Question

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

Short answer

The balanced equation for the given redox reaction using the half-reaction method is: \(\mathrm{Ag}(s) + \mathrm{CN}^{-}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) + 2\mathrm{OH}^{-}(a q)\)

Step-by-step solution

Identify the oxidation and reduction half-reactions

In the given reaction: \(\mathrm{Ag}(s)+\mathrm{CN}^{-}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) \) Silver (Ag) is changing from solid state to combined state as Ag(CN)₂⁻ which indicates the Ag (s) is being oxidized. On the other hand, the Oxygen (O₂) molecule gains electrons which indicates reduction. Therefore, we have the oxidation half-reaction: \(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) \) And the reduction half-reaction: \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) \)

Balance atoms other than Hydrogen and Oxygen in half-reactions

There is no need to balance any atoms in this step, as all the atoms other than Silver and Oxygen are already balanced in both half-reactions.

Balance the oxygen atoms

In the first half-reaction, there are no oxygen atoms; so no need to balance the oxygen atoms. In the second half-reaction, there are 2 oxygen atoms on the left side, and none on the right side. We can balance the oxygen atoms by adding 2 hydroxide (OH⁻) ions on the right side: \(\mathrm{O}_{2}(g) \longrightarrow 2\mathrm{OH}^{-}(a q)\)

Balance the hydrogen atoms

In the first half-reaction, there are no hydrogen atoms; so we have nothing to balance for hydrogen atoms in the first half-reaction. In the second half-reaction, there are 2 hydrogen atoms on the right side (in the form of 2 OH⁻). To balance the hydrogen atoms on the left side, we will add 2 water (H₂O) molecules: \(2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{O}_{2}(g) \longrightarrow 2\mathrm{OH}^{-}(a q)\)

Balance the charge in each half-reaction

For the oxidation half-reaction: \(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) + e^{-} \) For the reduction half-reaction: \(2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{O}_{2}(g) + 2e^{-} \longrightarrow 2\mathrm{OH}^{-}(a q)\)

Combine the balanced half-reactions to get the overall balanced equation

In order to combine the balanced half-reactions, we'll first need to make sure that the electrons gained and lost in each half-reaction are equal. In this case, both half-reactions involve one electron, so there's no need to multiply either half-reaction by a constant. Now we can add the two half-reactions together: \(\mathrm{Ag}(s) + \mathrm{CN}^{-}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) + 2\mathrm{OH}^{-}(a q)\) The balanced equation is: \(\mathrm{Ag}(s) + \mathrm{CN}^{-}(a q) + 2\mathrm{H}_{2}\mathrm{O}(l) + \mathrm{O}_{2}(g) \longrightarrow \mathrm{Ag}(\mathrm{CN})_{2}^{-}(a q) + 2\mathrm{OH}^{-}(a q)\)

Key concepts

Half-Reaction Method

The half-reaction method is a systematic process used to balance complex chemical equations, particularly those involving redox reactions, which are composed of oxidation and reduction processes. This method breaks down the overall chemical reaction into two separate half-reactions: one for oxidation and one for reduction.

To effectively use the half-reaction method, follow these steps:

• Identify the substances being oxidized and reduced.
• Write down the separate half-reactions for oxidation and reduction.
• Balance the atoms in each half-reaction, first the atoms apart from oxygen and hydrogen, followed by oxygen (with water molecules) and then hydrogen (with hydrogen ions in acidic solutions or hydroxide ions in basic solutions).
• Balance the charges by adding electrons. For oxidation, electrons are a product, and for reduction, electrons are a reactant.
• Equalize the number of electrons in each half-reaction if necessary by multiplying the half-reactions by appropriate coefficients.
• Add the balanced half-reactions back together, canceling out electrons and any other species that appear on both sides of the equation.
• Confirm that the mass and the charge are balanced, meaning the total charges and number of atoms for each element are the same on both sides of the equation.

Understanding this method allows students to tackle complex redox reactions and is fundamental to many chemical processes.

Oxidation and Reduction

Oxidation and reduction are chemical processes that involve the transfer of electrons between substances. They always occur together in a redox reaction, hence the term 'redox' representing the combination of reduction (gain of electrons) and oxidation (loss of electrons).

In oxidation, an element loses electrons, which increases its oxidation state. This process often involves adding oxygen or removing hydrogen from a substance. Conversely, in reduction, an element gains electrons, reducing its oxidation state, often associated with adding hydrogen or removing oxygen.

For instance, in the provided exercise, silver (Ag) is oxidized as it binds with cyanide (CN⁻) to form Ag(CN)₂⁻. Oxygen (O₂), on the other hand, is reduced when it combines with hydrogen to form hydroxide ions (OH⁻). By recognizing these changes in the oxidation state, we can correctly categorize the processes and understand their role in the chemical reaction.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. Therefore, the amount of each element must be the same on both sides of a balanced equation.

The stoichiometry of a reaction can be used to calculate how much product will form from a given amount of reactant and vice versa. This requires a balanced chemical equation, which provides the mole ratio of reactants to products. By using the coefficients from the balanced equation as conversion factors, we can solve for unknown quantities of substances involved in the reaction.

For example, completing the provided exercise includes ensuring that the same number of each type of atom appears on both sides of the equation. It also involves making sure that the total charge is the same on both sides, which reflects the conservation of charge. Accurate stoichiometry is critical for various applications, including pharmaceuticals, chemical manufacturing, and environmental science.