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 are carried out in acid solution.) (a) \(\mathrm{I}_{2}(\mathrm{s})+\mathrm{Br}^{-}(\mathrm{aq}) \rightarrow \mathrm{I}^{-}(\mathrm{aq})+\mathrm{Br}_{2}(\ell)\) (b) \(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Fe}^{3+}(\mathrm{aq})\) (c) \(\mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Cr}^{3+}(\mathrm{aq})\) (d) \(\mathrm{MnO}_{4}^{-}(\mathrm{aq})+\mathrm{HNO}_{2}(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq})+\mathrm{NO}_{3}^{-}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) Not product-favored (-0.53 V); (b) Not product-favored (-0.43 V); (c) Product-favored (0.56 V); (d) Product-favored (0.55 V).

Step 1: Balance Reaction (a) - Iodine and Bromine

First, balance the given reaction by identifying and balancing the change in oxidation states. For reaction (a):\[\mathrm{I}_2(\mathrm{s}) + \mathrm{Br}^-(\mathrm{aq}) \rightarrow \mathrm{I}^-(\mathrm{aq}) + \mathrm{Br}_2(\ell)\]1. Oxidation: \(\mathrm{Br}^- \rightarrow \mathrm{Br}_2\) involves losing electrons.2. Reduction: \(\mathrm{I}_2 \rightarrow \mathrm{I}^-\) involves gaining electrons.3. The balanced equation in acidic solution is:\[\mathrm{I}_2 + 2 \mathrm{Br}^- \rightarrow 2\mathrm{I}^- + \mathrm{Br}_2\]

Step 2: Calculate Standard Potential for Reaction (a)

Use standard reduction potentials from tables: \(E^∘(\mathrm{I}_2/\mathrm{I}^-) = 0.54 \text{ V}\) and \(E^∘(\mathrm{Br}_2/\mathrm{Br}^-) = 1.07 \text{ V}\).Calculate:\[E^{∘}_{\text{cell}} = E^{∘}_{\text{reduction}} - E^{∘}_{\text{oxidation}} = 0.54 - 1.07 = -0.53 \text{ V}\]Since the potential is negative, the reaction is not product-favored as written.

Step 3: Balance Reaction (b) - Iron and Copper

For reaction (b):\[\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}(\mathrm{s}) + \mathrm{Fe}^{3+}(\mathrm{aq})\]1. Oxidation: \(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}\) involves losing electrons.2. Reduction: \(\mathrm{Cu}^{2+} \rightarrow \mathrm{Cu}\) involves gaining electrons.3. This equation is already balanced.

Step 4: Calculate Standard Potential for Reaction (b)

Use standard reduction potentials: \(E^{∘}(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = 0.77 \text{ V}\) and \(E^{∘}(\mathrm{Cu}^{2+}/\mathrm{Cu}) = 0.34 \text{ V}\).Calculate:\[E^{∘}_{\text{cell}} = 0.34 - 0.77 = -0.43 \text{ V}\]The reaction is not product-favored as written due to the negative potential.

Step 5: Balance Reaction (c) - Iron and Dichromate

For reaction (c):\[\mathrm{Fe}^{2+}(\mathrm{aq}) + \mathrm{Cr}_2\mathrm{O}_7^{2-}(\mathrm{aq}) \rightarrow \mathrm{Fe}^{3+}(\mathrm{aq}) + \mathrm{Cr}^{3+}(\mathrm{aq})\]1. Balance the electrons transferred: \[6\mathrm{Fe}^{2+} + 1\mathrm{Cr}_2\mathrm{O}_7^{2-} + 14\mathrm{H}^+ \rightarrow 6\mathrm{Fe}^{3+} + 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}\]

Step 6: Calculate Standard Potential for Reaction (c)

Use potential values: \(E^{∘}(\mathrm{Fe}^{3+}/\mathrm{Fe}^{2+}) = 0.77 \text{ V}\) and \(E^{∘}(\mathrm{Cr}^{3+}/\mathrm{Cr}_2\mathrm{O}_7^{2-}) = 1.33 \text{ V}\).Calculate:\[E^{∘}_{\text{cell}} = 1.33 - 0.77 = 0.56 \text{ V}\]This reaction is product-favored as written due to a positive potential.

