Question

Write balanced equations for the following reactions in acid solution. (a) \(\mathrm{Ni}^{2+}(a q)+\mathrm{IO}_{4}^{-}(a q) \longrightarrow \mathrm{Ni}^{3+}(a q)+\mathrm{I}^{-}(a q)\) (b) \(\mathrm{O}_{2}(g)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{Br}_{2}(l)\) (c) \(\mathrm{Ca}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{I}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

Short answer

Question: Balance the following redox reactions in acid solution: (a) \(Ni^{2+}(aq)+IO_4^-(aq) \rightarrow Ni^{3+}(aq)+I^-(aq)\) (b) \(O_2(g)+Br^-(aq) \rightarrow H_2O+Br_2(l)\) (c) \(Ca(s)+Cr_2O_7^{2-}(aq) \rightarrow Ca^{2+}(aq)+Cr^{3+}(aq)\) (d) \(IO_3^-(aq)+Mn^{2+}(aq) \rightarrow I^-(aq)+MnO_2(s)\) Answer: (a) \(4Ni^{2+}(aq) + IO_4^-(aq) \rightarrow 4Ni^{3+}(aq) + I^-(aq) + 12H_2O(l)\) (b) \(O_2(g) + 2Br^-(aq) + 4H^+(aq) \rightarrow 2H_2O(l) + Br_2(l)\) (c) \(Ca(s) + Cr_2O_7^{2-}(aq) + 14H^+(aq) \rightarrow Ca^{2+}(aq) + 2Cr^{3+}(aq) + 7H_2O(l)\) (d) \(5IO_3^-(aq) + 6Mn^{2+}(aq) + 30H^+(aq) \rightarrow 5I^-(aq) + 6MnO_2(s) + 30H_2O(l)\)

Step-by-step solution

Write half-reactions

For reduction: \(\mathrm{I}^{-}(\text (a q))\) For oxidation: \(\mathrm{Ni}^{2+}(a q)\)

Balance each half-reaction

For reduction: \(\mathrm{IO}_4^-(aq) \longrightarrow \mathrm{I}^-(aq) + 4e^-\) For oxidation: \(\mathrm{Ni}^{2+}(aq) + e^- \longrightarrow \mathrm{Ni}^{3+}(aq)\)

Combine half-reactions and check for balance

Multiplying the reduction reaction by 1 and oxidation reaction by 4 to equalize the electrons, we get: \(\mathrm{4Ni}^{2+}(aq) + \mathrm{IO}_4^-(aq) \longrightarrow 4\mathrm{Ni}^{3+}(aq) + \mathrm{I}^-(aq) + 12\mathrm{H}_2\mathrm{O}(l)\) (b) \(\mathrm{O}_{2}(g)+\mathrm{Br}^{-}(a q) \longrightarrow \mathrm{H}_{2}\mathrm{O}+\mathrm{Br}_{2}(l)\)

Write half-reactions

For reduction: \(\mathrm{O}_2(g)\) For oxidation: \(\mathrm{Br}^-(aq)\)

Balance each half-reaction

For reduction: \(\mathrm{O}_2(g) + 4\mathrm{H}^+(aq) + 4e^- \longrightarrow 2\mathrm{H}_2\mathrm{O}(l)\) For oxidation: \(2\mathrm{Br}^-(aq) \longrightarrow \mathrm{Br}_2(l) + 2e^-\)

Combine half-reactions and check for balance

Multiplying the reduction reaction by 1 and oxidation reaction by 2, we get: \(\mathrm{O}_2(g) + 2\mathrm{Br}^-(aq) + 4\mathrm{H}^+(aq) \longrightarrow 2\mathrm{H}_2\mathrm{O}(l) + \mathrm{Br}_2(l)\) (c) \(\mathrm{Ca}(s)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{Cr}^{3+}(a q)\)

Write half-reactions

For reduction: \(\mathrm{Cr}_2\mathrm{O}_7^{2-}(aq)\) For oxidation: \(\mathrm{Ca}(s)\)

Balance each half-reaction

For reduction: \(\mathrm{Cr}_2\mathrm{O}_7^{2-}(aq) + 14\mathrm{H}^+(aq) + 6e^- \longrightarrow 2\mathrm{Cr}^{3+}(aq) + 7\mathrm{H}_2\mathrm{O}(l)\) For oxidation: \(\mathrm{Ca}(s) \longrightarrow \mathrm{Ca}^{2+}(aq) + 2e^-\)

