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Chapter 21: Problem 71

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: $$\mathrm{Ag}(s)+\mathrm{CN}^{-}(a q)+\mathrm{O}_{2}(g) \stackrel{\mathrm{Basic}}{\longrightarrow} \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q)$$ Balance this equation by using the half-reaction method.

Short Answer

Expert verified
The balanced chemical equation for the given reaction using the half-reaction method in a basic solution is: \(2 \mathrm{Ag}(s) + \mathrm{O}_{2}(g) + \mathrm{CN}^{-}(a q) + 4 \mathrm{OH}^{-}(a q) \rightarrow 2 \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) + 2 \mathrm{H}_{2}\mathrm{O}(l)\)

Step by step solution

01

1. Identify the oxidation and reduction half-reactions

Oxidation half-reaction: Ag(s) → Ag(CN)₂⁻(aq) Reduction half-reaction: O₂(g) + CN⁻(aq) → Ag(CN)₂⁻(aq)
02

2. Balance atoms other than O and H in each half-reaction

Both half-reactions are already balanced for atoms other than O and H: Oxidation half-reaction: Ag(s) → Ag(CN)₂⁻(aq) Reduction half-reaction: O₂(g) + CN⁻(aq) → Ag(CN)₂⁻(aq)
03

3. Balance O atoms by adding H2O molecules

In the reduction half-reaction, we will add 2 H₂O molecules in the product side to balance the O atoms: Reduction half-reaction: O₂(g) + CN⁻(aq) → Ag(CN)₂⁻(aq) + 2 H₂O(l)
04

4. Balance H atoms by adding H+ ions (but since it's a basic solution, we need to use OH- ions)

Since the reaction occurs in a basic solution, we will balance H atoms using OH⁻ ions instead of H⁺ ions. In the reduction half-reaction, add 4 OH⁻ ions to the reactant side to balance the H atoms: Reduction half-reaction: O₂(g) + CN⁻(aq) + 4 OH⁻(aq) → Ag(CN)₂⁻(aq) + 2 H₂O(l)
05

5. Balance charges by adding electrons

Balance charges by adding electrons to the respective half-reactions: Oxidation half-reaction: Ag(s) → Ag(CN)₂⁻(aq) + \(\underline{e⁻}\) Reduction half-reaction: O₂(g) + CN⁻(aq) + 4 OH⁻(aq) + \(\underline{2 e⁻}\) → Ag(CN)₂⁻(aq) + 2 H₂O(l)
06

6. Equalize the electrons in both half-reactions

Since the oxidation half-reaction has 1 electron and the reduction half-reaction has 2 electrons, we need to multiply the oxidation half-reaction by 2 to have the same number of electrons in both half-reactions: Oxidation half-reaction: \(2\) [Ag(s) → Ag(CN)₂⁻(aq) + \(e⁻\)] Resulting half-reaction: 2 Ag(s) → 2 Ag(CN)₂⁻(aq) + 2 \(e⁻\) Now, both half-reactions have the same number of electrons: Oxidation half-reaction: 2 Ag(s) → 2 Ag(CN)₂⁻(aq) + 2 \(e⁻\) Reduction half-reaction: O₂(g) + CN⁻(aq) + 4 OH⁻(aq) + 2 \(e⁻\) → Ag(CN)₂⁻(aq) + 2 H₂O(l)
07

7. Add the balanced half-reactions and cancel any common terms

Add both half-reactions and cancel common terms (2 \(e⁻\), Ag(CN)₂⁻): 2 Ag(s) + O₂(g) + CN⁻(aq) + 4 OH⁻(aq) → 2 Ag(CN)₂⁻(aq) + 2 H₂O(l) This is the balanced chemical equation for the given reaction.

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