Step 7: Balance Reaction (d) - Permanganate and Nitrous Acid

For reaction (d):\[\mathrm{MnO}_4^- (\mathrm{aq}) + \mathrm{HNO}_2(\mathrm{aq}) \rightarrow \mathrm{Mn}^{2+}(\mathrm{aq}) + \mathrm{NO}_3^-(\mathrm{aq})\]1. Balance the reaction: \[2\mathrm{MnO}_4^- + 5\mathrm{HNO}_2 + 6\mathrm{H}^+ \rightarrow 5\mathrm{NO}_3^- + 2\mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}\]

Step 8: Calculate Standard Potential for Reaction (d)

Use values: \(E^{∘}(\mathrm{MnO}_4^-/\mathrm{Mn}^{2+}) = 1.51 \text{ V}\) and \(E^{∘}(\mathrm{NO}_3^-/\mathrm{HNO}_2) = 0.96 \text{ V}\).Calculate:\[E^{∘}_{\text{cell}} = 1.51 - 0.96 = 0.55 \text{ V}\]This reaction is also product-favored as written due to a positive potential.

Key Concepts

Standard Electrode Potentials

Standard electrode potentials, often symbolized as \(E^{∘}\), are essential in understanding redox reactions. They are measured relative to the standard hydrogen electrode, which is assigned a potential of 0.00 V. This standard provides a common reference for evaluating the tendency of a given half-reaction to occur as a reduction.

• If the \(E^{∘}\) value is positive, the reaction is likely to proceed as a reduction reaction under standard conditions.
• If the \(E^{∘}\) value is negative, it denotes a less favorable tendency for reduction.

Standard electrode potentials are useful in predicting the spontaneity of redox reactions. The overall standard potential \(E^{∘}_{\text{cell}}\) for a full reaction is calculated by subtracting the standard potential of the oxidation reaction from that of the reduction reaction. When calculating \(E^{∘}_{\text{cell}}\), a positive value indicates a spontaneous, product-favored reaction under standard conditions, while a negative value suggests the opposite.
In our example, for reaction (a), the calculated \(E^{∘}_{\text{cell}}\) is -0.53 V, predicting non-favorability. Understanding these potentials is crucial as it allows chemists to predict whether a chemical reaction can produce energy or requires input to proceed.

Oxidation and Reduction

Oxidation and reduction are two fundamental concepts tied together, often summarized by the mnemonic OIL RIG: Oxidation is Loss, Reduction is Gain (of electrons). In each redox reaction, one compound undergoes oxidation while another undergoes reduction. These paired processes involve the transfer of electrons between chemical species.

• In oxidation, a substance loses electrons, increasing its oxidation state.
• In reduction, a substance gains electrons, decreasing its oxidation state.

For example, in reaction (a), iodide ions \(\mathrm{I}^-\) are reduced from iodine gas \(\mathrm{I}_2\) while bromide \(\mathrm{Br}^-\) undergoes oxidation to form bromine \(\mathrm{Br}_2\). The electrons lost by the oxidized species are gained by the reduced species, thus conserving charge.
These redox reactions must be balanced in terms of both mass and charge. This might involve adding electrons, hydrogen ions, or water molecules in acidic or basic solutions to ensure the reaction complies with conservation laws. Identifying oxidation and reduction reactions enables understanding of how substances interact, forming the basis for predicting chemical changes and energy production in a reaction.

Product-Favored Reactions

When considering redox reactions, it's crucial to determine whether a reaction is product-favored or reactant-favored. A product-favored reaction under standard conditions means the conversion of reactants into products is spontaneous. This can be effectively determined by the sign of \(E^{∘}_{\text{cell}}\).

• If \(E^{∘}_{\text{cell}} > 0\), the reaction is product-favored, indicating a spontaneous process with available energy released.
• If \(E^{∘}_{\text{cell}} < 0\), it is reactant-favored, implying the need for input energy for the reaction to occur as written.

In our provided solutions, reactions (c) and (d) have positive \(E^{∘}_{\text{cell}}\) values of 0.56 V and 0.55 V respectively, thus being product-favored. This means the conditions under which these reactions are carried out naturally encourage the formation of products.
On the contrary, reactions with a negative \(E^{∘}_{\text{cell}}\), like (a) and (b), would tend to revert to reactants unless external energy is applied. Understanding whether a reaction is product-favored is vital in practical applications, such as designing batteries, fuel cells, and industrial chemical reactions, to ensure efficient energy utilization and conversion.