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Chapter 20: Problem 95

Metallic gold is collected from below the anode when crude copper metal is refined by electrolysis. Explain this behavior.

Short Answer

Expert verified
During the electrolysis of crude copper, metallic gold is collected below the anode because it has a lower reactivity and oxidation potential than copper. Gold does not participate in the oxidation or reduction reactions during electrolysis and remains unoxidized in the sludge that forms below the anode. As the process continues, concentrations of gold and other less reactive metals increase and eventually precipitate out for collection.

Step by step solution

01

Understanding Electrolysis

Electrolysis is a process where electrical energy is used to drive a non-spontaneous redox reaction. In this process, an electric current is passed through an electrolyte, causing the oxidation and reduction to occur at the anode and cathode, respectively. The anode is the positive electrode where oxidation occurs, while the cathode is the negative electrode where reduction occurs.
02

Electrolysis of Copper

During the electrolysis of crude copper, the copper is oxidized at the anode and then reduced at the cathode to form pure copper. During this process, impurities in the crude copper anode dissolve into the electrolytic solution and accumulate in the form of sludge. Gold, being one of the impurities in the crude copper, also dissolves during the electrolysis process.
03

Gold Accumulation below the Anode

Gold has a lower reactivity than copper and therefore, it does not preferentially oxidize in the electrolytic cell. So, it remains unoxidized in the sludge that forms below the anode during electrolysis. As the process continues, the concentrations of gold and other less reactive metals increase, and eventually, they will precipitate out and get collected at the bottom of the anode. In other words, gold does not participate in the oxidation or reduction reactions during electrolysis and gathers below the anode because it has a lower oxidation potential than copper. This behavior of gold allows for its extraction and purification from the crude copper through the electrolysis process.

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Most popular questions from this chapter

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) - (a) \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2}+(a q)+2 \mathrm{Ag}(s)\) (b) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+\) \(\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\) \(5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4}(a q)\)

Given the following reduction half-reactions: $$ \begin{gathered} \mathrm{Fe}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) \quad E_{\text {red }}^{\infty}=+0.77 \mathrm{~V} \\ \mathrm{~S}_{2} \mathrm{O}_{6}^{2-}(a q)+4 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \quad E_{\mathrm{ret}}^{\circ}=+0.60 \mathrm{~V} \\ \mathrm{~N}_{2} \mathrm{O}(g)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(I) \quad \mathrm{E}_{\mathrm{red}}^{\circ}=-1.77 \mathrm{~V} \\ \mathrm{VO}_{2}^{+}(a q)+2 \mathrm{H}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{VO}^{2+}+\mathrm{H}_{2} \mathrm{O}(l) \quad E_{\mathrm{red}}^{0}=+1.00 \mathrm{~V} \end{gathered} $$ (a) Write balanced chemical equations for the oxidation of \(\mathrm{Fe}^{2+}(a q)\) by \(\mathrm{S}_{2} \mathrm{O}_{6}^{2-}(a q)\), by \(\mathrm{N}_{2} \mathrm{O}(a q)\), and by \(\mathrm{VO}_{2}^{+}\)(aq). (b) Calculate \(\Delta G^{\circ}\) for each reaction at \(298 \mathrm{~K}\). (c) Calculate the equilibrium constant \(K\) for each reaction at \(298 \mathrm{~K}\).

A voltaic cell similar to that shown in Figure \(20.5\) is constructed. One electrode half-cell consists of a silver strip placed in a solution of \(\mathrm{AgNO}_{\mathrm{g}}\) and the other has an iron strip placed in a solution of \(\mathrm{FeCl}_{2}\). The overall cell reaction is $$ \mathrm{Fe}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+2 \mathrm{Ag}(s) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the silver electrode to the iron electrode or from the iron to the silver? (f) In which directions do the cations and anions migrate through the solution?

In a Li-ion battery the composition of the cathode is \(\mathrm{LiCoO}_{2}\) when completely discharged. On charging approximately \(50 \%\) of the \(\mathrm{Li}^{+}\)ions can be extracted from the cathode and transported to the graphite anode where they intercalate between the layers. (a) What is the composition of the cathode when the battery is fully charged? (b) If the \(\mathrm{LiCoO} 2\) cathode has a mass of \(10 \mathrm{~g}\) (when fully discharged), how many coulombs of electricity can be delivered on completely discharging a fully charged battery?

\mathrm{~A}\( voltaic cell is based on the reaction $$ \mathrm{Sn}(s)+\mathrm{I}_{2}(s) \longrightarrow \mathrm{Sn}^{2+}(a q)+2 \mathrm{I}^{-}(a q) $$ Under standard conditions, what is the maximum electrical work, in joules, that the cell can accomplish if \)75.0 \mathrm{~g}\( of \)\mathrm{Sn}$ is consumed?
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