Problem 16 - 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) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{Br}^{-}(\mathrm{aq}) \longrightarrow \mathrm{I}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(\ell)\) (b) \(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq})\) (c) \(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \longrightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq})\) (d) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{HNO}_{2}(\mathrm{aq}) \longrightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) Not favored, (b) Not favored, (c) Favored, (d) Favored.

Step 1: Balance Equation (a) I2(s) + Br-(aq) → I-(aq) + Br2(l)

First, assign oxidation states: Iodine in I2 is 0, and in I- is -1; Bromine in Br- is -1 and in Br2 is 0. Iodine is reduced while bromine is oxidized. Therefore, the half-reactions are: 1. \( ext{I}_2 + 2 ext{e}^- \rightarrow 2 ext{I}^-\)2. \(2 ext{Br}^- \rightarrow ext{Br}_2 + 2 ext{e}^-\) The electrons cancel out, so the balanced equation is \( ext{I}_2 + 2 ext{Br}^- \rightarrow 2 ext{I}^- + ext{Br}_2\).

Step 2: Calculate Standard Potential for Reaction (a)

The standard reduction potential for I2/I- is +0.54 V, and for Br2/Br- it is +1.07 V.The overall cell potential \(E^{\circ}\) is calculated by \(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}\).Therefore, \(E^{\circ}_{cell} = 0.54 \, ext{V} - 1.07 \, ext{V} = -0.53 \, ext{V}.\)

Step 3: Determine if Reaction (a) is Product-Favored

Since \(E^{\circ}\) is negative, reaction (a) is not product-favored as written.

Step 4: Balance Equation (b) Fe2+ + Cu2+ → Cu(s) + Fe3+

Assign oxidation states: Fe2+ to Fe3+ (oxidized), Cu2+ to Cu (reduced). The half-reactions are: 1. \( ext{Fe}^{2+} \rightarrow ext{Fe}^{3+} + ext{e}^-\)2. \( ext{Cu}^{2+} + 2 ext{e}^- \rightarrow ext{Cu}\)Multiply the Fe half-reaction by 2 to balance electrons: \(2 ext{Fe}^{2+} \rightarrow 2 ext{Fe}^{3+} + 2 ext{e}^-\)Overall: \(2 ext{Fe}^{2+} + ext{Cu}^{2+} \rightarrow ext{Cu} + 2 ext{Fe}^{3+}\).

Step 5: Calculate Standard Potential for Reaction (b)

The standard potentials are: Fe3+/Fe2+ = 0.77 V; Cu2+/Cu = 0.34 V.\(E^{\circ}_{cell} = 0.34 \, ext{V} - 0.77 \, ext{V} = -0.43 \, ext{V}.\)

Step 6: Determine if Reaction (b) is Product-Favored

With a negative \(E^{\circ}\), reaction (b) is not product-favored.

Step 7: Balance Equation (c) Fe2+ + Cr2O7(2-) → Fe3+ + Cr3+

Oxidation states: Fe2+ to Fe3+ (oxidized), Cr in Cr2O7(2-) is reduced to Cr3+. Balance the reactions: 1. \( ext{6 Fe}^{2+} \rightarrow ext{6 Fe}^{3+} + 6 ext{e}^-\)2. \( ext{Cr}_2 ext{O}_7^{2-} + 14 ext{H}^+ + 6 ext{e}^- \rightarrow 2 ext{Cr}^{3+} + 7 ext{H}_2 ext{O}\)Overall balanced equation: \(6 ext{Fe}^{2+} + ext{Cr}_2 ext{O}_7^{2-} + 14 ext{H}^+ \rightarrow 6 ext{Fe}^{3+} + 2 ext{Cr}^{3+} + 7 ext{H}_2 ext{O}\).

Step 8: Calculate Standard Potential for Reaction (c)

Standard potentials: Fe3+/Fe2+ = 0.77 V; Cr2O7(2-)/Cr3+ = 1.33 V.\(E^{\circ}_{cell} = 1.33 \, ext{V} - 0.77 \, ext{V} = 0.56 \, ext{V}.\)

Step 9: Determine if Reaction (c) is Product-Favored

Since \(E^{\circ}\) is positive, the reaction is product-favored.

