Question

Nitric acid is a very strong acid, but is also a very strong oxidizing agent, and generally behaves as the latter. It will dissolve many metals. Balance the following oxidation-reduction reactions of nitric acid. a. \(\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}_{2}(g)\) b. \(\mathrm{Mg}(s)+\mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{H}_{2}(g)\)

Short answer

The balanced equations for the given reactions are: a. \(2 \mathrm{Cu}(s) + 8 \mathrm{HNO}_{3}(a q) \rightarrow 2 \mathrm{Cu}^{2+}(a q) + 4\mathrm{NO}_{2}(g) + 4 \mathrm{H}_{2}\mathrm{O}(l)\) b. \(\mathrm{Mg}(s) + 2 \mathrm{HNO}_{3}(a q) \rightarrow \mathrm{Mg}^{2+}(a q) + \mathrm{H}_{2}(g) + 2 \mathrm{NO}_{3}^-(a q)\)

Step-by-step solution

Assign oxidation states

To begin, determine the oxidation state of each element in the reactants and products of the given chemical equations. For reaction a: Cu(s): 0 HNO₃(aq): H(+1) + N(+5) + 3×O(-2) Cu²⁺(aq): +2 NO₂(g): N(+4) + 2×O(-2) For reaction b: Mg(s): 0 HNO₃(aq): H(+1) + N(+5) + 3×O(-2) Mg²⁺(aq): +2 H₂(g): 2×H(+1)

Identify oxidation and reduction processes

Now, look for the elements that change their oxidation states from reactants to products. For reaction a: Cu is oxidized from 0 to +2. N is reduced from +5 to +4. For reaction b: Mg is oxidized from 0 to +2. H is reduced from +1 to 0.

Balance the chemical equations

Balance the chemical equations by ensuring an equal number of atoms on both sides of the reactions. a. Balancing Cu and N: 2 Cu(s) + 8 HNO₃(aq) → 2 Cu²⁺(aq) + 4 NO₂(g) + 8 H⁺(aq) Now balance O and H by adding water and hydrogen ions: 2 Cu(s) + 8 HNO₃(aq) → 2 Cu²⁺(aq) + 4 NO₂(g) + 8 H⁺(aq) + 4 H₂O(l) The balanced equation for reaction a is: 2 Cu(s) + 8 HNO₃(aq) → 2 Cu²⁺(aq) + 4 NO₂(g) + 4 H₂O(l) b. Balancing Mg and H: Mg(s) + 2 HNO₃(aq) → Mg²⁺(aq) + H₂(g) + 2 NO₃⁻(aq) Now balance O and N by adding nitrate ions: Mg(s) + 2 HNO₃(aq) → Mg²⁺(aq) + H₂(g) + 2 NO₃⁻(aq) The balanced equation for reaction b is: Mg(s) + 2 HNO₃(aq) → Mg²⁺(aq) + H₂(g) + 2 NO₃⁻(aq)

Key concepts

Oxidation States

Understanding oxidation states is fundamental when studying redox (reduction-oxidation) reactions. An oxidation state, or oxidation number, is a theoretical charge that an atom would have if the compound were purely ionic. By following specific rules, oxidation states are assigned to each atom in a chemical species.

The most important rules include that the sum of oxidation states for all atoms in a neutral molecule must equal zero. For a polyatomic ion, the sum must equal the charge on the ion. Metals have positive oxidation numbers, and non-metals vary depending on the compound. Hydrogen is typically +1 (except when bonded to metals where it's -1), and oxygen is usually -2 (except in peroxides).

Using these guidelines, assigning oxidation states to atoms in the provided exercise helps us identify what is being oxidized and what is being reduced. Once you get familiar with these rules and practice on several examples, determining oxidation states becomes significantly easier and very intuitive.

Oxidation and Reduction Processes

The terms 'oxidation' and 'reduction' describe the transfer of electrons between reactants in a chemical reaction. Oxidation refers to the loss of electrons by a molecule, atom, or ion, which increases its oxidation state indicating that it becomes more positive. Meanwhile, reduction refers to the gain of electrons, leading to a decrease in the oxidation state.

In other words, whenever an atom undergoes oxidation, another atom must be getting reduced at the same time, since the electrons are not just destroyed but are transferred. These processes are what make up a redox reaction. In the exercise, copper (Cu) and magnesium (Mg) are oxidized as their oxidation states increase, and nitrogen (N) in nitric acid and hydrogen (H) are reduced. Remember the mnemonic 'OIL RIG': Oxidation Is Loss, Reduction Is Gain, to keep track of what process is occurring. Proper identification of these processes is the key to successfully balancing redox reactions.

Chemical Equation Balancing

Balancing chemical equations is essential to satisfy the law of conservation of matter, which states that matter cannot be created or destroyed in an isolated system. Therefore, we must have the same number of each type of atom on both sides of a chemical equation.

The balancing of equations involves adjusting the coefficients, which are the numbers placed in front of compounds or elements in a reaction. We do not change the subscripts because changing subscripts would change the substances involved in the reaction. In a redox reaction, it is crucial to balance the elements that undergo oxidation and reduction first, then balance the remaining atoms.

With the exercise given, once we have accounted for the changes in oxidation states, we adjust coefficients to ensure that atoms are conserved, and finally, we add water (H2O) and hydrogen ions (H+) if necessary to balance the hydrogen and oxygen atoms. Getting comfortable with balancing equations is a process that initially might require a systematic approach but becomes quicker with practice.