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Chapter 20: Problem 95

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

Expert verified
Gold remains below the anode because it has a higher reduction potential than copper, preventing it from oxidizing during electrolysis.

Step by step solution

01

Understanding Electrolysis

In electrolysis, electricity is passed through a solution or molten compound to cause a chemical reaction. In the case of refining metals, it is used to separate various metals from impure metal mixtures using the principles of oxidation and reduction.
02

Breakdown of the Electrolysis Process

During the electrolysis of a mixture containing copper and gold, the impure metal acts as the anode (positive electrode) while a pure metal layer is formed at the cathode (negative electrode). The process involves the transfer of electrons from the anode to the cathode.
03

Understanding Metal Reduction Potentials

Gold has a higher reduction potential than copper. This means gold is less likely to undergo oxidation compared to copper. Therefore, during electrolysis, copper ions easily dissolve into the solution, while gold remains solid at the anode.
04

Behavior of Copper and Gold during Electrolysis

When electricity passes through the metal mixture, copper atoms lose electrons and go into the solution as copper ions. Since gold has a lower tendency to oxidize, it doesn't dissolve but instead falls from the anode as a solid and gets deposited below it.
05

Conclusion of the Process

The mixture is refined by electrolysis, with gold remaining as a solid below the anode due to its high reduction potential, while copper ions travel to the cathode and form a pure copper layer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Metal Refinement
Metal refinement is a process used to purify metals and extract valuable components from a mixture. One of the most effective ways to achieve refinement is through electrolysis. In electrolysis, an electric current facilitates the separation of metals by using two electrodes placed in a solution. Here's how it works:
  • The impure metal acts as the anode, the positive electrode.
  • A sheet of pure metal serves as the cathode, the negative electrode.
  • When electric current flows, ions from the anode dissolve into the solution.
  • These ions can then travel to the cathode where they deposit as pure metal.
This technique is particularly useful because it allows selective refinement, depending on the properties of the metals involved, such as their reduction potential.
Oxidation and Reduction
Oxidation and reduction are chemical reactions that involve the transfer of electrons between substances. In an electrochemical process like electrolysis:
  • Oxidation occurs at the anode, where a substance loses electrons.
  • Reduction occurs at the cathode, where a substance gains electrons.
During the electrolysis of mixed metals, the different tendencies of metals to oxidize or reduce play a crucial role. Metals with lower oxidation potential readily lose electrons and go into solution, while those with higher reduction potential resist oxidation, remaining solid or depositing onto the cathode.
Understanding these reactions is vital, as they determine which metals get dissolved and which remain unchanged during the refinement process.
Reduction Potential
Reduction potential is a measure of the tendency of a substance to gain electrons and undergo reduction. It is usually expressed in volts and indicates how easily a metal can be reduced.
The higher the reduction potential, the less likely the metal is to oxidize, meaning it prefers to stay in its solid state or be deposited as pure metal on the cathode. For example:
  • Gold, with a high reduction potential, does not oxidize easily.
  • Copper, with a lower reduction potential compared to gold, oxidizes and dissolves more readily.
This concept explains why, during the electrolysis of copper and gold, gold remains at or below the anode as solid metal, while copper dissolves to form ions that migrate to the cathode.
Copper and Gold Separation
The separation of copper and gold through electrolysis involves understanding each metal's behavior under electric current. During the process:
  • Copper atoms at the anode lose electrons and dissolve into the solution as ions.
  • These ions move toward the cathode, where they gain electrons again, depositing as pure copper.
  • Gold, however, remains at the anode due to its high reduction potential, resisting oxidation.
  • As a result, gold forms a residue below the anode as a solid.
This selective separation allows for effective refinement in electrolysis, as gold, with its significant economic value, is preserved in its metallic form, while the copper is purified through deposition.

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Most popular questions from this chapter

Elemental calcium is produced by the electrolysis of molten \(\mathrm{CaCl}_{2}\). (a) What mass of calcium can be produced by this process if a current of \(7.5 \times 10^{3} \mathrm{~A}\) is applied for \(48 \mathrm{~h}\) ? Assume that the electrolytic cell is \(68 \%\) efficient. (b) What is the minimum voltage needed to cause the electrolysis?

You may have heard that "antioxidants" are good for your health. Is an "antioxidant" an oxidizing agent or a reducing agent? [Sections 20.1 and 20.2\(]\)

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are $$ \begin{array}{l} \mathrm{HgO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Hg}(l)+2 \mathrm{OH}^{-}(a q) \\ \mathrm{Zn}(s)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \end{array} $$ (a) Write the overall cell reaction. (b) The value of \(E_{\mathrm{red}}^{\circ}\) for the cathode reaction is \(+0.098 \mathrm{~V}\). The overall cell potential is \(+1.35 \mathrm{~V}\). Assuming that both half- cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

Magnesium is obtained by electrolysis of molten \(\mathrm{MgCl}_{2}\). (a) Why is an aqueous solution of \(\mathrm{MgCl}_{2}\) not used in the electrolysis? (b) Several cells are connected in parallel by very large copper bars that convey current to the cells. Assuming that the cells are \(96 \%\) efficient in producing the desired products in electrolysis, what mass of \(\mathrm{Mg}\) is formed by passing a current of 97,000 A for a period of 24 h?

A voltaic cell utilizes the following reaction: $$ \mathrm{Al}(s)+3 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Al}^{3+}(a q)+3 \mathrm{Ag}(s) $$ What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution. (b) The size of the aluminum electrode is increased. (c) A solution of \(\mathrm{AgNO}_{3}\) is added to the cathode half-cell, increasing the quantity of \(\mathrm{Ag}^{+}\) but not changing its concentration. (d) HCl is added to the \(\mathrm{AgNO}_{3}\) solution, precipitating some of the \(\mathrm{Ag}^{+}\) as \(\mathrm{AgCl}\).
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