Problem 18 - Chemistry The Central Science

Question

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. $$ \begin{array}{l}{\text { (a) } 2 \mathrm{MnO}_{4}^{-}(a q)+3 \mathrm{S}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{S}(s)+} \\ {\quad 2 \mathrm{MnO}_{2}(s)+8 \mathrm{OH}^{-}(a q)} \\ {\text { (b) } 4 \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{Cl}_{2} \mathrm{O}_{7}(g)+2 \mathrm{OH}^{-}(a q) \longrightarrow 2 \mathrm{ClO}_{2}^{-}(a q)+} \\ {\quad 5 \mathrm{H}_{2} \mathrm{O}(l)+4 \mathrm{O}_{2}(g)} \\\\{\text { (c) } \mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q) \longrightarrow} \\ {\quad \mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)}\end{array} $$

Step-by-Step Solution

Answer

In the given redox reactions: (a) 6 electrons are transferred, with S(-II) -> S(0) having a change in oxidation number of +2, and Mn(+VII) -> Mn(+IV) having a change of -3. (b) 4 electrons are transferred, with O(-I) in \(H_2O_2\) -> O(0) in \(O_2\) having a change in oxidation number of +1, and Cl(+VII) -> Cl(+III) having a change of -4. (c) 1 electron is transferred, with O(-I) in \(H_2O_2\) -> O(0) in \(O_2\) having a change in oxidation number of +1, and no electrons transferred for Cl(+IV) -> Cl(+IV).

Step 1: (a) Assign Oxidation Numbers

For reaction (a), we assign the oxidation numbers for each element in both reactants and products: \(MnO_4^-\): Mn has an oxidation number of +7, while O has -2. \(S^{2-}\): S has an oxidation number of -2. \(H_2O\): H has an oxidation number of +1, while O has -2. \(S(s)\): S has an oxidation number of 0. \(MnO_2\): Mn has an oxidation number of +4, while O has -2. \(OH^-\): O has an oxidation number of -2, and H has +1.

Step 2: (a) Calculate Electron Transfer

For reaction (a), we determine the change in oxidation numbers for the species involved in redox reactions: For S(-II) -> S(0), the change in oxidation number is +2. For Mn(+VII) -> Mn(+IV), the change in oxidation number is -3. Now, multiply the change in oxidation numbers by the stoichiometric coefficients to determine the total number of electrons transferred: The total number of electrons gained by S is \(3 \times 2 = 6\). The total number of electrons lost by Mn is \(2 \times 3 = 6\). In reaction (a), 6 electrons are transferred.

Step 3: (b) Assign Oxidation Numbers

For reaction (b), we assign the oxidation numbers for each element: \(H_2O_2\): H has an oxidation number of +1, while O has -1. \(Cl_2O_7\): Cl has an oxidation number of +7, while O has -2. \(OH^-\): O has an oxidation number of -2, and H has +1. \(ClO_2^-\): Cl has an oxidation number of +3, while O has -2. \(H_2O\): H has an oxidation number of +1, while O has -2. \(O_2\): O has an oxidation number of 0.

Step 4: (b) Calculate Electron Transfer

For reaction (b), we determine the change in oxidation numbers for the species involved in redox reactions: For O(-I) in \(H_2O_2\) -> O(0) in \(O_2\), the change in oxidation number is +1. For Cl(+VII) -> Cl(+III), the change in oxidation number is -4. Multiply the change in oxidation numbers by the stoichiometric coefficients to determine the total number of electrons transferred: The total number of electrons gained by O is \(4 \times 1 = 4\). The total number of electrons lost by Cl is \(1 \times 4 = 4\). In reaction (b), 4 electrons are transferred.

Step 5: (c) Assign Oxidation Numbers

For reaction (c), we assign the oxidation numbers: \(Ba^{2+}\): Ba has an oxidation number of +2. \(OH^-\): O has an oxidation number of -2, and H has +1. \(H_2O_2\): H has an oxidation number of +1, while O has -1. \(ClO_2\): Cl has an oxidation number of +4, while O has -2. \(Ba(ClO_2)_2\): Ba has an oxidation number of +2, Cl has +4 and O has -2. \(H_2O\): H has an oxidation number of +1, while O has -2. \(O_2\): O has an oxidation number of 0.

Step 6: (c) Calculate Electron Transfer

For reaction (c), we determine the change in oxidation numbers for the species involved in redox reactions: For O(-I) in \(H_2O_2\) -> O(0) in \(O_2\), the change in oxidation number is +1. For Cl(+IV) -> Cl(+IV) (no change), no electrons are transferred. Multiply the change in oxidation numbers by the stoichiometric coefficients to determine the total number of electrons transferred: The total number of electrons gained by O is \(1 \times 1 = 1\). In reaction (c), 1 electron is transferred.

Key Concepts

Oxidation Numbers

Understanding oxidation numbers is key to identifying oxidation-reduction reactions. An oxidation number is a value assigned to an element in a chemical compound that represents the number of electrons lost or gained by an atom of that element during the formation of the compound. These numbers help track how electrons are distributed among atoms. Here’s how oxidation numbers are assigned to elements in various compounds:

• For a pure element like diatomic oxygen ( O_2 ), the oxidation number is always 0 because the element is not combined with anything else.
• In compounds, oxygen generally has an oxidation number of -2. However, in peroxides like hydrogen peroxide ( H_2O_2 ), oxygen has an oxidation number of -1.
• Hydrogen typically has an oxidation number of +1 when combined with non-metals and -1 when combined with metals.
• The sum of oxidation numbers in a neutral compound is zero, while in polyatomic ions, it equals the charge of the ion.

Using oxidation numbers, we can find out which substances are oxidized and which are reduced during the reaction. For example, manganese ( Mn ) in permanganate ( MnO_4^− ) has an oxidation number of +7, which decreases to +4 in manganese dioxide ( MnO_2 ), indicating a reduction.

Electron Transfer

Electron transfer is the process where electrons move from one atom to another during a chemical reaction. This transfer is what constitutes oxidation-reduction or redox reactions. In such reactions, oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Let’s see how electron transfer takes place using reaction (a) as an example:
For sulfur ( S ), the oxidation number goes from -2 in S^{2-} to 0 in S , implying that sulfur is oxidized. This indicates it loses electrons and its oxidation number increases. The loss of electrons (oxidation) by sulfur is matched by the gain of electrons (reduction) by manganese in MnO_4^− , where the oxidation number of Mn changes from +7 to +4, signifying it is reduced. For each sulfur atom that loses 2 electrons, manganese receives 3 electrons. So when observing the stoichiometry, 6 electrons in total are transferred as multiple sulfur atoms interact with permanganate ions.

Redox Reactions

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between two substances. During redox processes, one element gets oxidized, losing electrons, while another gets reduced, gaining electrons.
Here are key points to understand redox reactions:

• Oxidation is the increase in oxidation state due to electron loss.
• Reduction is the decrease in oxidation state through electron gain.
• The substances taking part in oxidation and reduction are respectively called the reducing agent and the oxidizing agent. The reducing agent is oxidized, while the oxidizing agent is reduced.

Consider reaction (b) for example: In the reaction involving hydrogen peroxide ( H_2O_2 ) and dichlorine heptoxide ( Cl_2O_7 ), the oxidation number for oxygen in H_2O_2 increases from -1 to 0 when forming O_2 . Hence, H_2O_2 acts as a reducing agent, losing electrons in the process. Cl, going from +7 in Cl_2O_7 to +3 in ClO_2^− , is reduced, acting as the oxidizing agent. This balanced change in oxidation number through electron transfer makes the reaction a redox process.