Step 10: Balance Equation (d) MnO4- + HNO2 → Mn2+ + NO3-

Assign oxidation states: Mn (in MnO4-) to Mn2+ (reduced), N (in HNO2) to NO3- (oxidized). The half-reactions:1. \( ext{MnO}_4^- + 8 ext{H}^+ + 5 ext{e}^- \rightarrow ext{Mn}^{2+} + 4 ext{H}_2 ext{O}\)2. \( ext{HNO}_2 + ext{H}_2 ext{O} \rightarrow ext{NO}_3^- + 4 ext{H}^+ + 2 ext{e}^-\)Balance electrons, multiply first by 2 and second by 5:Overall balanced equation: \(2 ext{MnO}_4^- + 5 ext{HNO}_2 + 6 ext{H}^+ \rightarrow 2 ext{Mn}^{2+} + 5 ext{NO}_3^- + 3 ext{H}_2 ext{O}\).

Step 11: Calculate Standard Potential for Reaction (d)

Standard potentials: MnO4-/Mn2+ = 1.51 V; NO3-/HNO2 = 0.94 V.\(E^{\circ}_{cell} = 1.51 \, ext{V} - 0.94 \, ext{V} = 0.57 \, ext{V}.\)

Step 12: Determine if Reaction (d) is Product-Favored

With a positive \(E^{\circ}\), reaction (d) is product-favored.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is an essential skill in understanding redox reactions. This process ensures that the number of atoms and charges are conserved on both sides of the reaction. When you balance equations, you make certain that all mass goes where it needs to be, based on the Law of Conservation of Mass.

Here's a simple way to balance equations:

• Identify the oxidation states of all elements involved.
• Determine which species are oxidized and reduced.
• Write the half-reactions for oxidation and reduction.
• Balance the electrons lost in oxidation and gained in reduction by multiplying the reactions by appropriate coefficients.
• Add the half-reactions, ensuring all atoms and charges balance.

This rule of balancing is required when equations involve intricate reactions, especially in redox processes, ensuring the equations accurately reflect the reaction taking place.

Standard Electrode Potentials

Standard electrode potentials, denoted as \(E^{ ext{°}}\), are used to gauge the tendency of a chemical species to be reduced. These potentials provide an insight into the energy involved when a specific reaction happens under standard conditions.

This potential is measured against the standard hydrogen electrode, which is assigned a potential of 0 volts. When assessing a reaction, you identify the cathode (where reduction occurs) and the anode (where oxidation occurs) and determine their standard potentials from tables. The overall cell potential \(E^{ ext{°}}_{cell}\) is calculated using:

• \(E^{ ext{°}}_{cell} = E^{ ext{°}}_{cathode} - E^{ ext{°}}_{anode}\)

A positive \(E^{ ext{°}}_{cell}\) indicates a spontaneous, or product-favored, reaction, while a negative value signifies that the reaction is not favorable in the forward direction.

Oxidation States

Oxidation states play a pivotal role in understanding the electron transfer during reactions, especially in redox reactions. An oxidation state is a theoretical charge that an atom would have if all bonds to atoms of different elements were completely ionic.

Here’s how oxidation states can help:

• They allow you to track which chemical species are oxidized and which are reduced. Oxidation involves an increase in oxidation state, while reduction involves a decrease.
• They are useful for determining if electrons are being gained or lost in a reaction.

Assigning correct oxidation states enables one to write balanced equations and calculate the standard electrode potentials efficiently, as these depend significantly on knowing how valence electrons are rearranged.

Acidic Solution

Redox reactions can occur in different environments, with acidic solutions being quite common. When working with reactions in acidic media, you'll often use hydrogen ions (H+) to balance the reaction.

Here's how acidity influences balancing equations:

• In acidic solutions, aside from balancing the atoms and charges, you might need to add \(H^+\) ions to balance hydrogen atoms.
• Water molecules (\(H_2O\)) may be added to balance oxygen atoms.

Balancing redox reactions in an acidic environment requires careful attention to the changes in pH and how \(H^+\) ions participate in maintaining charge balance, ultimately affecting the solubility and reactivity of the species involved.