Question

Many oxidation-reduction reactions can be balanced by inspection. Try to balance the following reactions by inspection. In each reaction, identify the substance reduced and the substance oxidized. a. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{S}(g)\) c. \(C_{3} H_{8}(g)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(l)\) d. \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)\)

Short answer

The balanced oxidation-reduction reactions and the substances oxidized and reduced are as follows: a. \(Al(s)+3HCl(aq) \rightarrow AlCl_{3}(aq)+1.5H_{2}(g)\); Al is oxidized, and H in HCl is reduced. b. \(CH_{4}(g)+2S(s) \rightarrow CS_{2}(l)+2H_{2}S(g)\); C in CH4 is oxidized, and S is reduced. c. \(C_{3}H_{8}(g)+5O_{2}(g) \rightarrow 3CO_{2}(g)+4H_{2}O(l)\); C in C3H8 is oxidized, and O in O2 is reduced. d. \(Cu(s)+Ag^{+}(aq) \rightarrow Ag(s)+Cu^{2+}(aq)\); Cu is oxidized, and Ag in Ag+ is reduced.

Step-by-step solution

Determine the oxidation states

First, we need to determine the oxidation states of the elements in each reactant and product. Al has an oxidation state of 0, and in AlCl3, its oxidation state is +3. H in HCl has an oxidation state of +1, while its oxidation state in H2 is 0.

Balance the atoms

Now we can balance the equation. We can already see that the Al and Cl atoms are balanced on both sides of the equation. We just need to balance the hydrogen atoms by placing a coefficient of 3 in front of HCl and a coefficient of 1.5 in front of H2. Balanced equation: $\mathrm{Al}(s)+3\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+1.5\mathrm{H}_{2}(g)$

Identify the substance reduced and the substance oxidized

Since Al goes from an oxidation state of 0 to +3, it is oxidized. H in HCl goes from an oxidation state of +1 to 0, so it is reduced. b. $\mathrm{CH}_{4}(g)+\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+\mathrm{H}_{2} \mathrm{S}(g)$

Determine the oxidation states

First, we need to determine the oxidation states of the elements in each reactant and product. The oxidation state of C in CH4 is -4, and its oxidation state in CS2 is +4. The oxidation state of S in its elemental form is 0, and in H2S, its oxidation state is -2.

Balance the atoms

Now we can balance the equation. We just need to place a coefficient of 2 in front of H2S to balance the hydrogen and sulfur atoms on both sides. Balanced equation: $\mathrm{CH}_{4}(g)+2\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+2\mathrm{H}_{2} \mathrm{S}(g)$

Identify the substance reduced and the substance oxidized

Since C in CH4 goes from an oxidation state of -4 to +4, it is oxidized. S goes from an oxidation state of 0 to -2, so it is reduced. c. \(C_{3} H_{8}(g)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(l)\)

Determine the oxidation states

First, we need to determine the oxidation states of the elements in each reactant and product. The oxidation state of C in C3H8 is -2, and its oxidation state in CO2 is +4. The oxidation state of O in O2 is 0, and in H2O, its oxidation state is -2.

Balance the atoms

Now we can balance the equation. We can place coefficients of 3 in front of CO2 and 4 in front of H2O. To balance the O atoms on both sides, we need to place a coefficient of 5 in front of O2. Balanced equation: \(C_{3} H_{8}(g)+5O_{2}(g) \rightarrow 3CO_{2}(g)+4H_{2} O(l)\)

Identify the substance reduced and the substance oxidized

Since C in C3H8 goes from an oxidation state of -2 to +4, it is oxidized. O in O2 goes from an oxidation state of 0 to -2, so it is reduced. d. $\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)$

Determine the oxidation states

First, we need to determine the oxidation states of the elements in each reactant and product. Cu has an oxidation state of 0, and in Cu2+, its oxidation state is +2. Ag in Ag+ has an oxidation state of +1, while its oxidation state in its elemental form is 0.

Balance the atoms

The equation is already balanced with respect to the number of atoms on each side. Balanced equation: $\mathrm{Cu}(s)+\mathrm{Ag}^{+}(a q) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(a q)$

Identify the substance reduced and the substance oxidized

Since Cu goes from an oxidation state of 0 to +2, it is oxidized. Ag in Ag+ goes from an oxidation state of +1 to 0, so it is reduced.

Key concepts

Determining Oxidation States

Understanding oxidation states is crucial for mastering redox reactions. An element's oxidation state signifies its electronic status in a compound and is crucial for identifying which atoms undergo oxidation or reduction. To determine the oxidation state, we follow a set of rules: free elements have an oxidation state of 0, the sum of oxidation states in a neutral compound must be zero, and in ions, it should equal the charge of the ion.

For instance, in \(\mathrm{Al}(s)+3\mathrm{HCl}(aq) \rightarrow \mathrm{AlCl}_{3}(aq)+1.5\mathrm{H}_{2}(g)\), aluminum (Al) in its elemental form has an oxidation state of 0, while after reacting, in \(\mathrm{AlCl}_{3}\), it has an oxidation state of +3. This shift indicates a loss of electrons, meaning Al is oxidized.

Identifying Oxidized and Reduced Substances

The core of redox reactions is electron transfer, with oxidation involving a loss and reduction involving a gain of electrons. To identify the substances that have been oxidized or reduced, we compare their oxidation states before and after the reaction.

In \(\mathrm{Cu}(s)+\mathrm{Ag}^{+}(aq) \rightarrow \mathrm{Ag}(s)+\mathrm{Cu}^{2+}(aq)\), copper (Cu) changes from an oxidation state of 0 to +2, losing electrons and thus is oxidized. Conversely, silver (Ag), which initially has an oxidation state of +1 and changes to 0 upon receiving electrons, is the substance reduced. Recognizing these changes enables us to track the flow of electrons during the reaction.

Stoichiometry of Redox Reactions

Stoichiometry in redox reactions involves correctly balancing the atoms and charge to reflect the conservation of mass and charge. While balancing a redox reaction, equal numbers of electrons must be exchanged between the oxidized and reduced species. When balancing the stoichiometry of the reaction \(C_{3}H_{8}(g)+5O_{2}(g) \rightarrow 3CO_{2}(g)+4H_{2}O(l)\), we ensure that the atoms of carbon and hydrogen from propane (\(C_{3}H_{8}\)) are equal to those in carbon dioxide (\(CO_{2}\)) and water (\(H_{2}O\)), respectively.

Furthermore, the total charges before and after the reaction should be consistent—a key factor when electrons are transferred, as seen in ionic redox reactions.

Chemical Equations Balancing

Balancing chemical equations is a fundamental skill in chemistry. It ensures that the Law of Conservation of Mass is adhered to, with no atoms disappearing or appearing spontaneously. For redox reactions, this also means balancing the electrons gained and lost.

For the reaction \(\mathrm{CH}_{4}(g)+2\mathrm{S}(s) \rightarrow \mathrm{CS}_{2}(l)+2\mathrm{H}_{2}S(g)\), the stoichiometric coefficients of methane (\(\mathrm{CH}_{4}\)) and sulfur (S) are adjusted to balance carbon and sulfur atoms, while hydrogen is balanced by altering the coefficient of hydrogen sulfide (\(\mathrm{H}_{2}S\)). This balanced equation depicts a clear stoichiometric relation between reactants and products, maintaining the conservation of both mass and charge.