Question

In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) \(\mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g)-\mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow\) \(2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)\) (c) \(3 \mathrm{H}_{2} \mathrm{~S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)-\cdots\) \(3 \mathrm{~S}(s)+2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Ba}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)+\) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{ClO}_{2}(a q)-\mathrm{-} \rightarrow\) \(\mathrm{Ba}\left(\mathrm{ClO}_{2}\right)_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(g)\)

Short answer

In each of the following balanced oxidation-reduction equations: (a) Iodine undergoes a change in oxidation number from +5 to 0, with a magnitude of change of 5. Carbon undergoes a change from +2 to +4, with a magnitude of change of 2. (b) Mercury undergoes a change in oxidation number from +2 to 0, with a magnitude of change of 2. Nitrogen undergoes a change from -2 to 0, with a magnitude of change of 2. (c) Sulfur undergoes a change in oxidation number from -2 to 0, with a magnitude of change of 2. Nitrogen undergoes a change from +5 to +2, with a magnitude of change of 3. (d) Oxygen undergoes a change in oxidation number from -1 in H2O2 to 0 in O2, with a magnitude of change of 1. Barium and Chlorine do not undergo any changes in oxidation number.

Step-by-step solution

Identify oxidation states in reactants

For each element in the reactants, assign an oxidation state: - In I2O5: Iodine has an oxidation state of +5, Oxygen has an oxidation state of -2 - In CO: Carbon has an oxidation state of +2, Oxygen has an oxidation state of -2

Identify oxidation states in products

For each element in the products, assign an oxidation state: - In I2: Iodine has an oxidation state of 0 - In CO2: Carbon has an oxidation state of +4, Oxygen has an oxidation state of -2

Determine changes in oxidation states

Compare the oxidation states of elements in the reactants and products: - Iodine: changes from +5 to 0, with a magnitude of change of 5. - Carbon: changes from +2 to +4, with a magnitude of change of 2. (b) Mercury, Nitrogen

Identify oxidation states in reactants

For each element in the reactants, assign their oxidation states: - In Hg2+: Mercury has an oxidation state of +2 - In N2H4: Nitrogen has an oxidation state of -2, Hydrogen has an oxidation state of +1

Identify oxidation states in products

For each element in the products, assign an oxidation state: - In Hg: Mercury has an oxidation state of 0 - In N2: Nitrogen has an oxidation state of 0 - In H+: Hydrogen has an oxidation state of +1

Determine changes in oxidation states

Compare the oxidation states of elements in the reactants and products: - Mercury: changes from +2 to 0, with a magnitude of change of 2. - Nitrogen: changes from -2 to 0, with a magnitude of change of 2. (c) Sulfur, Nitrogen

Identify oxidation states in reactants

For each element in the reactants, assign their oxidation states: - In H2S: Sulfur has an oxidation state of -2, Hydrogen has an oxidation state of +1 - In H+: Hydrogen has an oxidation state of +1 - In NO3-: Nitrogen has an oxidation state of +5, Oxygen has an oxidation state of -2

Identify oxidation states in products

For each element in the products, assign an oxidation state: - In S: Sulfur has an oxidation state of 0 - In NO: Nitrogen has an oxidation state of +2, Oxygen has an oxidation state of -2 - In H2O: Hydrogen has an oxidation state of +1, Oxygen has an oxidation state of -2

Determine changes in oxidation states

Compare the oxidation states of elements in the reactants and products: - Sulfur: changes from -2 to 0, with a magnitude of change of 2. - Nitrogen: changes from +5 to +2, with a magnitude of change of 3. (d) Barium, Oxygen, and Chlorine

Identify oxidation states in reactants

For each element in the reactants, assign an oxidation state: - In Ba2+: Barium has an oxidation state of +2 - In OH-: Oxygen has an oxidation state of -2, Hydrogen has an oxidation state of +1 - In H2O2: Hydrogen has an oxidation state of +1, Oxygen has an oxidation state of -1 - In ClO2: Chlorine has an oxidation state of +4, Oxygen has an oxidation state of -2

Identify oxidation states in products

For each element in the products, assign an oxidation state: - In Ba(ClO2)2: Barium has an oxidation state of +2, Chlorine has an oxidation state of +4, Oxygen has an oxidation state of -2 - In H2O: Hydrogen has an oxidation state of +1, Oxygen has an oxidation state of -2 - In O2: Oxygen has an oxidation state of 0

Determine changes in oxidation states

Compare the oxidation states of elements in the reactants and products: - Barium: No change, remains +2. - Oxygen: Changes from -1 in H2O2 to 0 in O2, with a magnitude of change of 1. - Chlorine: No change, remains +4.

Key concepts

Changes in Oxidation Number

Understanding changes in oxidation number is crucial when studying oxidation-reduction (redox) reactions. An oxidation number is a value that represents the number of electrons an atom can gain, lose, or share during a chemical reaction. When there's a change in the oxidation number, it means that a transfer of electrons has occurred, indicating either oxidation (loss of electrons) or reduction (gain of electrons).

• Oxidation involves an increase in oxidation number and is associated with a loss of electrons.
• Reduction involves a decrease in oxidation number and signifies a gain of electrons.

These alterations can be small or large, affecting the chemical properties of the elements involved. To identify these changes, compare the oxidation numbers before and after the reaction. If the oxidation number increases, the element has been oxidized; if it decreases, the element has been reduced.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is upheld in a reaction. Here's the gist: the number of atoms of each element must be equal on both sides of the equation. To balance an equation, start by counting the number of atoms of each element in both reactants and products. Adjust coefficients, never the subscripts of the compounds, to get the same number of atoms of each element on both sides.

In a redox reaction, it is also vital to ensure that the total increase and decrease in oxidation numbers are balanced. This not only assures the conservation of mass but also the conservation of charge. By balancing changes in oxidation numbers, you account for the electrons lost and gained during the reaction, enabling accurate predictions of the products formed.

Identifying Oxidation States

To solve redox problems effectively, identifying the oxidation states of different elements in a chemical compound is essential. Oxidation states, often known as oxidation numbers, provide insight into the electronic structure of the compound. Here are a few guidelines to help determine oxidation states:

• The oxidation state of a free element (not combined with other elements) is always zero.
• For ions composed of a single atom, the oxidation state is equal to the charge on the ion.
• In a compound, Group 1 elements have an oxidation state of +1; Group 2 elements have +2. Oxygen typically has an oxidation state of -2, except in peroxides where it's -1.
• The sum of oxidation states in a neutral compound is zero; in a polyatomic ion, it's equal to the ion's charge.

By applying these rules and comparing the oxidation states of each element in the reactants and products, you can determine the changes that occur during the reaction. Identifying these states is pivotal for understanding and predicting the behavior of different elements in chemical reactions.