Question

Calcium oxalate is one of the minerals found in kidney stones. If a strong acid is added to calcium oxalate, the compound will dissolve and the oxalate ion will be changed to oxalic acid (a weak acid). Oxalate ion is a moderately strong reducing agent. Write a balanced net ionic equation for the oxidation of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) by \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) in an acidic solution. The reaction yields \(\mathrm{Cr}^{3+}\) and \(\mathrm{CO}_{2}\) among the products.

Short answer

The balanced net ionic equation for the oxidation of oxalic acid by dichromate in acidic solution is \(3\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{CO}_{2} + 7\mathrm{H}_{2}\mathrm{O}\).

Step-by-step solution

Write the unbalanced half-reactions

Oxidation half-reaction: \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow \mathrm{CO}_{2}\) \ewline Reduction half-reaction: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} \rightarrow \mathrm{Cr}^{3+}\)

Balance the atoms other than H and O

Oxidation half-reaction: \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow 2\mathrm{CO}_{2}\) \ewline Reduction half-reaction stays unchanged as Cr and O are already balanced.

Balance the oxygen atoms with water

Oxidation half-reaction needs no additional balancing for oxygen. \ewline Reduction half-reaction: Add 7 water molecules to the right: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\)

Balance the hydrogen atoms with H+ ions

Oxidation half-reaction: Add 4 H+ to the right: \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow 2\mathrm{CO}_{2} + 4\mathrm{H}^{+}\) \ewline Reduction half-reaction is already balanced for hydrogen.

Balance the charges by adding electrons

Oxidation half-reaction: Add 2 electrons to the right to balance the charge: \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow 2\mathrm{CO}_{2} + 4\mathrm{H}^{+} + 2\mathrm{e}^{-}\) \ewline Reduction half-reaction: Add 6 electrons to the left: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_{2}\mathrm{O}\)

Multiply the half-reactions to cancel out the electrons

Oxidation half-reaction is multiplied by 3: \(3\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} \rightarrow 6\mathrm{CO}_{2} + 12\mathrm{H}^{+} + 6\mathrm{e}^{-}\) \ewline Reduction half-reaction is multiplied by 1 (no change needed as 6 electrons are already there).

Add the half-reactions

Combine the half-reactions, making sure to add the reactants together and the products together, cancelling out terms that appear on both sides of the equation.

Verify the final equation

Check that all the atoms and the charges are balanced. The final equation should be: \(3\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4} + \mathrm{Cr}_{2}\mathrm{O}_{7}^{2-} + 14\mathrm{H}^{+} \rightarrow 2\mathrm{Cr}^{3+} + 6\mathrm{CO}_{2} + 7\mathrm{H}_{2}\mathrm{O}\)

Key concepts

Oxidation-Reduction Reactions

Understanding oxidation-reduction reactions, often referred to as redox reactions, is foundational in chemistry. Redox reactions involve the transfer of electrons between two species. An easy way to remember this is 'OIL RIG': oxidation is loss, and reduction is gain of electrons. In these reactions, one substance gets oxidized by losing electrons, while another gets reduced by gaining electrons.

Let's apply this to the exercise: Oxalic acid, (H_{2}C_{2}O_{4}), reduces potassium dichromate (K_{2}Cr_{2}O_{7}) as it gets oxidized to carbon dioxide (CO_{2}), and (Cr_{2}O_{7}^{2-}) gets reduced to (Cr^{3+}). The ability to identify which half-reaction is oxidation and which is reduction is crucial in writing balanced net ionic equations.

Balancing Chemical Equations

Balancing chemical equations ensures the principle of conservation of mass is upheld. This means the number of atoms for each element should be equal on both the reactant and the product sides of the equation. In the context of redox reactions, careful attention is also given to the balancing of charge to ensure that the electrons lost in the oxidation half-reaction equal the electrons gained in the reduction half-reaction.

In our exercise, this balancing act involves adjusting coefficients and adding (H^{+}) ions and water molecules to balance hydrogen and oxygen atoms, respectively. The final step involves making sure that the total number of electrons lost equals the number of electrons gained, achieving charge balance alongside mass balance.

Chemical Stoichiometry

Chemical stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction. Stoichiometry is based on the balanced chemical equation and uses mole ratios to convert between masses or volumes of reactants and products.

In our example, stoichiometry would come into play after balancing the net ionic equation, to predict how much oxalic acid is needed to reduce a specified amount of potassium dichromate. Through stoichiometry, the importance of a balanced equation becomes even more apparent, as it provides the mole ratios necessary to carry out these calculations.