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Chapter 20: Problem 69

Silver is sometimes found in nature as large nuggets; more often it is found mixed with other metals and their ores. Cyanide ion is often used to extract the silver by the following reaction that occurs in basic solution: $$ \operatorname{Ag}(s)+\mathrm{CN}^{-}(a q)+\mathrm{O}_{2}(g) \stackrel{\text { Basic }}{\longrightarrow} \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) $$

Short Answer

Expert verified
In the given reaction, silver (Ag, solid) reacts with cyanide ion (CN⁻, aqueous) and oxygen gas (O₂) in a basic environment containing excess hydroxide ions (OH⁻). The cyanide ion forms a coordination complex with silver, producing a soluble silver cyanide complex (Ag(CN)₂⁻) that can be isolated from the mixture. Oxygen promotes oxidation, driving the reaction forward. After isolation, pure silver can be recovered using methods like zinc cementation or electrolysis to remove the cyanide ion.

Step by step solution

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1. Understanding the reaction

In the given reaction, silver (Ag, in solid form) reacts with cyanide ion (CN⁻, in aqueous form) and oxygen gas (O₂) to produce a complex ion of silver cyanide (Ag(CN)₂⁻, in aqueous form).
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2. Basic environment

It is mentioned that the reaction occurs in a basic solution. It means the solution contains an excess of hydroxide ions (OH⁻) which makes the environment favorable for the reaction to take place. The basic environment helps to maintain the stability of the cyanide ion, ensuring efficient extraction of silver.
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3. Role of cyanide ion

Cyanide ion (CN⁻) plays the key role in this reaction. It forms a coordination complex with silver (Ag) atoms, which allows the silver to be separated from other metals and ores. The resulting complex ion (Ag(CN)₂⁻) is soluble in water, facilitating the isolation of silver from its mixtures.
04

4. Role of oxygen

Oxygen gas (O₂) participates in the reaction by promoting the oxidation of silver ions. The addition of O₂ helps drive the reaction forward, making it more efficient at extracting silver from its mixtures.
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5. Isolation of silver cyanide complex

After the formation of the soluble silver cyanide complex (Ag(CN)₂⁻), the solution can then be separated from the mixture. This can be achieved through processes such as filtration or decantation.
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6. Recovery of pure silver

The final step in this exercise involves recovering the pure silver from the silver cyanide complex. This can be done by methods like zinc cementation or electrolysis, which allow for the removal of the cyanide ion and retrieval of the elemental silver.

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Most popular questions from this chapter

A compound related to acetylacetone is 1,1,1 -trifluoroacetylacetone (abbreviated Htfa): Htfa forms complexes in a manner similar to acetylacetone. (See Exercise \(47 .\) Both \(\mathrm{Be}^{2+}\) and \(\mathrm{Cu}^{2+}\) form complexes with tfa \(^{-}\) having the formula \(\mathrm{M}\) (tfa) \(_{2}\). Two isomers are formed for each metal complex. a. The \(\mathrm{Be}^{2+}\) complexes are tetrahedral. Draw the two isomers of Be(tfa)_2. What type of isomerism is exhibited by \(\mathrm{Be}(\mathrm{tfa})_{2} ?\) b. The \(\mathrm{Cu}^{2+}\) complexes are square planar. Draw the two isomers of \(\mathrm{Cu}(\mathrm{tfa})_{2} .\) What type of isomerism is exhibited by $\mathrm{Cu}(\mathrm{tfa})_{2} ?

Use standard reduction potentials to calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) (at \(298 \mathrm{K}\) ) for the reaction that is used in production of gold: \(2 \mathrm{Au}(\mathrm{CN})_{2}^{-}(a q)+\mathrm{Zn}(s) \longrightarrow 2 \mathrm{Au}(s)+\mathrm{Zn}(\mathrm{CN})_{4}^{2-}(a q)\) The relevant half-reactions are \(\begin{aligned} \mathrm{Au}(\mathrm{CN})_{2}^{-}+\mathrm{e}^{-} \longrightarrow \mathrm{Au}+2 \mathrm{CN}^{-} & \mathscr{E}^{\circ}=-0.60 \mathrm{~V} \\ \mathrm{Zn}(\mathrm{CN})_{4}^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn}+4 \mathrm{CN}^{-} & \mathscr{E}^{\circ}=-1.26 \mathrm{~V} \end{aligned}\)

Name the following complex ions. a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}\) b. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) c. \(\mathrm{Mn}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}^{2+}\) d. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}^{2+}\) a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Cl}^{2+}\) b. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) c. \(\mathrm{Mn}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\right)_{3}^{2+}\) d. \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{NO}_{2}^{2+}\)

Tetrahedral complexes of \(\mathrm{Co}^{2+}\) are quite common. Use a \(d\) -orbital splitting diagram to rationalize the stability of \(\mathrm{Co}^{2+}\) tetrahedral complex ions.

Draw all geometrical isomers of \(\mathrm{Pt}(\mathrm{CN})_{2} \mathrm{Br}_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} .\) Which of these isomers has an optical isomer? Draw the various optical isomers.
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