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. $$ \begin{array}{l}{\text { (a) } \mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)} \\\ {\text { (b) } 2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)} \\\ {\text { (c) } 3 \mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow 3 \mathrm{S}(s)+} \\\\{\quad\quad 2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$

Short answer

In the given reactions: (a) In I2O5, iodine has an oxidation state of +5, and carbon in CO has an oxidation state of +2. In the products, iodine in I2 has an oxidation state of 0, and carbon in CO2 has an oxidation state of +4. In this reaction, 10 electrons are transferred. (b) In the reactants, mercury in Hg2+ has an oxidation state of +2, and nitrogen in N2H4 has an oxidation state of -2. In the products, mercury in Hg has an oxidation state of 0, and nitrogen in N2 has an oxidation state of 0. In this reaction, 4 electrons are transferred. (c) In the reactants, sulfur in H2S has an oxidation state of -2, and nitrogen in NO3- has an oxidation state of +5. In the products, sulfur as an element has an oxidation state of 0, and nitrogen in NO has an oxidation state of +2. In this reaction, 6 electrons are transferred.

Step-by-step solution

Identify oxidation states

In the reactants: Iodine (I) in I2O5: \[ \mathrm{+5} \] Oxygen (O) in I2O5: \[ \mathrm{-2} \] Carbon (C) in CO: \[ \mathrm{+2} \] Oxygen (O) in CO: \[ \mathrm{-2} \] In the products: Iodine (I) in I2: \[ \mathrm{0} \] Carbon (C) in CO2: \[ \mathrm{+4} \] Oxygen (O) in CO2: \[ \mathrm{-2} \]

Determine change in oxidation states

For Iodine (I): \( +5 \rightarrow 0 \), the change is -5. For Carbon (C): \( +2 \rightarrow +4 \), the change is +2.

Calculate total number of electrons transferred

Since there are two atoms of Iodine, the total change for Iodine is -10. Since there are five molecules of CO, the total change for Carbon is +10. Therefore, the total number of electrons transferred in this reaction is 10. ## Reaction (b) ##

Identify oxidation states

In the reactants: Mercury (Hg) in Hg2+: \[ \mathrm{+2} \] Nitrogen (N) in N2H4: \[ \mathrm{-2} \] Hydrogen (H) in N2H4: \[ \mathrm{+1} \] In the products: Mercury (Hg) in Hg: \[ \mathrm{0} \] Nitrogen (N) in N2: \[ \mathrm{0} \] Hydrogen (H) in H+: \[ \mathrm{+1} \]

Determine change in oxidation states

For Mercury (Hg): \( +2 \rightarrow 0 \), the change is -2. For Nitrogen (N): \( -2 \rightarrow 0 \), the change is +2.

Calculate total number of electrons transferred

Since there are two atoms of Mercury, the total change for Mercury is -4. Since there are two atoms of Nitrogen, the total change for Nitrogen is +4. Therefore, the total number of electrons transferred in this reaction is 4. ## Reaction (c) ##

Identify oxidation states

In the reactants: Hydrogen (H) in H2S: \[ \mathrm{+1} \] Sulfur (S) in H2S: \[ \mathrm{-2} \] Hydrogen (H) in H+: \[ \mathrm{+1} \] Nitrogen (N) in NO3-: \[ \mathrm{+5} \] Oxygen (O) in NO3-: \[ \mathrm{-2} \] In the products: Sulfur (S) in S: \[ \mathrm{0} \] Nitrogen (N) in NO: \[ \mathrm{+2} \] Oxygen (O) in NO: \[ \mathrm{-2} \] Hydrogen (H) in H2O: \[ \mathrm{+1} \] Oxygen (O) in H2O: \[ \mathrm{-2} \]

Determine change in oxidation states

For Sulfur (S): \( -2 \rightarrow 0 \), the change is +2. For Nitrogen (N): \( +5 \rightarrow +2 \), the change is -3.

Calculate total number of electrons transferred

Since there are three atoms of Sulfur, the total change for Sulfur is +6. Since there are two atoms of Nitrogen, the total change for Nitrogen is -6. Therefore, the total number of electrons transferred in this reaction is 6.

Key concepts

Oxidation Numbers

Understanding oxidation numbers is crucial to mastering redox reactions. An oxidation number, sometimes referred to as oxidation state, is a figure that represents the total number of electrons that an atom either gains or loses to form a chemical bond with another atom. Each atom in a molecule is assigned an oxidation number, which can be positive, negative, or zero, indicating the atom's electronic status when it forms part of a compound.

For instance, in a compound like water (H₂O), hydrogen (H) has an oxidation number of +1, while oxygen (O) has an oxidation number of -2. This assignment is based on electronegativity and standardized rules, such as the fact that oxygen generally has an oxidation number of -2 in its compounds, except in peroxides. By calculating how these numbers change in a redox reaction, we can determine which species are oxidized and which are reduced.

Electron Transfer

In the realm of redox reactions, the movement of electrons from one substance to another is a defining feature. The term electron transfer refers to this process. When a substance loses electrons, it is oxidized; when a substance gains electrons, it is reduced. This electron transfer is fundamental to the chemical changes in redox reactions, often resulting in changes in chemical properties and energy release.

For example, when sodium (Na) reacts with chlorine (Cl) to form sodium chloride (NaCl), sodium loses one electron becoming Na⁺, and chlorine gains that electron to become Cl^-. Here, we see electron transfer resulting in the formation of ions, which is indicative of oxidation and reduction.

Chemical Reactions

The core of chemistry lies in chemical reactions, where reactants interact and transform into products. These transformations occur due to the breaking and forming of chemical bonds, which involve energy changes. Reactions are classified into different types, such as synthesis, decomposition, single displacement, double displacement, and combustion. Importantly, redox reactions are a type of chemical reaction that includes the transfer of electrons between species.

Redox reactions are essential in both biological and industrial processes. For instance, cellular respiration is a series of redox reactions that convert glucose and oxygen into carbon dioxide, water, and energy. Understanding these reactions enables us to comprehend how energy is generated and utilized in living organisms and how it can be harnessed for various technological applications.

Balancing Redox Equations

To fully represent a redox reaction, it is necessary to balance the equation, which makes sure that the same number of atoms for each element is present on both the reactant and product sides. Moreover, balancing redox equations requires that the charge should also be balanced, ensuring the same number of electrons are lost and gained during the reaction.

The method often used to balance redox equations is the 'half-reaction method', where the equation is separated into two half-reactions, one for oxidation and one for reduction. Each half-reaction is balanced separately and then combined to form the balanced overall equation. For instance, in our earlier examples from the textbook, the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction, resulting in a balanced overall redox reaction.

Properly balancing redox equations is fundamental in understanding stoichiometry and reaction stoichiometry, which are significant in predicting the amounts of reactants needed and products formed in chemical reactions.