Problem 15 - Chemistry and Chemical Reactivity

Question

Balance each of the following unbalanced equations, then calculate the standard potential, \(E^{\circ},\) and decide whether each is product-favored as written. (All reactions occur in acid solution.) (a) \(\mathrm{Sn}^{2+}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s}) \longrightarrow \mathrm{Sn}(\mathrm{s})+\mathrm{Ag}^{+}(\mathrm{aq})\) (b) \(\mathrm{Al}(\mathrm{s})+\mathrm{Sn}^{4+}(\mathrm{aq}) \longrightarrow \mathrm{Sn}^{2+}(\mathrm{aq})+\mathrm{Al}^{3+}(\mathrm{aq})\) (c) \(\mathrm{ClO}_{3}^{-}(\mathrm{aq})+\mathrm{Ce}^{3+}(\mathrm{aq}) \longrightarrow \mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Ce}^{4+}(\mathrm{aq})\) (d) \(\mathrm{Cu}(\mathrm{s})+\mathrm{NO}_{3}^{-}(\mathrm{aq}) \longrightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\)

Step-by-Step Solution

Answer

All reactions are product-favored as they all have positive standard potentials, and are properly balanced accordingly.

Step 1: Balance the equation for Reaction (a)

First, write the half-reactions for oxidation and reduction:- **Oxidation half-reaction**: \( \mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}^{+}(\mathrm{aq}) + \mathrm{e}^{-} \)- **Reduction half-reaction**: \( \mathrm{Sn}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Sn}(\mathrm{s}) \)Notice that electrons are not balanced. Multiply the oxidation half-reaction by 2 to balance the electrons:- **Balanced oxidation half-reaction**: \( 2\mathrm{Ag}(\mathrm{s}) \rightarrow 2\mathrm{Ag}^{+}(\mathrm{aq}) + 2\mathrm{e}^{-} \)Combine the balanced half-reactions:\[ \mathrm{Sn}^{2+}(\mathrm{aq}) + 2\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Sn}(\mathrm{s}) + 2\mathrm{Ag}^{+}(\mathrm{aq}) \]

Step 2: Calculate the standard potential, \(E^{\circ}\), for Reaction (a)

Use the standard reduction potential values:- **\( \mathrm{Ag}^{+}/\mathrm{Ag} \)**: \( +0.80 \mathrm{V} \)- **\( \mathrm{Sn}/\mathrm{Sn}^{2+} \)**: \( -0.14 \mathrm{V} \)Standard potential calculation:\[ E^{\circ} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = (+0.80\ \mathrm{V}) - (-0.14\ \mathrm{V}) = +0.94\ \mathrm{V} \]

Step 3: Determine if Reaction (a) is product-favored

Since the calculated \(E^{\circ} = +0.94\ \mathrm{V}\) is positive, Reaction (a) is product-favored.

Step 4: Balance the equation for Reaction (b)

Write the half-reactions for oxidation and reduction:- **Oxidation half-reaction**: \( \mathrm{Al}(\mathrm{s}) \rightarrow \mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{e}^{-} \)- **Reduction half-reaction**: \( \mathrm{Sn}^{4+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Sn}^{2+}(\mathrm{aq}) \)Multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3 to balance electrons:- **Balanced oxidation half-reaction**: \( 2\mathrm{Al}(\mathrm{s}) \rightarrow 2\mathrm{Al}^{3+}(\mathrm{aq}) + 6\mathrm{e}^{-} \)- **Balanced reduction half-reaction**: \( 3\mathrm{Sn}^{4+}(\mathrm{aq}) + 6\mathrm{e}^{-} \rightarrow 3\mathrm{Sn}^{2+}(\mathrm{aq}) \)Combined balanced equation:\[ 2\mathrm{Al}(\mathrm{s}) + 3\mathrm{Sn}^{4+}(\mathrm{aq}) \rightarrow 2\mathrm{Al}^{3+}(\mathrm{aq}) + 3\mathrm{Sn}^{2+}(\mathrm{aq}) \]

