Question

A mixture of copper and gold metals that is subjected to electrorefining contains tellurium as an impurity. The standard reduction potential between tellurium and its lowest common oxidation state, \(\mathrm{Te}^{4+},\) is $$ \mathrm{Te}^{4+}(a q)+4 \mathrm{e}^{-} \longrightarrow \mathrm{Te}(s) \quad E_{\mathrm{red}}^{\circ}=0.57 \mathrm{V} $$ Given this information, describe the probable fate of tellurium impurities during electrorefining. Do the impurities fall to the bottom of the refining bath, unchanged, as copper is oxidized, or do they go into solution as ions? If they go into solution, do they plate out on the cathode?

Short answer

During the electrorefining process, tellurium impurities will go into the solution as Te^(4+) ions, and they'll likely plate out on the cathode as they have a higher standard reduction potential than copper (0.57 V) but lower than gold (1.50 V).

Step-by-step solution

Understand the electrorefining process

Electrorefining is a process that involves applying an electrical current to dissolve and purify metals. An anode (positive electrode) made of impure metal is put into an electrolyte solution containing metal ions. A cathode (negative electrode) is placed in the same solution. The metal ions in the solution are attracted to the cathode, where they gain electrons and are reduced to metal atoms. These metal atoms build up on the cathode, while the impurities either fall to the bottom or stay in the solution.

Compare the standard reduction potentials

To determine the behavior of tellurium during electrorefining, we need to compare the standard reduction potentials (E°) of copper (Cu), gold (Au), and tellurium (Te). The given standard reduction potential of tellurium is: \[ \mathrm{Te}^{4+}(a q)+4 \mathrm{e}^{-} \longrightarrow \mathrm{Te}(s) \quad E_{\mathrm{red}}^{\circ}=0.57 \mathrm{V} \] Now let's look up the standard reduction potentials for copper and gold. They are: Copper: \[ \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \quad E_{\mathrm{red}}^{\circ}=0.34 \mathrm{V} \] Gold: \[ \mathrm{Au}^{3+}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s) \quad E_{\mathrm{red}}^{\circ}=1.50 \mathrm{V} \]

Determine the fate of tellurium impurities

During the electrorefining process, metals with lower reduction potentials (more negative) are easier to oxidize and dissolve, while metals with higher reduction potentials (more positive) are more readily reduced and plated out on the cathode. To determine the fate of tellurium, it's important to compare its reduction potential with those of copper and gold: - Te: E° = 0.57 V - Cu: E° = 0.34 V - Au: E° = 1.50 V Since Te has a higher standard reduction potential than Cu but lower than Au, Te is more readily oxidized (goes into solution as ions) than Au and more likely to be reduced (plated out) than Cu. Thus, we can conclude that tellurium impurities will go into the solution as Te^(4+) ions, and they'll likely plate out on the cathode during the electrorefining process.

Key concepts

Standard Reduction Potential

When studying electrochemistry, the standard reduction potential (E°) is a key concept, as it helps us predict the behavior of elements in an electrolytic process.

Standard reduction potential is a measure of the tendency of a chemical species to be reduced, and it's represented by a voltage value. When comparing different substances, those with more positive E° values have a greater likelihood of gaining electrons and becoming reduced. In contrast, substances with more negative E° values tend to lose electrons and undergo oxidation.

In the context of electrorefining, by comparing the E° values of different metals, we can determine which metal will be deposited, or plated, onto the cathode. Metals with higher E° values plate onto the cathode preferentially, as they are more likely to gain electrons from the cathode.

Oxidation States

Understanding oxidation states is essential for explaining chemical reactions, especially in electrorefining. An oxidation state indicates the degree of oxidation of an atom within a compound. It is represented by a numerical value that can be positive, zero, or negative.

In the given exercise, we are considering Tellurium (Te) in its +4 oxidation state, denoted as Te^(4+). Oxidation states allow us to write and balance chemical equations accurately. They also help in determining how electrons are transferred in a redox reaction, which is the basis of the electrorefining process. An atom's oxidation state helps predict how it will interact when electrical current is applied: whether it will lose electrons and dissolve into the solution or gain electrons and precipitate onto the electrode.

Metal Purification

The process of refining and purifying metals is fundamental in obtaining materials with the desired levels of purity for industrial applications. Metal purification, like electrorefining, employs the principles of electrochemistry to separate pure metals from their impurities.

During electrorefining, an electrical current is used to transfer metal ions from an impure anode to a pure cathode submerged in an electrolyte solution. Impurities with different reduction potentials either remain in the electrolyte, dissolve, or collect at the bottom as sludges, depending on their nature and their reaction to the electrical current.

The efficiency of this process is influenced by various factors, including the relative standard reduction potentials of the metals involved, ensuring that the desired metal is deposited at the cathode, while impurities are removed, resulting in a high-purity final product.