Problem 20 - 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 at equilibrium 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; (b) Not product-favored; (c) Product-favored; (d) Product-favored.

Step 1: Balance the equation for reaction (a)

Let's start with reaction (a): \[\text{I}_{2}(s) + \text{Br}^{-}(aq) \rightarrow \text{I}^{-}(aq) + \text{Br}_{2}(l)\]1. Calculate the half-reactions: - Reduction half-reaction: \(\text{I}_{2}(s) + 2e^- \rightarrow 2\text{I}^{-}(aq)\) - Oxidation half-reaction: \(2\text{Br}^{-}(aq) \rightarrow \text{Br}_{2}(l) + 2e^-\)2. Combine the two half-reactions to cancel the electrons: - \(\text{I}_{2}(s) + 2\text{Br}^{-}(aq) \rightarrow 2\text{I}^{-}(aq) + \text{Br}_{2}(l)\)

Step 2: Calculate standard potential for reaction (a)

1. Standard reduction potentials are: - \(E^{\circ}(\text{I}_{2}(s)/\text{I}^{-}) = +0.54 \text{ V}\) - \(E^{\circ}(\text{Br}_{2}(l)/\text{Br}^{-}) = +1.07 \text{ V}\)2. Calculate \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\): - \(E^{\circ}_{\text{cell}} = 0.54\text{ V} - 1.07 \text{ V} = -0.53 \text{ V}\)3. The reaction is not product-favored at equilibrium since \(E^{\circ}_{\text{cell}} < 0\).

Step 3: Balance the equation for reaction (b)

Let's focus on reaction (b):\[\text{Fe}^{2+}(aq) + \text{Cu}^{2+}(aq) \rightarrow \text{Cu}(s) + \text{Fe}^{3+}(aq)\]1. Calculate the half-reactions: - Reduction half-reaction: \(\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)\) - Oxidation half-reaction: \(\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-\)2. Adjust the number of electrons and combine: - Multiply oxidation by 2: \(2\text{Fe}^{2+}(aq) \rightarrow 2\text{Fe}^{3+}(aq) + 2e^-\) - Combine: \(2\text{Fe}^{2+}(aq) + \text{Cu}^{2+}(aq) \rightarrow \text{Cu}(s) + 2\text{Fe}^{3+}(aq)\)

Step 4: Calculate standard potential for reaction (b)

1. Standard reduction potentials are: - \(E^{\circ}(\text{Cu}^{2+}/\text{Cu}) = +0.34 \text{ V}\) - \(E^{\circ}(\text{Fe}^{3+}/\text{Fe}^{2+}) = +0.77 \text{ V}\)2. Calculate \(E^{\circ}_{\text{cell}}\): - \(E^{\circ}_{\text{cell}} = 0.34\text{ V} - 0.77\text{ V} = -0.43 \text{ V}\)3. The reaction is not product-favored at equilibrium since \(E^{\circ}_{\text{cell}} < 0\).

Step 5: Balance the equation for reaction (c)

For reaction (c):\[\text{Fe}^{2+}(aq) + \text{Cr}_{2}\text{O}_{7}^{2-}(aq) \rightarrow \text{Fe}^{3+}(aq) + \text{Cr}^{3+}(aq)\]1. Calculate the half-reactions: - Oxidation: \(\text{Fe}^{2+}(aq) \rightarrow \text{Fe}^{3+}(aq) + e^-\) - Reduction: \(\text{Cr}_{2}\text{O}_{7}^{2-}(aq) + 14H^+ + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7H_2O\)2. Balance electrons by multiplying the oxidation half-reaction by 6: - Combine: \(6\text{Fe}^{2+}(aq) + \text{Cr}_{2}\text{O}_{7}^{2-}(aq) + 14H^+ \rightarrow 6\text{Fe}^{3+}(aq) + 2\text{Cr}^{3+}(aq) + 7H_2O\)

Step 6: Calculate standard potential for reaction (c)

1. Standard reduction potentials: - \(E^{\circ} = \text{for Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+} = +1.33 \text{ V}\) - \(E^{\circ} = \text{for Fe}^{3+}/\text{Fe}^{2+} = +0.77 \text{ V}\)2. Calculate \(E^{\circ}_{\text{cell}}\): - \(E^{\circ}_{\text{cell}} = 1.33\text{ V} - 0.77\text{ V} = +0.56 \text{ V}\)3. The reaction is product-favored at equilibrium since \(E^{\circ}_{\text{cell}} > 0\).

