Question

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}$$ (b) If a \(1.00-\mathrm{g}\) sample of plant matter requires \(24.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?

Short answer

The balanced equation is \[2\mathrm{MnO}_4^- + 16 \mathrm{H}^+ + 5 \mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}\] and the percent by mass of oxalic acid is 5.40%.

Step-by-step solution

Determine Oxidation States

Identify the oxidation states of each element in the reactants and products. For \(\mathrm{MnO}_4^-\), \(\mathrm{Mn}\) has an oxidation state of +7. For \(\mathrm{C}_2\mathrm{O}_4^{2-}\), \(\mathrm{C}\) has an oxidation state of +3. In the products, \(\mathrm{Mn}^{2+}\) has an oxidation state of +2, and \(\mathrm{CO}_2\) has carbon with an oxidation state of +4.

Identify Redox Half-Reactions

The reduction half-reaction involves \(\mathrm{MnO}_4^-\) being reduced to \(\mathrm{Mn}^{2+}\) and the oxidation half-reaction involves \(\mathrm{C}_2\mathrm{O}_4^{2-}\) being oxidized to \(\mathrm{CO}_2\).

Balance Oxidation Half-Reaction

For the half-reaction \(\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow \mathrm{CO}_2\), balance the carbon and charge: \(\mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{CO}_2 + 2e^-\).

Balance Reduction Half-Reaction

For the half-reaction \(\mathrm{MnO}_4^- \rightarrow \mathrm{Mn}^{2+}\), balance oxygen with water and charge with \(\mathrm{H}^+\): \(\mathrm{MnO}_4^- + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}\).

Equalize Electrons and Combine Half-Reactions

Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to equalize electrons (10 electrons exchanged). Combine them: \[2\mathrm{MnO}_4^- + 16 \mathrm{H}^+ + 5 \mathrm{C}_2\mathrm{O}_4^{2-} \rightarrow 2\mathrm{Mn}^{2+} + 10\mathrm{CO}_2 + 8\mathrm{H}_2\mathrm{O}\]

Calculate Moles of KMnO4 Used

Use the volume and molarity of \(\mathrm{KMnO}_4\) to find moles. \(\text{Moles of } \mathrm{KMnO}_4 = 0.0100 \mathrm{M} \times 0.0240 \mathrm{L} = 2.40 \times 10^{-4} \text{ moles}\) which is also moles of \(\mathrm{MnO}_4^-\) used.

Calculate Moles of Oxalic Acid

According to the balanced chemical equation, 2 moles of \(\mathrm{MnO}_4^-\) react with 5 moles of \(\mathrm{C}_2\mathrm{O}_4^{2-}\). Use the stoichiometry to find: \(\text{Moles of } \mathrm{C}_2\mathrm{O}_4^{2-} = \left(\frac{5}{2}\right) \times 2.40 \times 10^{-4} = 6.00 \times 10^{-4} \text{ moles}\).

Calculate Mass of Oxalic Acid

Convert moles of \(\mathrm{C}_2\mathrm{O}_4^{2-}\) to grams using its molar mass (90.03 g/mol): \(\text{Mass of } \mathrm{H}_2\mathrm{C}_2\mathrm{O}_4 = 6.00 \times 10^{-4} \text{ moles} \times 90.03 \frac{\text{g}}{\text{mol}} = 0.0540 \text{ g}\).

Calculate Percent by Mass of Oxalic Acid

Find the percent by mass of \(\mathrm{H}_2\mathrm{C}_2\mathrm{O}_4\) in the plant sample: \(\text{Percent by mass} = \left(\frac{0.0540 \text{ g}}{1.00 \text{ g}}\right) \times 100\% = 5.40\%\).

Key concepts

Oxalic Acid

Oxalic acid, with the chemical formula \( \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \), is an organic compound that naturally occurs in several plants and vegetables.
It's known for its role in various metabolic processes within plant structures.
Oxalic acid appears as a white solid and can behave as a diacid, meaning it has two acidic hydrogens.
In the context of redox reactions, it acts as a reducing agent.
Reducing agents are substances that donate electrons in chemical reactions. The role of oxalic acid in redox reactions is significant due to its tendency to donate electrons and convert into carbon dioxide \( \mathrm{CO}_{2} \).
Understanding this property of oxalic acid helps in balancing redox reactions, as will be explored in the next section on balancing chemical equations.
This conversion is exemplified in the exercise where oxalic acid undergoes oxidation, a key process in understanding how it interacts with other chemical species.

Balancing Chemical Equations

Balancing chemical equations is essential to ensuring the law of conservation of mass is obeyed.
In these equations, the same number of each type of atom must exist on both sides of the reaction.
When dealing with redox reactions, balancing becomes more complex due to the transfer of electrons.Redox reactions involve two half-reactions: oxidation and reduction.
Oxidation involves the loss of electrons while reduction involves the gain of electrons.
To balance these, we often use the half-reaction method, which separates the oxidation and reduction processes.
Once separated, you can balance elements besides \( \mathrm{O} \) and \( \mathrm{H} \), add water molecules to balance oxygen, and use hydrogen ions (\( \mathrm{H}^+ \)) for balancing hydrogen.
After balancing charges with electrons, these half-reactions are combined to present a full, balanced redox equation.
In acidic solutions, like the equation from the exercise, \( \mathrm{H}^+ \) ions are used extensively for balancing.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction.
Understanding stoichiometry is crucial for determining how much of one substance will react with a given quantity of another.
This involves the mole ratio, which is derived from the coefficients in a balanced chemical equation.In the given redox reaction involving oxalic acid and permanganate ions (\( \mathrm{MnO}_4^- \)), stoichiometry plays a key role.
The reaction shows that 2 moles of permanganate ions react with 5 moles of oxalate ions.
This ratio helps in calculating the amount of oxalic acid present in the sample from the permanganate solution used for the reaction.
Performing such calculations accurately allows chemists to understand the concentration and mass of compounds in a given sample.

Oxidation States

Oxidation states, or oxidation numbers, provide insight into the electron transfer in a redox reaction.
They reflect the degree of oxidation of an atom, signifying either an increase (oxidation) or decrease (reduction) in oxidation state.To identify oxidation states in a compound, follow these rules:

• The oxidation state for any pure element is zero.
• For monoatomic ions, the oxidation state equals the ion's charge.
• In compounds, hydrogen is usually +1, and oxygen is usually -2.
• The sum of oxidation states in a neutral compound is zero.
• For polyatomic ions, the sum equals the ion charge.

In the exercise, manganese in \( \mathrm{MnO}_4^- \) has an oxidation state of +7 and is reduced to +2 in \( \mathrm{Mn}^{2+} \).
Carbon in \( \mathrm{C}_2\mathrm{O}_4^{2-} \) starts at +3 and oxidizes to +4 in \( \mathrm{CO}_2 \).
Writing and balancing equations with these states provides a clear understanding of the shifts and helps in the portrayal of electron flow in reactions.