Question

Classify each half-reaction occurring in acidic aqueous solution as an oxidation or a reduction and balance the half-reaction. (a) \(\mathrm{S}(s) \rightarrow \mathrm{H}_{2} \mathrm{~S}(g)\) (b) \(\mathrm{S}_{2} \mathrm{O}_{8}{ }^{2-}(a q) \longrightarrow 2 \mathrm{SO}_{4}{ }^{2-}(a q)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{3}^{-}(a q)\)

Short answer

(a) Reduction: \(S(s) + 2H^+ + 2e^- \rightarrow H_{2}S(g)\). (b) Reduction: \(S_2O_8^{2-} + 2e^- \rightarrow 2SO_4^{2-}\). (c) Reduction: \(Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\). (d) Oxidation: \(NO(g) + 2H_2O \rightarrow NO_3^{-} + 4H^+ + 3e^-\).

Step-by-step solution

Classify and Balance (a) - S(s) to H2S(g)

The half-reaction is \(\mathrm{S}(s) \rightarrow \mathrm{H}_{2}\mathrm{S}(g)\).It can be classified as a reduction because sulfur is gaining hydrogen atoms. To balance it in acidic solution, add 2 H+ to the right side to account for the hydrogen atoms in H2S and then balance charge by adding 2 electrons to the right side:\(\mathrm{S}(s) + 2\mathrm{H}^+ (aq) + 2e^- \rightarrow \mathrm{H}_{2}\mathrm{S}(g)\).

Classify and Balance (b) - S2O82- to 2SO42-

The half-reaction is \(\mathrm{S}_{2}\mathrm{O}_{8}^{2-}(aq) \longrightarrow 2\mathrm{SO}_{4}^{2-}(aq)\).It can be classified as a reduction since the oxidation state of sulfur is decreasing from +6 in the peroxodisulfate ion to +6 in the sulfate ion, but the oxygen-oxygen bond is broken. To balance it add 2 electrons to the left side to balance the charges: \(\mathrm{S}_{2}\mathrm{O}_{8}^{2-}(aq) + 2e^- \rightarrow 2\mathrm{SO}_{4}^{2-}(aq)\).

Classify and Balance (c) - Cr2O72- to Cr3+

The half-reaction is \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) \longrightarrow \mathrm{Cr}^{3+}(aq)\).It is a reduction since the oxidation state of chromium decreases from +6 in dichromate to +3 in chromium ion. To balance, add 14 H+ to the right side to combine with the oxygen atoms forming 7 H2O and then balance the chromium atoms and charge by adding 6 electrons to the left side:\(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}(aq) + 14\mathrm{H}^+(aq) + 6e^- \rightarrow 2\mathrm{Cr}^{3+}(aq) + 7\mathrm{H}_{2}\mathrm{O}(l)\).

Classify and Balance (d) - NO to NO3-

The half-reaction is \(\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{3}^{-}(aq)\).It is an oxidation since the nitrogen atom increases its oxidation state from +2 in NO to +5 in NO3-. To balance it in acidic solution, first add water to balance the oxygen atoms, then add the appropriate number of H+ ions to balance the hydrogen atoms from water, and finally add electrons to balance the charge. This gives us:\(\mathrm{NO}(g) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{NO}_{3}^{-}(aq) + 4\mathrm{H}^+(aq) + 3e^-\).

Key concepts

Oxidation and Reduction

Oxidation and reduction are fundamental concepts in the field of redox chemistry. An easy analogy to remember these concepts is by the acronym 'OIL RIG' - Oxidation Is Loss, Reduction Is Gain. It refers to the loss and gain of electrons, respectively. In redox reactions, one species will give up electrons (oxidized) while another accepts those electrons (reduced).

In the given exercise, the sulfur atom in example (a) goes from an elemental form, to being part of which has hydrogen atoms bonded to it. This process is a reduction because sulfur gains electrons to form the stable bonds with hydrogen found in A critical aspect to remember is that in any redox process, conservation of charge is a must. Therefore, the overall charge before and after the reaction must remain the same. This is achieved by adding electrons to balance the half-reaction, as seen in the solutions provided.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions. This is particularly useful because it allows us to focus on each part of the reaction independently, simplifying the overall process. In essence, the method involves separating the entire redox reaction into two half-reactions: one for oxidation and one for reduction.

Each half-reaction is balanced separately, first for mass and then for charge. For mass balance, we add coefficients to get the same number of each type of atom on both sides. In acidic solution, should be added to balance for the hydrogen atoms, while is used for oxygen. Once the atoms are balanced, we use electrons ( to balance the charge. The balanced half-reactions in the exercise examples effectively demonstrate this method. When both half-reactions are balanced, they can be combined to form the complete balanced redox equation.

Redox Chemistry in Acidic Solution

Balancing redox reactions in acidic solutions involves a specific set of steps due to the presence of ions. These hydrogen ions play a significant role in maintaining charge neutrality. The solutions in the exercise highlight that when a half-reaction is combined with it often requires the addition of to balance any oxygen atoms that might be part of the reactants or products.

After ensuring the atoms are balanced, additional ions might be added to balance the excess hydrogens coming from the This approach is clearly detailed in the solutions for and where hydrogen ions are added to balance both mass and charge. Subsequently, in the redox chemistry of acidic solutions, water ( can act as both a reactant and a product, reflecting its role as a solvent and a participant in the redox process.