Problem 96 - Chemistry The Science in Context

Question

Silverware Tarnish Low concentrations of hydrogen sulfide in air react with silver to form \(\mathrm{Ag}_{2} \mathrm{S}\), more familiar to us as tarnish. Silver polish contains aluminum metal powder in a basic suspension. a. Write a balanced net ionic equation for the redox reaction between \(\mathrm{Ag}_{2} \mathrm{S}\) and \(\mathrm{Al}\) metal that produces \(\mathrm{Ag}\) metal and \(\mathrm{Al}(\mathrm{OH})_{3}\) b. Calculate \(E^{\circ}\) for the reaction.

Step-by-Step Solution

Answer

a. Write the balanced net ionic equation for the redox reaction between Ag₂S and Al metal that produces Ag metal and Al(OH)₃. The balanced net ionic equation for the redox reaction is: $$2\mathrm{Al}(s) + 3\mathrm{Ag}_{2} \mathrm{S}(s) + 6\mathrm{OH}^{-}(aq) \rightarrow 6\mathrm{Ag}(s) + 2\mathrm{Al(OH)_{3}}(s) + 3\mathrm{S}^{2-}(aq)$$ b. Calculate the standard electrode potential, 𝐸°, for the reaction. The standard electrode potential for the reaction is \(+2.461\,\text{V}\).

Step 1: a. Writing the balanced net ionic equation

First, let's write the half-reactions for both the oxidation and the reduction process. Oxidation half-reaction (Aluminum is being oxidized): $$\mathrm{Al}(s) \rightarrow \mathrm{Al}^{3+}(aq) + 3e^{-}$$ Reduction half-reaction (Silver sulfide is being reduced): $$\mathrm{Ag}_{2} \mathrm{S}(s) + 2e^{-} \rightarrow 2\mathrm{Ag}(s) + \mathrm{S}^{2-}(aq)$$ Next, we need to balance the electrons transferred in both half-reactions. We will multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3 in order to have 6 electrons in both reactions: Oxidation half-reaction (multiplied by 2): $$2\mathrm{Al}(s) \rightarrow 2\mathrm{Al}^{3+}(aq) + 6e^{-}$$ Reduction half-reaction (multiplied by 3): $$3\mathrm{Ag}_{2} \mathrm{S}(s) + 6e^{-} \rightarrow 6\mathrm{Ag}(s) + 3\mathrm{S}^{2-}(aq)$$ Now, let's add both half-reactions together and include the \(\mathrm{Al(OH)_3}\) from the problem statement: $$2\mathrm{Al}(s) + 3\mathrm{Ag}_{2} \mathrm{S}(s) + 6\mathrm{OH}^{-}(aq) \rightarrow 2\mathrm{Al}^{3+}(aq) + 6\mathrm{Ag}(s) + 3\mathrm{S}^{2-}(aq) + 6\mathrm{OH}^{-}(aq)$$ Finally, since aluminum ions and hydroxide ions are combined in the final product \(\mathrm{Al(OH)_3}\), let's rewrite the equation with this compound: $$2\mathrm{Al}(s) + 3\mathrm{Ag}_{2} \mathrm{S}(s) + 6\mathrm{OH}^{-}(aq) \rightarrow 6\mathrm{Ag}(s) + 2\mathrm{Al(OH)_{3}}(s) + 3\mathrm{S}^{2-}(aq)$$ This is the balanced net ionic equation for the redox reaction.

Step 2: b. Calculating \(E^{\circ}\) for the reaction.

To calculate the standard electrode potential (\(E^{\circ}\)) for the reaction, we will use the standard reduction potentials for both half-reactions: $$E^{\circ}_{\text{red}}(\text{Ag}) = +0.799\,\text{V}$$ $$E^{\circ}_{\text{red}}(\text{Al}) = -1.662\,\text{V}$$ Since aluminum is being oxidized, we need to use the standard oxidation potential for the aluminum half-reaction, which is the negative of the standard reduction potential: $$E^{\circ}_{\text{ox}}(\text{Al}) = +1.662\,\text{V}$$ Now, we can calculate the overall standard electrode potential for the reaction: $$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{red}}(\text{Ag}) + E^{\circ}_{\text{ox}}(\text{Al})$$ $$E^{\circ}_{\text{cell}} = +0.799\,\text{V} + (+1.662)\,\text{V} = +2.461\,\text{V}$$ The standard electrode potential for the reaction is \(+2.461\,\text{V}\).

Key Concepts

Redox Reactions

Redox reactions are a type of chemical reaction where oxidation and reduction occur simultaneously. In simple terms, one substance gains electrons (reduction), while another loses electrons (oxidation). These electron transfers are essential for processes like energy production and corrosion removal.

In the exercise, silver sulfide (\(\mathrm{Ag}_2 \mathrm{S}\)) is reduced, and aluminum (\(\mathrm{Al}\)) is oxidized. This specific reaction restores silver from its tarnished state. Understanding how electrons transfer between substances helps in predicting reaction outcomes and balancing equations correctly.

In any redox process:

• The substance that loses electrons is oxidized.
• The substance that gains electrons is reduced.
• Redox reactions balance these electron shifts.

Grasping redox reactions aids in learning about batteries, metabolism, and even industrial processes.

Electrochemistry

Electrochemistry explores the interaction between electrical energy and chemical change. It's crucial for understanding batteries, electrolysis, and plating applications. In the textbook problem, the calculation of the standard electrode potential \( (E^{\circ}) \) relates to electrochemistry.

This potential \( (E^{\circ}_{\text{cell}}) \) helps predict if a reaction is spontaneous. Positive values often signal spontaneous reactions. Here, the standard electrode potential is calculated as \(+2.461\,\text{V}\), showing that the silver-polishing reaction will occur without additional energy input.

Electrochemical concepts include:

• Identifying oxidation and reduction half-reactions.
• Using standard electrode potentials to find the overall reaction potential.
• Applications in devices like electrochemical cells and galvanic cells.

Understanding these principles is vital for studying modern technology, like rechargeable batteries and fuel cells.

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom appears on both sides of the equation, adhering to the conservation of mass. This is crucial for accurately representing chemical reactions.

In the given exercise, both oxidation and reduction half-reactions are first written, with emphasis on balancing electron transfer.

Steps include:

• Determining individual half-reactions.
• Balancing elements and charges.
• Multiplying reactions to balance electrons.
• Combining half-reactions for the complete balanced equation.

For instance, equalizing electrons with multipliers (like in the problem: using 2 for oxidation and 3 for reduction) achieves balance, giving insight into the precise amounts of substances interacting. Mastery of equation balancing aids in predicting the quantities of reactants/products in reactions, which is fundamental in chemical engineering and laboratory work.