Step 5: Calculate the standard potential, \(E^{\circ}\), for Reaction (b)

Use the standard potentials:- **\( \mathrm{Sn}^{4+}/\mathrm{Sn}^{2+} \)**: \( +0.15 \mathrm{V} \)- **\( \mathrm{Al}^{3+}/\mathrm{Al} \)**: \( -1.66 \mathrm{V} \)Standard potential calculation:\[ E^{\circ} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = (+0.15\ \mathrm{V}) - (-1.66\ \mathrm{V}) = +1.81\ \mathrm{V} \]

Step 6: Determine if Reaction (b) is product-favored

Since \(E^{\circ} = +1.81\ \mathrm{V}\), Reaction (b) is product-favored.

Step 7: Balance the equation for Reaction (c)

Write the half-reactions for oxidation and reduction:- **Oxidation half-reaction**: \( \mathrm{Ce}^{3+}(\mathrm{aq}) \rightarrow \mathrm{Ce}^{4+}(\mathrm{aq}) + \mathrm{e}^{-} \)- **Reduction half-reaction**: \( \mathrm{ClO}_{3}^{-}(\mathrm{aq}) + 6\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow \mathrm{Cl}^{-}(\mathrm{aq}) + 3\mathrm{H}_{2}\mathrm{O} \)Multiply the oxidation half-reaction by 6 to balance electrons:- **Balanced oxidation half-reaction**: \( 6\mathrm{Ce}^{3+}(\mathrm{aq}) \rightarrow 6\mathrm{Ce}^{4+}(\mathrm{aq}) + 6\mathrm{e}^{-} \)Combined balanced equation:\[ \mathrm{ClO}_{3}^{-}(\mathrm{aq}) + 6\mathrm{Ce}^{3+}(\mathrm{aq}) + 6\mathrm{H}^{+} \rightarrow \mathrm{Cl}^{-}(\mathrm{aq}) + 3\mathrm{H}_{2}\mathrm{O} + 6\mathrm{Ce}^{4+}(\mathrm{aq}) \]

Step 8: Calculate the standard potential, \(E^{\circ}\), for Reaction (c)

Use the standard potentials:- **\( \mathrm{Cl}^{-}/\mathrm{ClO}_{3}^{-} \)** \(= +1.47 \mathrm{V} \)- **\( \mathrm{Ce}^{4+}/\mathrm{Ce}^{3+} \)**: \( +1.61 \mathrm{V} \)Standard potential calculation:\[ E^{\circ} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = (+1.61\ \mathrm{V}) - (+1.47\ \mathrm{V}) = +0.14\ \mathrm{V} \]

Step 9: Determine if Reaction (c) is product-favored

Since \(E^{\circ} = +0.14\ \mathrm{V}\), Reaction (c) is product-favored.

Step 10: Balance the equation for Reaction (d)

Write the half-reactions for oxidation and reduction:- **Oxidation half-reaction**: \( \mathrm{Cu}(\mathrm{s}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \)- **Reduction half-reaction**: \( \mathrm{NO}_{3}^{-}(\mathrm{aq}) + 4\mathrm{H}^{+} + 3\mathrm{e}^{-} \rightarrow \mathrm{NO}(\mathrm{g}) + 2\mathrm{H}_{2}\mathrm{O} \)Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to balance electrons:- **Balanced oxidation half-reaction**: \( 3\mathrm{Cu}(\mathrm{s}) \rightarrow 3\mathrm{Cu}^{2+}(\mathrm{aq}) + 6\mathrm{e}^{-} \)- **Balanced reduction half-reaction**: \( 2\mathrm{NO}_{3}^{-}(\mathrm{aq}) + 8\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow 2\mathrm{NO}(\mathrm{g}) + 4\mathrm{H}_{2}\mathrm{O} \)Combined balanced equation:\[ 3\mathrm{Cu}(\mathrm{s}) + 2\mathrm{NO}_{3}^{-}(\mathrm{aq}) + 8\mathrm{H}^{+} \rightarrow 3\mathrm{Cu}^{2+}(\mathrm{aq}) + 2\mathrm{NO}(\mathrm{g}) + 4\mathrm{H}_{2}\mathrm{O} \]

