Problem 29 - Chemistry: Principles and Reactions

Question

Use the following half-equations to write three spontaneous reactions. Justify your answers by calculating \(E^{\circ}\) for the cells. (1) \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}\) \(E^{\circ}=+1.512 \mathrm{~V}\) (2) \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=+1.229 \mathrm{~V}\) (3) \(\mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s) \quad E^{\circ}=-0.282 \mathrm{~V}\)

Step-by-Step Solution

Answer

The standard cell potentials for the spontaneous reactions are: 1. For the reaction combining half-equations (1) and (3): \(E^{\circ}_{\text{cell}} = +1.230 \mathrm{~V}\) 2. For the reaction combining half-equations (2) and (3): \(E^{\circ}_{\text{cell}} = +0.947 \mathrm{~V}\) There is no valid third spontaneous reaction.

Step 1: Choose the half-equations that result in a positive \(E^{\circ}\) value

When combining half-reactions, a spontaneous reaction will occur if the sum of the half-reactions' \(E^{\circ}\) values is positive. So, we should choose two half-equations with \(E^{\circ}\) values that, when added, result in a positive value.

Step 2: Write the first spontaneous reaction

To write the first spontaneous reaction, we will combine half-equations (1) and (3), ensuring that electrons are balanced: (1) \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}\) (3) \(2(\mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s))\) Now, we add the two half-equations to create the spontaneous reaction: \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} + 2(\mathrm{Co}^{2+}(a q)+2 e^{-}) \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O} + 2\mathrm{Co}(s)\) Notice that the number of electrons in the reactions (5 and 4) are balanced.

Step 3: Calculate the \(E^{\circ}\) value for the first reaction

To calculate the standard cell potential for this reaction, add the values of the two half-equations: \(E^{\circ}_{\text{cell}} = E^{\circ}_{1} + E^{\circ}_{3} = (+1.512 \mathrm{~V}) + (-0.282 \mathrm{~V}) = +1.230 \mathrm{~V}\) The positive value of \(E^{\circ}_{\text{cell}}\) indicates a spontaneous reaction.

Step 4: Write the second spontaneous reaction

To write the second spontaneous reaction, we will combine half-equations (2) and (3), ensuring that electrons are balanced: (2) \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}\) (3) \(2(\mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s))\) Now, we add the two half-equations to create the spontaneous reaction: \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} + 2(\mathrm{Co}^{2+}(a q)+2 e^{-}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} + 2\mathrm{Co}(s)\) The number of electrons in the reactions (4 and 4) are balanced.

Step 5: Calculate the \(E^{\circ}\) value for the second reaction

To calculate the standard cell potential for this reaction, add the values of the two half-equations: \(E^{\circ}_{\text{cell}} = E^{\circ}_{2} + E^{\circ}_{3} = (+1.229 \mathrm{~V}) + (-0.282 \mathrm{~V}) = +0.947 \mathrm{~V}\) The positive value of \(E^{\circ}_{\text{cell}}\) indicates a spontaneous reaction.

Step 6: Write the third spontaneous reaction

Since there are only three half-equations, our third spontaneous reaction will combine half-equations (1) and (2): (1) \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}\) (2) \(1.25(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O})\) Now, we add the two half-equations to create the spontaneous reaction: \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} + 1.25(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-}) \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O} + 2.5 \mathrm{H}_{2} \mathrm{O}\) The number of electrons in the reactions (5 and 4) are not balanced. Therefore, we cannot create a third spontaneous reaction by combining these two half-equations.

Key Concepts

Half-equations

In electrochemistry, reactions are frequently broken down into simpler steps, known as half-equations. These half-equations represent the oxidation and reduction processes that occur during the overall reaction. By considering these processes separately, we can more easily understand and balance the chemical reaction.

• The oxidation half-equation describes the loss of electrons by a species. This occurs when an element increases its oxidation state. A classic example is the conversion of metallic zinc to zinc ions.
• The reduction half-equation represents the gain of electrons by a species, resulting in a decrease in its oxidation state. For instance, the conversion of copper ions to copper metal.

To create a balanced chemical equation from half-equations, it is important to ensure that the electrons lost in the oxidation process are equal to the electrons gained in the reduction process. Only when accurate electron balance is achieved can the half-equations be combined to illustrate the complete reaction. This approach provides clarity and insight into how electrons are shuffled during chemical changes.

Standard cell potential

The standard cell potential, denoted as \(E^{\circ}\), is a measure of the potential energy difference between two electrodes in an electrochemical cell under standard conditions. The standard conditions imply a concentration of 1 M for solutions, a pressure of 1 atm for gases, and a temperature of 25°C (298 K).

Calculating the standard cell potential involves examining the two half-reactions that constitute the electrochemical cell. Each half-reaction has an associated standard reduction potential, indicating the relative tendency of a species to gain electrons.

• If the standard reduction potential is positive, the species is more likely to gain electrons and be reduced. Such reactions are favorable.
• If negative, the species is less inclined to gain electrons, indicating an unfavored reduction process.

The overall standard cell potential \(E^{\circ}_{\text{cell}}\) can be determined by adding the standard potentials of the two half-reactions. If \(E^{\circ}_{\text{cell}}\) is positive, the overall reaction is spontaneous, proceeding without external energy input. This potential difference is the driving force behind the flow of electric current in electrochemical cells.

Spontaneous reactions

A reaction is considered spontaneous if it occurs naturally without the need for additional energy input beyond its activation energy. In electrochemistry, the spontaneity of a reaction is closely associated with the sign of the standard cell potential \(E^{\circ}_{\text{cell}}\).

Spontaneous reactions are identified by their positive \(E^{\circ}_{\text{cell}}\) values. But why does this matter? Here’s a simple breakdown:

• A positive \(E^{\circ} \) indicates that the energy released by the system can drive the reaction forward. This inherent favorable potential means that the reactants can convert to products without needing an external power source.
• Conversely, a negative \(E^{\circ} \) suggests that the reaction is non-spontaneous under standard conditions, necessitating external energy to proceed.

When evaluating whether reactions are spontaneous, it’s important to consider the sum of the potential energies of the involved half-reactions. The broader implication of spontaneous reactions in electrochemistry is seen in batteries and galvanic cells, where chemical energy is converted into electrical energy to power devices efficiently.