Question

Balance the following equations for reactions occurring in an acidic solution. (a) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+\mathrm{HNO}_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{NO}\) (b) \(\mathrm{HNO}_{2}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{NO}_{3}^{-}\) (c) \(\mathrm{H}_{3} \mathrm{PO}_{2}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} \longrightarrow \mathrm{H}_{3} \mathrm{PO}_{4}+\mathrm{Cr}^{3+}\) (d) \(\mathrm{XeF}_{2}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Xe}+\mathrm{F}^{-}+\mathrm{Cl}_{2}\)

Short answer

The balanced equations in an acidic solution are: (a) \(\text{H}_2\text{C}_2\text{O}_4 + \text{HNO}_2 \longrightarrow 2 \text{CO}_2 + \text{NO} + 2\text{H}_2\text{O}\), (b) \(\text{HNO}_2 + \text{MnO}_4^- + 4\text{H}_2\text{O} \longrightarrow \text{Mn}^{2+} + 3\text{NO}_3^- + 8\text{H}^+ + 5\text{e}^-\), (c) \(\text{H}_3\text{PO}_2 + \text{Cr}_2\text{O}_7^{2-} + 10\text{H}^+ + 6\text{e}^- \longrightarrow \text{H}_3\text{PO}_4 + 2\text{Cr}^{3+} + 5\text{H}_2\text{O}\), (d) \(\text{XeF}_2 + 2\text{Cl}^- \longrightarrow \text{Xe} + 2\text{F}^- + \text{Cl}_2 + 2\text{e}^-\).

Step-by-step solution

Balance (a) for Non-Hydrogen and Non-Oxygen Elements

Count atoms of non-hydrogen and non-oxygen elements on both sides. In reaction (a), there is 1 nitrogen atom on both sides. For carbon, we have 2 on the left and 1 on the right, so we put a coefficient of 2 in front of CO2. The updated equation is: \(\text{H}_2\text{C}_2\text{O}_4 + \text{HNO}_2 \longrightarrow 2 \text{CO}_2 + \text{NO}\).

Balance Oxygen Atoms in (a)

Now balance the oxygen atoms. The left side has 6 oxygens (4 from oxalic acid and 2 from nitrous acid), and the right side has 5 (4 from CO2 and 1 from NO). To balance, add 2 water molecules on the right side. The equation now is: \(\text{H}_2\text{C}_2\text{O}_4 + \text{HNO}_2 \longrightarrow 2 \text{CO}_2 + \text{NO} + 2\text{H}_2\text{O}\).

Balance Hydrogen Atoms in (a)

Finally, balance the hydrogen atoms. The left side has 4 hydrogen atoms (2 from oxalic acid), and the right side has now 4 (2 from NO and 4 from water). The hydrogens are balanced, so no further changes are needed.

Balance Charge in (a)

Since this is an acidic solution, we can add H+ ions to balance the charge. The left side currently has a charge of 0, and the right side has a charge of 0 as well. Therefore, no additional H+ ions are needed, and the equation is already balanced.

Balance (b) for Non-Hydrogen and Non-Oxygen Elements

First, balance the manganese atoms. There is 1 Mn on both sides of the equation. Next, balance nitrogen by putting a coefficient of 3 in front of NO3-. The updated equation is: \(\text{HNO}_2 + \text{MnO}_4^- \longrightarrow \text{Mn}^{2+} + 3\text{NO}_3^-\).

Balance the Oxygen Atoms in (b)

The left side has 5 oxygens (2 from nitrous acid and 4 from permanganate minus 1 because of the charge) and the right side has 9 (from the nitrate ions). We can add 4 water molecules to the left side to balance this. The equation now is: \(\text{HNO}_2 + \text{MnO}_4^- + 4\text{H}_2\text{O} \longrightarrow \text{Mn}^{2+} + 3\text{NO}_3^-\).

Balance Hydrogen Atoms in (b)

The left side now has 11 hydrogen atoms and the right side has none. We add 8 H+ ions to the right side to balance hydrogens. Our balanced equation is: \(\text{HNO}_2 + \text{MnO}_4^- + 4\text{H}_2\text{O} \longrightarrow \text{Mn}^{2+} + 3\text{NO}_3^- + 8\text{H}^+\).

Balance Charge in (b)

Charge on the left is +1 (from HNO2) and charge on the right is +7 (from Mn2+ and 8H+). To balance charges, we need 5 electrons. Add 5 e- on the right to reach the charge balance: \(\text{HNO}_2 + \text{MnO}_4^- + 4\text{H}_2\text{O} \longrightarrow \text{Mn}^{2+} + 3\text{NO}_3^- + 8\text{H}^+ + 5\text{e}^-\).

