Question

Balance each redox reaction occurring in acidic aqueous solution. $$\begin{array}{l}{\text { a. } \mathrm{I}^{-}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{I}_{2}(s)+\mathrm{NO}(g)} \\ {\text { b. } \mathrm{ClO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\mathrm{Cl}_{2}(g)} \\ {\text { c. } \mathrm{NO}_{3}^{-}(a q)+\mathrm{Sn}^{2+}(a q) \rightarrow \mathrm{Sn}^{4+}(a q)+\mathrm{NO}(g)}\end{array}$$

Short answer

a. Balancing I- and NO2- yields \(2 \mathrm{I}^-(aq) + 2 \mathrm{NO}_2^-(aq) + 4\mathrm{H}^+ \rightarrow \mathrm{I}_2(s) + 2 \mathrm{NO}(g) + 2\mathrm{H}_2\mathrm{O}(l)\). b. Balancing ClO4- and Cl- yields \(\mathrm{ClO}_4^-(aq) + 8\mathrm{H}^++ 8\mathrm{e}^- + \mathrm{Cl}^-(aq) \rightarrow \mathrm{ClO}_3^-(aq) + \mathrm{Cl}_2(g) + 4\mathrm{H}_2\mathrm{O}(l)\). c. Balancing NO3- and Sn2+ yields \(\mathrm{NO}_3^-(aq) + 4\mathrm{H}^+ + 3\mathrm{e}^- + \mathrm{Sn}^{2+}(aq) \rightarrow \mathrm{Sn}^{4+}(aq) + \mathrm{NO}(g) + 2\mathrm{H}_2\mathrm{O}(l)\).

Step-by-step solution

Separate into Half-Reactions

Split each reaction into two half-reactions, one for oxidation and one for reduction. For example, in reaction a, write the I- oxidation and NO2- reduction reactions separately.

Balance Atoms Other Than O and H

Balance all atoms except for oxygen and hydrogen in each half-reaction. For reaction a, ensure that the number of I and N atoms is the same on both sides of each half-reaction.

Balance Oxygen by Adding H2O

For each half-reaction, if needed, balance oxygen atoms by adding H2O molecules. For the reactions in question b and c, balance the oxygens now.

Balance Hydrogen by Adding H+

For each half-reaction, balance the hydrogen atoms by adding H+ ions. Add these H+ ions to the side that needs hydrogen in each half-reaction.

Balance Charge by Adding Electrons

Balance the charges in each half-reaction by adding electrons (e-). Add electrons to the more positive side of each half-reaction to equalize the charges on both sides.

Make Electron Transfer Equal

Multiply the half-reactions by appropriate coefficients so that the number of electrons gained in the reduction half-reaction equals the number lost in the oxidation half-reaction.

Combine the Half-Reactions

Add the two half-reactions together, canceling out electrons and any other species that appear on both sides, such as H2O or H+ if they are the same number on opposite sides.

Confirm Mass and Charge Balance

Check that both the mass and charge are balanced in the final equation for all reactions. This confirms that the law of conservation of mass and charge is satisfied.

Key concepts

Half-Reactions in Redox Reactions

Understanding half-reactions is crucial when balancing redox reactions, especially in an acidic aqueous solution. A redox reaction is composed of two half-reactions: oxidation and reduction. In the process of oxidation, an element loses electrons, whereas in reduction, an element gains electrons. To clearly see these electron shifts, we split the overall chemical reaction into its respective half-reactions.

Let's take the reaction \(\mathrm{I}^{-}(aq) + \mathrm{NO}_{2}^{-}(aq) \longrightarrow \mathrm{I}_{2}(s) + \mathrm{NO}(g)\) as an example. The half-reaction for oxidation may be written as \(\mathrm{I}^{-} \longrightarrow \mathrm{I}_{2}\), which indicates iodine atoms going from a -1 to 0 oxidation state. Meanwhile, the reduction half-reaction is \(\mathrm{NO}_{2}^{-} \longrightarrow \mathrm{NO}\), where the nitrogen atom reduces from a +3 to a +2 oxidation state.

The separation into half-reactions is a powerful tool because it simplifies balancing the overall redox reaction. It allows us to balance the atoms and charges involved separately, making it more manageable to ensure the law of conservation of mass and charge are satisfied.

Oxidation and Reduction Balancing

Once half-reactions have been written, it's important to balance them in terms of both mass and charge. This is not always straightforward as it requires taking into account the different environments the reaction might occur in, such as an acidic solution in this case. The key to this balancing act is a systematic approach that involves several steps.

Initially, we balance all atoms other than oxygen and hydrogen. After that, we then balance oxygen atoms by adding water \(\mathrm{H}_{2}\mathrm{O}\) molecules to the appropriate side of the half-reaction. This is often required in chemical reactions involving oxyanions or oxyacids. Following this, hydrogen atoms are balanced by adding hydrogen ions \(\mathrm{H}^{+}\) to the corresponding side of the half-reaction. Next, to balance the charges, we add electrons \(\mathrm{e}^{-}\) to one side of the half-reaction.

For our reaction \(\mathrm{ClO}_{4}^{-}(aq) + \mathrm{Cl}^{-}(aq) \longrightarrow \mathrm{ClO}_{3}^{-}(aq) + \mathrm{Cl}_{2}(g)\), you’d balance the chlorines first, add water to balance the oxygens, and then add hydrogen ions to balance the hydrogens. Finally, you’d add electrons to ensure the charges on both sides of the half-reaction are equal. The careful attention to detail in this step-wise method ensures that the balances of mass and charge are upheld.

Chemical Equations Balancing

After we've balanced the individual half-reactions, the next step is to combine them to form the complete balanced redox equation. To ensure the electrons cancel out, we may need to multiply the half-reactions by appropriate coefficients, so the number of electrons gained in reduction equals the number lost in oxidation.

In the example \(\mathrm{NO}_{3}^{-}(aq) + \mathrm{Sn}^{2+}(aq) \rightarrow \mathrm{Sn}^{4+}(aq) + \mathrm{NO}(g)\), if the oxidation half-reaction requires less electrons than the reduction half-reaction, then it must be multiplied by a suitable factor to balance the electron exchange.

Combining the half-reactions includes careful cancellation of species that appear on both sides, such as water molecules or hydrogen ions, which often are present in reactions in acidic solutions. The final check to confirm that the chemical equation is balanced correctly involves verifying that both mass and charge are balanced. This is a crucial step to ensure that the equation obeys conservation laws and is reflective of a reaction that could feasibly occur under the stipulated conditions. Balancing chemical equations is not just a procedural task – it’s a fundamental principle reflecting the underlying stoichiometry and physics of the reaction.