Step 7: Balance the equation for reaction (d)

For reaction (d):\[\text{MnO}_{4}^{-}(aq) + \text{HNO}_{2}(aq) \rightarrow \text{Mn}^{2+}(aq) + \text{NO}_{3}^{-}(aq)\]1. Calculate the half-reactions: - Reduction: \(\text{MnO}_{4}^{-}(aq) + 8H^{+} + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4H_2O\) - Oxidation: \(\text{HNO}_{2}(aq) \rightarrow \text{NO}_{3}^{-}(aq) + 2H^+ + 2e^-\)2. Balance electrons by multiplying the oxidation by 5 and the reduction by 2: - Combine: \(2\text{MnO}_{4}^{-} + 5\text{HNO}_{2} + 6H^+ \rightarrow 2\text{Mn}^{2+} + 4\text{NO}_{3}^{-} + 4H_2O\)

Step 8: Calculate standard potential for reaction (d)

1. Standard reduction potentials: - \(E^{\circ} = \text{for MnO}_4^{-}/\text{Mn}^{2+} = +1.51 \text{ V}\) - \(E^{\circ} = \text{for NO}_3^{-}/\text{HNO}_2 = +0.94 \text{ V}\)2. Calculate \(E^{\circ}_{\text{cell}}\): - \(E^{\circ}_{\text{cell}} = 1.51\text{ V} - 0.94\text{ V} = +0.57 \text{ V}\)3. The reaction is product-favored at equilibrium since \(E^{\circ}_{\text{cell}} > 0\).

Key Concepts

Standard Reduction Potential

The standard reduction potential is a measure that helps us understand how easily a chemical species can gain electrons. In simpler words, it indicates the tendency of a molecule to be reduced, or to accept electrons. When you see the standard reduction potential value written as a positive number, it means that the reaction is more likely to occur because it has a strong tendency to accept electrons. On the other hand, a negative value indicates that the species is less likely to gain electrons.

Each element or compound involved in a reaction has its own standard reduction potential, measured in volts (V). For example, in the reaction between iodine and bromide, the value for iodine is +0.54 V. This means that iodine is reasonably good at accepting electrons. By comparing the standard reduction potentials of two substances, we can figure out which one will undergo a reduction and which will act as the oxidizing agent. Therefore, standard reduction potentials are critical for balancing and predicting the direction of redox reactions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two chemical species. They occur when one species loses electrons (oxidation) and another gains those electrons (reduction). This process always happens simultaneously, which is why we call it a coupled process. Let's break down the terminology:

• Oxidation: Loss of electrons.
• Reduction: Gain of electrons.
• Oxidizing agent: The species that gets reduced.
• Reducing agent: The species that gets oxidized.

In the exercise, for instance, iron and chromium(III) are part of a redox reaction where iron gets oxidized from Fe²⁺ to Fe³⁺ by losing an electron, and chromium(III) ions are the reduced counterpart, mitigating electrons as they transform from dichromate to chromium ions. Balancing these reactions involves ensuring that mass and charge are conserved, often by splitting the reaction into half-reactions for clarity.

Electrochemical Cells

Electrochemical cells are fascinating and practical applications of redox reactions. They convert the chemical energy resulting from the spontaneous redox reactions into electrical energy. By coupling the oxidation reaction and the reduction reaction in separate compartments, we can facilitate a flow of electrons through an external circuit, essentially creating electricity.

An electrochemical cell typically consists of two halves: the anode and the cathode. Let's mark out some details:

• Anode: Where oxidation occurs and electrons are released. It's the negative side of a galvanic (voltaic) cell.
• Cathode: Where reduction occurs and electrons are accepted. It's the positive side.

In these cells, the standard reduction potential gives us an idea of the spontaneous nature of reactions. A positive cell potential means the reaction in question will have the potential to produce electricity. In exercises where we calculate the standard cell potential, we typically subtract the standard reduction potential of the anode from the cathode. This helps predict whether the reaction will be spontaneous, like in some of the examples from the initial exercise.