Problem 16 - Chemistry The Science in Context

Question

In the redox reaction below, how many electrons are transferred for each molecule of \(\mathrm{H}_{2} \mathrm{O}_{2}\) consumed? \(2 \mathrm{MnO}_{4}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow\) $$2 \mathrm{MnO}_{2}(s)+3 \mathrm{O}_{2}(g)+2 \mathrm{OH}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$$

Step-by-Step Solution

Answer

Answer: 2 electrons

Step 1: Identify the oxidation states of reactants and products

To determine the change in oxidation states, we must first identify the initial oxidation states of all the elements in the reactants and products of the given reaction. The redox reaction is: $2 \mathrm{MnO}_{4}^{-}(a q)+3 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \rightarrow 2 \mathrm{MnO}_{2}(s)+3 \mathrm{O}_{2}(g)+2 \mathrm{OH}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell)$ Oxidation states: - Oxygen is usually -2 except in peroxides (e.g., \(\mathrm{H}_{2}\mathrm{O}_{2}\)) where it is -1. - Hydrogen is +1. - Mn in \(\mathrm{MnO}_{4}^{-}\) is +7 (since there are four oxygen atoms with a charge of -2 each and a single negative charge on the ion). Now we can assign oxidation states to the elements in reaction. Initial oxidation states: - Mn in \(\mathrm{MnO}_{4}^{-}\): +7 - O in \(\mathrm{H}_{2} \mathrm{O}_{2}\): -1 Final oxidation states: - Mn in \(\mathrm{MnO}_{2}\): +4 (as Mn is bonded to two oxygen atoms, each with -2) - O in \(\mathrm{O}_{2}\): 0 (each oxygen atom has no charge) - O in \(2 \mathrm{OH}^{-}\): -2 (as in hydroxyl ion)

Step 2: Determine the number of electrons transferred during the reaction

Next, we will calculate the change in oxidation states for Mn and O and identify the number of electrons transferred. For Mn: Change in oxidation state = initial oxidation state - final oxidation state = +7 - (+4) = +3 Since there are two Mn atoms in the reaction, the total change in oxidation state for Mn is +6. For O: Change in oxidation state for \(\mathrm{H}_{2} \mathrm{O}_{2}\) = initial oxidation state - final oxidation state = -1 - 0 = -1 Since there are six O atoms in 3 molecules of \(\mathrm{H}_{2} \mathrm{O}_{2}\), the total change in oxidation state for O is -6. Total number of electrons transferred for one molecule of \(\mathrm{H}_{2} \mathrm{O}_{2}\) consumed can be determined as follows: - For Mn (Mn changes its oxidation state from +7 to +4): +6 electrons (as there are two Mn atoms and they gain electrons) - For O (O changes its oxidation state from -1 to 0): -6 electrons (as there are six O atoms in 3 \(\mathrm{H}_{2} \mathrm{O}_{2}\) molecules and they lose electrons)

Step 3: Determine the number of electrons transferred for each molecule of \(\mathrm{H}_{2} \mathrm{O}_{2}\)

Since there are 3 \(\mathrm{H}_{2} \mathrm{O}_{2}\) molecules consumed in the reaction, we need to divide the total number of electrons transferred by 3 to find the number of electrons transferred for each molecule of \(\mathrm{H}_{2} \mathrm{O}_{2}\). Number of electrons transferred for each molecule of \(\mathrm{H}_{2} \mathrm{O}_{2}\) consumed = Total number of electrons transferred / Number of \(\mathrm{H}_{2} \mathrm{O}_{2}\) molecules consumed = -6 / 3 = -2 electrons Thus, 2 electrons are transferred for each molecule of \(\mathrm{H}_{2} \mathrm{O}_{2}\) consumed in the redox reaction.

Key Concepts

Oxidation States

In a redox reaction, understanding oxidation states is key to spotting how electrons move between reactants and products. Oxidation states indicate the degree of oxidation of an atom; it's essentially the charge that an atom would have if all bonds were ionic. Let's consider some general rules for assigning oxidation states:

• Pure elements have an oxidation state of 0. For example, oxygen in \(\mathrm{O}_{2}\) is 0.
• Oxygen typically has an oxidation state of -2, except in peroxides like hydrogen peroxide \(\mathrm{H}_{2}\mathrm{O}_{2}\), where it is -1.
• Hydrogen generally holds an oxidation state of +1.
• The sum of oxidation states in a neutral compound is 0, while in an ion it is equal to the ion's charge.

Knowing how oxidation states change helps us determine which species are oxidized and reduced in a reaction. In the example reaction, manganese (Mn) goes from an oxidation state of +7 in \(\mathrm{MnO}_{4}^{-}\) to +4 in \(\mathrm{MnO}_{2}\), indicating a reduction as it gains electrons. Likewise, oxygen in \(\mathrm{H}_{2}\mathrm{O}_{2}\) goes from -1 to 0, meaning it is oxidized and loses electrons.

Electron Transfer

Electron transfer is at the heart of redox reactions. These reactions involve either the gain or loss of electrons, dictating how substances are transformed during chemical reactions. In the example given, understanding electron transfer requires tracking changes in oxidation states.

The oxidation part involves the loss of electrons. When a species, like oxygen in \(\mathrm{H}_{2}\mathrm{O}_{2}\), moves from an oxidation state of -1 to 0, it indicates oxidation.

• Oxidation involves an increase in oxidation state due to electron loss.

Conversely, the reduction part involves the gain of electrons. Manganese in \(\mathrm{MnO}_{4}^{-}\) goes from an oxidation state of +7 to +4 in \(\mathrm{MnO}_{2}\). This change signifies reduction.

• Reduction involves a decrease in oxidation state due to electron gain.

In this reaction, electrons are exchanged between Mn and O atoms, showcasing the coordinated electron transfers that define redox reactions and ultimately balance the chemical equation.

Chemical Equations

Chemical equations represent how reactants transform into products in a reaction. They are symbolic representations where atoms' and molecules' identities and quantities are expressed. Balancing chemical equations ensures that the number of each type of atom is consistent on both sides of the reaction.

A balanced equation must accurately reflect:

• The conservation of mass, ensuring the same number of each atom exists on both sides.
• The conservation of charge, particularly in ionic reactions, where the total charge must also be balanced.

For the redox reaction given, the balanced equation is:

\(2 \, \mathrm{MnO}_{4}^{-}(aq) + 3 \, \mathrm{H}_{2} \mathrm{O}_{2}(aq) \rightarrow 2 \, \mathrm{MnO}_{2}(s) + 3 \, \mathrm{O}_{2}(g) + 2 \, \mathrm{OH}^{-}(aq) + 2 \, \mathrm{H}_{2} \mathrm{O}(\ell)\)

In balancing the equation, introduce coefficients to ensure atoms like Mn, O, and H appear equally on both sides. This makes sure the equation maintains consistency with the conservation principles. Each species is correctly placed by considering both matter and electron flow, particularly in redox reactions where oxidation and reduction processes must reconcile.