Balance (c) for Non-Hydrogen and Non-Oxygen Elements

Start with phosphorus, one on both sides, so it's balanced. Chromium has 2 on the left and we need 2 on the right. The partially balanced equation is: \(\text{H}_3\text{PO}_2 + \text{Cr}_2\text{O}_7^{2-} \longrightarrow \text{H}_3\text{PO}_4 + 2\text{Cr}^{3+}\).

Balance Oxygen Atoms in (c)

The left side has 9 oxygens (2 from phosphorous acid and 7 from dichromate), and the right side has 4 (from phosphoric acid). To balance, add 5 water molecules to the right side. Now we have: \(\text{H}_3\text{PO}_2 + \text{Cr}_2\text{O}_7^{2-} \longrightarrow \text{H}_3\text{PO}_4 + 2\text{Cr}^{3+} + 5\text{H}_2\text{O}\).

Balance Hydrogen Atoms in (c)

The left side has 3 hydrogens (from phosphorous acid), and the right side now has 13 (from phosphoric acid and water). Add 10 H+ ions to the left to balance hydrogens: \(\text{H}_3\text{PO}_2 + \text{Cr}_2\text{O}_7^{2-} + 10\text{H}^+ \longrightarrow \text{H}_3\text{PO}_4 + 2\text{Cr}^{3+} + 5\text{H}_2\text{O}\).

Balance Charge in (c)

Charge on the left is +8 (10 H+ minus 2 from dichromate) while on the right it's +6 (from 2 Cr3+). Six electrons are needed on the left to balance the charges: \(\text{H}_3\text{PO}_2 + \text{Cr}_2\text{O}_7^{2-} + 10\text{H}^+ + 6\text{e}^- \longrightarrow \text{H}_3\text{PO}_4 + 2\text{Cr}^{3+} + 5\text{H}_2\text{O}\).

Balance (d) for Xenon and Fluorine

First, balance xenon (Xe), which is already balanced. For fluorine, we have 2 on the left and 1 on the right, so we add another F- to the right side to get \(\text{XeF}_2 + 2\text{Cl}^- \longrightarrow \text{Xe} + 2\text{F}^- + \text{Cl}_2\).

Balance Chlorine Atoms in (d)

The left side has 2 chlorine atoms and the right side has 2 as well (from Cl2). The chlorine atoms are balanced with no further changes needed.

Balance Charge in (d)

The left side has a charge of -2 (from 2 Cl-), while the right side has a charge of 0. To balance the charges, add 2 electrons to the right side: \(\text{XeF}_2 + 2\text{Cl}^- \longrightarrow \text{Xe} + 2\text{F}^- + \text{Cl}_2 + 2\text{e}^-\).

Key concepts

Acidic Solution Reactions

Understanding reactions in acid solutions is crucial for mastering redox chemistry. Acidic solution reactions involve the transfer of electrons between substances, which change their oxidation states. When substances react in an acidic environment, H+ ions are present and often participate in the redox process.

These reactions require special attention because H+ ions can be added or removed to balance the hydrogen atoms in the equation. As such, it is common to see H2O or H+ appear in the balanced chemical equations for these reactions when they are in acidic solutions.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical equation. It is akin to a recipe that outlines how much of each ingredient you need to create a certain amount of product.

Balancing redox reactions, particularly in acidic solutions, requires careful stoichiometric calculations to ensure that the number of atoms of each element and the total charge is the same on both sides of the equation. This ensures that matter and charge are conserved, adhering to the principles of stoichiometry.

Oxidation States

Oxidation states, also known as oxidation numbers, are a measure of the degree of oxidation of an atom in a substance. They are very helpful for balancing redox reactions because they allow us to determine which atoms are gaining electrons (being reduced) and which atoms are losing electrons (being oxidized).

In acidic solution reactions, H+ does not alter the oxidation state of an individual element but is a balancing factor for hydrogen atoms, impacting the overall stoichiometry of the equation.

Chemical Equation Balancing

Balancing chemical equations ensures that the law of conservation of mass is respected in every chemical reaction. To balance a chemical equation, start by balancing atoms of elements that appear only once on each side. Then, proceed to balance atoms of more complex elements.

In redox reactions, after balancing the atoms, it's important to also balance the overall charge on each side of the equation. This is often done by adding electrons (e-) to one side of the equation.