Step 11: Calculate the standard potential, \(E^{\circ}\), for Reaction (d)

Use the standard potentials:- **\( \mathrm{NO}_{3}^{-}/\mathrm{NO} \)**: \( +0.96 \mathrm{V} \)- **\( \mathrm{Cu}^{2+}/\mathrm{Cu} \)**: \( +0.34 \mathrm{V} \)Standard potential calculation:\[ E^{\circ} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = (+0.96\ \mathrm{V}) - (+0.34\ \mathrm{V}) = +0.62\ \mathrm{V} \]

Step 12: Determine if Reaction (d) is product-favored

Since \(E^{\circ} = +0.62\ \mathrm{V}\), Reaction (d) is product-favored.

Key Concepts

Redox Reactions

Redox reactions are fascinating chemical processes where reduction and oxidation occur simultaneously. In these reactions, one species loses electrons (oxidation), while another gains electrons (reduction). This transfer of electrons is often accompanied by energy changes, making redox reactions fundamental to understanding chemical energy and electrode reactions.
To analyze a redox reaction, chemists break it into two half-reactions:

• The oxidation half-reaction: Electrons are removed from an element. For example, in the given problem where \( 2\mathrm{Ag}(\mathrm{s}) \rightarrow 2\mathrm{Ag}^{+}(\mathrm{aq}) + 2\mathrm{e}^{-}\)Ag is oxidized.
• The reduction half-reaction: Electrons are added to an element. For example, \(\mathrm{Sn}^{2+}(\mathrm{aq}) + 2\mathrm{e}^{-} \rightarrow \mathrm{Sn}(\mathrm{s})\)Sn is reduced.

The electrons lost in oxidation must equal the electrons gained in reduction to keep the charge and mass balanced. Hence, balancing redox reactions accurately is key, and each of these reactions must be paired properly.
It's crucial to note that a balanced redox reaction will tell you the stoichiometry of the reaction, how many moles of reactants are required, and how many moles of products are formed.

Standard Electrode Potential

Standard electrode potential, symbolized as \(E^{\circ}\), is a measure of the driving force behind an electrochemical reaction. It represents the voltage or potential difference between two electrodes when the concentration of all solutes is 1 M and the pressure is 1 atm, typically at 25 degrees Celsius. Each half-reaction of a redox pair has a standard electrode potential, which can be located in tables of electrochemical data. These potentials are used to determine the feasibility and spontaneity of electrochemical cells by calculating \(E^{\circ}\_{cell}\) for the overall reaction. This is found by taking the difference between the standard reduction potentials of the cathode and the anode:\[E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\]For instance, in the solution provided, the reaction with Ag and Sn gives a positive \(E^{\circ}\), suggesting the reaction is product-favored as written. A positive \(E^{\circ}_{\text{cell}}\) indicates a spontaneous process under standard conditions. Understanding \(E^{\circ}\) helps us predict the direction of electron flow in the electrochemical cell and whether the reaction will occur spontaneously.

Balancing Chemical Equations

Balancing chemical equations ensures that mass and charge are conserved in a chemical reaction. This is a fundamental skill when dealing with redox reactions in electrochemical cells.
To balance a redox equation, it is often broken into separate half-reactions—one for oxidation, one for reduction. We then balance both:

• Start by balancing the atoms in each half-reaction, except for O and H.
• Balance oxygen atoms by adding water (\(H_2O\)).
• Balance hydrogen atoms by adding protons (\(H^+\)). This is crucial in acidic solutions, as in our textbook exercise.
• Balance the charge by adding electrons to one side of the half-reaction.
• Finally, ensure the number of electrons in the oxidation half equals those in the reduction half by multiplying the entire half-reactions by appropriate coefficients.

Balancing is complete when the total number of atoms of each element and the total charge are equal on both sides. Only then do we combine the half-reactions, producing a net balanced equation. This step is essential because an unbalanced equation cannot accurately represent what happens in a chemical reaction.