Question

For the following oxidation-reduction reaction, (a) write out the two half- reactions and balance the equation, (b) calculate \(\Delta E^{\circ}\), and (c) determine whether the reaction will proceed spontaneously as written; \(\mathrm{Fe}^{2+}+\mathrm{MnO}^{-}{ }_{4}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (1) \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \leftrightarrows \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=0.77 \mathrm{eV}\) (2) \(\mathrm{MnO}^{-}{ }_{4}+8 \mathrm{H}^{+}+6 \mathrm{e}^{-} \leftrightarrows \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{eV}\)

Short answer

The balanced equation is: \(\mathrm{Fe}^{2+}+\mathrm{MnO}^{-}{ }_{4}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\). The calculated value for \(\Delta E^{\circ}\) is \(0.74 \, \mathrm{eV}\), indicating that the reaction is spontaneous as written.

Step-by-step solution

Identify and balance the half-reactions

We are given two half-reactions: (1) \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \leftrightarrows \mathrm{Fe}^{2+}, \mathrm{E}^{\circ}=0.77 \mathrm{eV}\) (2) \(\mathrm{MnO}^{-}{ }_{4}+8 \mathrm{H}^{+}+6 \mathrm{e}^{-} \leftrightarrows \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}, \mathrm{E}^{\circ}=1.51 \mathrm{eV}\) These half-reactions are already balanced, so we can proceed to the next step.

Combine the half-reactions

To combine the half-reactions, we need to make sure that the number of electrons lost in the oxidation half-reaction (1) equals the number of electrons gained in the reduction half-reaction (2). This involves multiplying the half-reaction (1) by 6 and then adding the two half-reactions: (3) \(6 \times (\mathrm{Fe}^{3+}+\mathrm{e}^{-} \leftrightarrows \mathrm{Fe}^{2+})\): \(6\mathrm{Fe}^{3+}+6\mathrm{e}^{-} \leftrightarrows 6\mathrm{Fe}^{2+}\) Add (2) and (3): \(\mathrm{Fe}^{2+}+\mathrm{MnO}^{-}{ }_{4}+\mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\)

Calculate \(\Delta E^{\circ}\)

To determine whether a reaction is spontaneous, use the following equation: \(\Delta E^\circ = E_{red}^\circ - E_{ox}^\circ\) Where \(E_{red}^\circ\) and \(E_{ox}^\circ\) represent the standard electrode potentials (given) for reduction and oxidation half-reactions, respectively. Here, \(E_{red}^\circ = 1.51 \, \mathrm{eV}\) (given in half-reaction 2) and \(E_{ox}^\circ = 0.77 \, \mathrm{eV}\) (given in half-reaction 1). Substitute these values into the equation: \(\Delta E^\circ = 1.51 \, \mathrm{eV} - 0.77 \, \mathrm{eV}\) \(\Delta E^\circ = 0.74 \, \mathrm{eV}\)

Determine spontaneity

Since \(\Delta E^{\circ}\) is positive, this reaction is spontaneous as written. In conclusion, we have found the balanced equation for the reaction, calculated \(\Delta E^{\circ}\), and determined that the reaction will proceed spontaneously as written.

Key concepts

Half-Reactions

In oxidation-reduction reactions, or redox reactions, electrons are transferred between atoms. To understand these reactions, it's useful to break them down into two separate half-reactions. A half-reaction focuses on either the oxidation process, where electrons are lost, or the reduction process, where electrons are gained. This separation makes it easier to see what is happening with the electrons in the reaction.

For example, in the reaction given, we have two half-reactions:

• The oxidation half-reaction is: \( \mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+} + \mathrm{e}^{-} \)
• The reduction half-reaction is: \( \mathrm{MnO}_4^{-} + 8 \mathrm{H}^{+} + 6 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_2\mathrm{O} \)

This process helps to identify and balance the atoms and charges separately before they are combined later.

Standard Electrode Potentials

Standard electrode potentials are crucial when working with redox reactions. They give us information about the tendency of a chemical species to be reduced or oxidized. Each half-reaction has its own electrode potential, measured under standard conditions (usually 1M concentration for solutions, 1 atm pressure for gases, and 25°C temperature).

In our example, the standard electrode potentials are:

• For the reduction half-reaction of \( \mathrm{MnO}_4^{-} \): \( E^{\circ} = 1.51 \text{ eV} \)
• For the oxidation half-reaction of \( \mathrm{Fe}^{2+} \): \( E^{\circ} = 0.77 \text{ eV} \)

By comparing these values, we can determine the direction of electron flow and gain insight into the energetics of the reaction.

Spontaneity of Reactions

Determining whether a reaction occurs spontaneously is essential in chemistry. For redox reactions, spontaneity can be predicted using the change in standard electrode potential, \( \Delta E^{\circ} \). This value is obtained by subtracting the standard potential of the oxidation half-reaction from the reduction half-reaction.

If the calculated \( \Delta E^{\circ} \) is positive, the reaction will occur spontaneously. In our exercise, \( \Delta E^{\circ} \) is calculated as follows:

• \( \Delta E^{\circ} = 1.51 \text{ eV} - 0.77 \text{ eV} = 0.74 \text{ eV} \)

A result of 0.74 eV, being positive, indicates that the redox reaction will indeed proceed without additional energy input.

Redox Balancing

Balancing redox reactions involves ensuring both mass and charge are conserved. Initially, this might appear challenging, but using half-reactions simplifies the process. We have already identified our half-reactions. Now, it's crucial to make sure the loss and gain of electrons are equal across both half-reactions.

For the given redox reaction:

• The oxidation reaction involves the loss of 1 electron, hence it needs to be multiplied by 6 to match the reduction, which involves the gain of 6 electrons.
• Resulting combined balanced forms: \( 6\mathrm{Fe}^{2+} + \mathrm{MnO}_4^{-} + 8\mathrm{H}^{+} \rightarrow 6\mathrm{Fe}^{3+} + \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O} \)

Verify that both atoms and charges are balanced to ensure the equation correctly represents the chemical process.