions (cations) migrating to the cathode, while negative ions (anions) move towards the anode.

Contrast gold and copper reactivity

Gold and copper are both metals with different reactivities. Gold is a noble metal, which means it is resistant to corrosion and oxidation in moist air. It is also one of the least reactive chemical elements and usually exists in a highly stable state. On the other hand, copper is more reactive and can form ions more easily. This difference in reactivity will play a role in the electrolytic refining process.

Explore the electrochemical series and standard electrode potentials

The electrochemical series is a list of elements sorted by their standard electrode potentials. Elements with higher electrode potentials are more likely to be reduced, while those with lower potentials are more likely to be oxidized. Gold has a standard electrode potential of \(+1.50\, V\) and copper has \(+0.34\, V\). Because of the difference in the standard electrode potentials, gold is more likely to be reduced than copper in the presence of an electric current during electrolysis.

Explain the behavior during electrolytic refining

When refining a mixture of copper and gold, electrolysis is used to separate the gold from the copper. At the anode, both copper and gold metal atoms lose electrons and become ions, moving into the electrolyte solution. However, due to the higher standard electrode potential of gold, it is more likely to be reduced back to its metallic form, while copper remains in the solution as ions. As a result, gold deposits below the anode, while copper ions continue to migrate towards the cathode. At the cathode, copper ions are reduced to form metallic copper, which is deposited. This selective reduction process enables the separation and collection of metallic gold during the electrolytic refining process.