Question

Balance each of the following oxidation-reduction reactions, which take place in acidic solution, by using the "half-reaction" method. a. \(\mathrm{I}^{-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{I}_{2}(a q)+\mathrm{Mn}^{2+}(a q)\) b. \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(a q)+\mathrm{Cr}^{3+}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)\) c. \(\mathrm{BiO}_{3}^{-}(a q)+\mathrm{Mn}^{2+}(a q) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}^{-}(a q)\)

Short answer

a. \(2\mathrm{I}^{-}(a q)+16\mathrm{H}^{+}(a q)+2\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{I}_{2}(a q)+2\mathrm{Mn}^{2+}(a q)+8\mathrm{H}_{2}\mathrm{O}(l)\) b. \(2\mathrm{S}_{2} \mathrm{O}_{8}^{2-}(a q)+2\mathrm{Cr}^{3+}(a q)+14\mathrm{H}^{+}(aq) \rightarrow 3\mathrm{SO}_{4}^{2-}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+7\mathrm{H}_{2}\mathrm{O}(l)\) c. \(2\mathrm{BiO}_{3}^{-}(a q)+10\mathrm{H}^{+}(a q)+2\mathrm{Mn}^{2+}(a q) \rightarrow 2\mathrm{Bi}^{3+}(a q)+2\mathrm{MnO}_{4}^{-}(a q)+3\mathrm{H}_{2}\mathrm{O}(l)\)

Step-by-step solution

Identify and Write the Half Reactions

By identifying which atoms undergo oxidation and reduction, the half reactions can be written as: \(I^{-}\rightarrow I_{2}\) (oxidation) and \(MnO_{4}^{-}\rightarrow Mn^{2+}\) (reduction).

Balance Atoms

In the oxidation half reaction, balance the I atoms by placing a coefficient of 2 in front of \(I^{-}\) to get: \(2I^{-}\rightarrow I_{2}\). In the reduction half reaction, balance the Mn atoms, which are already balanced. So, \(MnO_{4}^{-}\rightarrow Mn^{2+}\) remains the same.

Balance Charges

In the oxidation half reaction, since the net charge on the left is -2 and on the right is 0, 2 electrons are added to the right to balance the charge: \(2I^{-}\rightarrow I_{2} + 2e^{-}\). In the reduction half reaction, 5 water molecules are added to the right, and 8 H+ ions to the left side to balance the O atoms. Since the net charge is still -1 on the left and +8 on the right, add 7 electrons to the left to balance the charge: \(MnO_{4}^{-} + 8H^+ + 5e^{-} \rightarrow Mn^{2+} + 4H_2O\).

Equalize the electrons and Add the Half Reactions

Multiply the oxidation and reduction half reactions by suitable factors such that the number of electrons in the two half reactions are equal. Here, multiply the oxidation half reaction by 5 and the reduction half reaction by 2. Then add them up to get the final balanced redox reaction: \(2I^- + 16H^+ + 2MnO4^- \rightarrow I_{2} + 2Mn^{2+} + 8H_{2}O\). Proceed similarly for the reactions b and c.

Key concepts

Half-Reaction Method

The half-reaction method is a systematic way of balancing oxidation-reduction reactions. It breaks down a redox equation into two separate equations: one for oxidation and one for reduction. Each half-reaction is balanced independently before they are combined to form the complete balanced equation. This method is particularly useful for reactions occurring in aqueous solutions, including those in acidic or basic mediums.

To begin, you need to identify the species that are oxidized and reduced. Oxidation involves the loss of electrons, whereas reduction involves the gain of electrons. Once you identify these species, write the oxidation half-reaction and the reduction half-reaction separately.

The purpose of using the half-reaction method is to make it easier to balance the transfer of electrons. Electrons lost must equal electrons gained to ensure the charge is balanced. This method provides a clear path to handling complex redox reactions that cannot be balanced by inspection alone.

Balancing Chemical Equations

Balancing chemical equations is crucial in chemistry as it ensures that matter is conserved during a reaction. For a reaction to be balanced, the number of atoms of each element must be equal on both sides of the equation.

In oxidation-reduction reactions, balancing involves not just the atoms but also charges. Begin by balancing the main elements of interest, such as in your half-reactions. Then, balance the oxygen atoms by adding water ( H_2O ) molecules, and balance hydrogen atoms by adding hydrogen ions ( H^+ ) if working in an acidic solution.

Moreover, balance the charges by adding electrons (e⁻). Remember, the number of electrons lost in oxidation must equal the number of electrons gained in reduction. This is achieved by multiplying the half-reactions by suitable integers to equalize the electron transfer between oxidation and reduction processes.

Acidic Solution Reactions

Oxidation-reduction reactions in acidic solutions present a unique challenge due to the presence of H^+ ions, which play a crucial role in balancing the equation. The acidic conditions mean that we can readily use these H^+ ions to balance the hydrogen atoms without introducing any external species that don't belong in the acidic medium.

When balancing redox reactions in an acidic solution, once the main atoms are balanced, you are allowed to introduce H^+ ions to balance hydrogen. Similarly, use water molecules ( H_2O ) to balance oxygen atoms. This guided approach, capitalizing on the acidic environment, simplifies the balancing process greatly compared to if you were balancing without specifying the solution condition.

Always ensure that the stoichiometry of the hydrogen and oxygen atoms conforms to the acidic conditions. This ultimately leads to precision and accuracy in balancing complex chemical equations, paving the way to understanding and predicting chemical behaviors in various environments.