Combine half-reactions and check for balance

Multiplying the reduction reaction by 1 and oxidation reaction by 3, we get: \(\mathrm{Ca}(s) + \mathrm{Cr}_2\mathrm{O}_7^{2-}(aq) + 14\mathrm{H}^+(aq) \longrightarrow \mathrm{Ca}^{2+}(aq) + 2\mathrm{Cr}^{3+}(aq) + 7\mathrm{H}_2\mathrm{O}(l)\) (d) \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{I}^{-}(a q)+\mathrm{MnO}_{2}(s)\)

Write half-reactions

For reduction: \(\mathrm{IO}_3^-(aq)\) For oxidation: \(\mathrm{Mn}^{2+}(aq)\)

Balance each half-reaction

For reduction: \(\mathrm{IO}_3^-(aq) + 6\mathrm{H}^+(aq) + 5e^- \longrightarrow \mathrm{I}^-(aq) + 3\mathrm{H}_2\mathrm{O}(l)\) For oxidation: \(\mathrm{Mn}^{2+}(aq) + 4\mathrm{H}_2\mathrm{O}(l) \longrightarrow \mathrm{MnO}_2(s) + 8\mathrm{H}^+(aq) + 4e^-\)

Combine half-reactions and check for balance

Multiplying the reduction reaction by 4 and oxidation reaction by 5 to equalize the electrons, we get: \(\mathrm{5IO}_3^-(aq) + 6\mathrm{Mn}^{2+}(aq) + 30\mathrm{H}^+(aq) \longrightarrow 5\mathrm{I}^-(aq) + 6\mathrm{MnO}_2(s) + 30\mathrm{H}_2\mathrm{O}(l)\)

Key concepts

Oxidation Half-Reaction

In redox reactions, the oxidation half-reaction is where the loss of electrons occurs, signifying an increase in oxidation state. This process is crucial for balancing redox reactions because it's the counterpart to the reduction half-reaction. When identifying an oxidation half-reaction, look for the species that donates electrons. For example, in the given reaction \(\mathrm{Ni}^{2+}(aq) + e^- \longrightarrow \mathrm{Ni}^{3+}(aq)\), \(\mathrm{Ni}^{2+}\) loses an electron and becomes \(\mathrm{Ni}^{3+}\), thus undergoing oxidation.

It's important to ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction. When writing these out, always balance the atoms that change oxidation state first, followed by the oxygen atoms with water (in acidic or neutral solutions) or hydroxide ions (in basic solutions), then the hydrogen atoms with hydrogen ions or water, and finally, balance the charge with electrons. This provides a clear method to follow for balancing the equations accurately and efficiently.

Reduction Half-Reaction

Conversely, the reduction half-reaction involves the gain of electrons and a decrease in oxidation state. In the context of our redox reactions, a species is reduced when it acquires electrons. Taking the same set of reactions as an example, \(\mathrm{IO}_4^-\) is reduced to \(\mathrm{I}^-\) by gaining electrons: \(\mathrm{IO}_4^-(aq) \longrightarrow \mathrm{I}^-(aq) + 4e^-\).

In balancing a reduction half-reaction, the order of balancing differs if the reaction occurs in an acidic or a basic solution. In an acidic solution, which is our focus, after balancing the atoms undergoing a change in oxidation number and the oxygen atoms, the hydrogen atoms should be balanced using \(\ce{H^+}\) ions. Finally, the electrons are used to balance the charges. This methodical approach ensures that the reaction is stoichiometrically sound and accurately reflects the chemical process.

Acidic Solution Reactions

Balancing redox reactions in an acidic solution adds an additional layer of complexity. The presence of \(\ce{H^+}\) ions and potentially water molecules means that these species must be accounted for in the balanced equations. As in our exercise examples, acid is required to balance the hydrogen atoms that appear or disappear during the reaction. In the given reduction half-reaction \(\mathrm{IO}_3^-(aq) + 6\mathrm{H}^+(aq) + 5e^- \longrightarrow \mathrm{I}^-(aq) + 3\mathrm{H}_2\mathrm{O}(l)\), the \(\ce{H^+}\) ions are necessary for balancing the number of hydrogen atoms, and water is used to balance the oxygen atoms.

For an effective understanding and application, it's imperative to familiarize oneself with the concept of half-reactions and how to balance them independently. Then, the final step involves adjusting the stoichiometry by multiplying the half-reactions by appropriate factors to equalize the number of electrons transferred, ensuring both mass and charge balance across the full redox reaction. Remember, practice with varied problems is key to mastering balancing acidic solution redox reactions.