Question

Balance these equations for redox reactions occurring in acidic solution. (a) \(\mathrm{MnO}_{4}^{-}+\mathrm{I}^{-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{I}_{2}(\mathrm{s})\) (b) \(\mathrm{BrO}_{3}^{-}+\mathrm{N}_{2} \mathrm{H}_{4} \longrightarrow \mathrm{Br}^{-}+\mathrm{N}_{2}\) (c) \(\mathrm{VO}_{4}^{3-}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{VO}^{2+}+\mathrm{Fe}^{3+}\) (d) \(\mathrm{UO}^{2+}+\mathrm{NO}_{3}^{-} \longrightarrow \mathrm{UO}_{2}^{2+}+\mathrm{NO}(\mathrm{g})\)

Short answer

(a) 5MnO4- + 16H+ + 2I- > 5Mn2+ + 2I2(s) + 8H2O, \n(b) 3BrO3- + 10H+ + 2N2H4 > 3Br- + 2N2 + 6H2O, \n(c) VO4 3- + Fe2+ + H+ > VO2+ + Fe3+ + H2O, \n(d) UO + 2H+ + NO3- > UO2 2+ + NO + H2O

Step-by-step solution

Identify the oxidation and reduction species and assign oxidation numbers

\n(a) In the reaction MnO4- + I- > Mn2+ + I2(s), Mn is reduced from 7 to 2, and I- is oxidized to I2. \n(b) In the reaction BrO3- + N2H4 > Br- + N2, Br is reduced from 5 to -1 and N is oxidized from -2 to 0. \n(c) In the reaction VO4 3- + Fe2+ > VO2+ + Fe3+, V is reduced from 5 to 4 and Fe is oxidized from 2 to 3. \n(d) In the reaction UO+ + NO3- > UO2 2+ + NO, N is reduced from 5 in NO3- to 2 in NO and U is oxidized from 4 to 6.

Balance the atoms in the half reactions and add H2O and H+ where necessary

\n(a) Add 8H+ to the left side of the reduction equation (MnO4- > Mn2+) to balance with 4H2O on the right side. Add 2I- on the left side to balance with I2 on the right side in the oxidation equation. \n(b) Add 6H+ to the right side of the reduction equation and 2H2O to the left side to balance with 6H2O on the right side. \n(c) For both the reduction and oxidation equations, no further balancing is needed. \n(d) Add 2H+ to the right side of the reduction equation and H2O to the left side to balance with 2H2O on the right side.

Balance charges by adding electrons

\n(a) Add 5 electrons to the left side of the Mn half reaction and 2 to the right side of the I- equation. \n(b) Add 6 electrons to the left side of the Br half reaction and 2 to the left side of the N2H4 equation. \n(c) Add 1 electron to the left side of the Fe2+ equation and 1 to the right side of the VO4 3- equation. \n(d) Add 1 electron to the right side of the NO3- equation and 1 to the right side of the UO equation.

Cancel redundant entities and balance the equations

\n(a) Multiply the equations by factors of 5 and 2 respectively and then add them together, cancel out redundant entities. The balanced equation is: 5MnO4- + 16H+ + 2I- > 5Mn2+ + 2I2(s) + 8H2O. \n(b) Multiply the equations by factors of 3 and 2 respectively, the balanced equation will be: 3BrO3- + 10H+ + 2N2H4 > 3Br- + 2N2 + 6H2O. \n(c) There's no need to multiply the equations, the balanced equation is: VO4 3- + Fe2+ + H+ > VO2+ + Fe3+ + H2O. \n(d) There's also no need to multiply the equations, the balanced equation is: UO + 2H+ + NO3- > UO2 2+ + NO + H2O.

Key concepts

Oxidation States

Understanding oxidation states is crucial for identifying how elements change during redox reactions. To determine oxidation states, assign hypothetical charges to atoms in compounds based on certain rules:

• For an atom in its elemental form, the oxidation state is zero. For example, in Iodine (I₂), each iodine atom has an oxidation state of 0.
• For monoatomic ions, the oxidation state equals the ion charge. For instance, in \( ext{Mn}^{2+}\), manganese has an oxidation state of +2.
• In compounds, assign oxidation states by assuming the overall charge equals the sum of individual oxidation states. Common rules include \( ext{H}\) as +1 (except in metal hydrides where it is -1) and O as -2 (except in peroxides where it is -1).

In the provided reactions, manganese changes from an oxidation state of +7 in \( ext{MnO}_4^-\) to +2 in \( ext{Mn}^{2+}\), indicating reduction. Conversely, iodine changes from -1 in \( ext{I}^-\) to 0 in \( ext{I}_2\), indicating oxidation. Recognizing changes in oxidation states helps to identify which species are oxidized and reduced.

Half-Reaction Method

The half-reaction method is a systematic approach to balancing redox equations by separating them into oxidation and reduction parts. This method simplifies complex reactions by focusing on changes in oxidation states.

• First, divide the equation into two half-reactions: one for oxidation and one for reduction.
• Identify and balance all elements except hydrogen and oxygen. This ensures the stoichiometry of the central atoms is correct.
• Next, balance oxygen atoms by adding water molecules (H₂O).
• Balance hydrogen atoms by adding hydrogen ions (H⁺). This step is relevant especially in acidic solutions.
• Finally, balance charges by adding electrons (e^-) to one or both sides of each half-reaction.

Once both half-reactions are balanced in terms of atoms and charge, combine them to form the full balanced redox equation. This accounts for electron transfer and ensures the overall conversion makes sense both chemically and electrically. For example, in the given reaction, the manganese half-reaction involves transferring 5 electrons, while the iodine half-reaction involves 2 electrons.

Charge Balancing

Charge balancing ensures the net charge is the same on both sides of a redox equation. After balancing atoms and adding water and hydrogen ions, adjust the electron count to equalize charges on each half-reaction.

• Calculate the charge on each side of a half-reaction. The imbalance reveals how many electrons are necessary for charge balance.
• Add electrons to the more positive or less negative side to achieve neutrality.
• In our step-by-step solutions, consistent charge balancing is pivotal. For Manganese, adding 5 electrons to \( ext{MnO}_4^- ightarrow ext{Mn}^{2+}\) converts a +7 to +2 state, confirming reduction.
• For the iodine system \( ext{I}^- ightarrow ext{I}_2\), electrons appear on the right to match charges after oxidation occurs.

Once individual half-reactions are charge-balanced, collectively adjust them to zero net charge in the complete equation. Finally, combine and simplify to eliminate excess electrons and redundant species, confirming a balanced redox reaction.