Question

Indicate the coefficient in front of \(\mathrm{H}_{2} \mathrm{O}_{2}\) when the following redox equation is balanced in an acidic medium. \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+\mathrm{MnO}_{4}^{-}(\mathrm{aq}) \rightarrow \mathrm{O}_{2}(\mathrm{~g})+\mathrm{Mn}^{2+}(\mathrm{aq})\) (a) 4 (b) 5 (c) 6 (d) 7

Short answer

5.

Step-by-step solution

Identify Oxidation and Reduction Half Reactions

In the given redox reaction, identify the species being oxidized and reduced. Hydrogen peroxide (\(\mathrm{H}_{2}\mathrm{O}_{2}\)) is oxidized to \(\mathrm{O}_{2}\), and \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\).

Write Half Reactions

Separate the original equation into two half reactions. The oxidation half reaction is: \(\mathrm{H}_{2}\mathrm{O}_{2} \rightarrow \mathrm{O}_{2} + 2\mathrm{H}^{+} + 2e^{-}\). The reduction half reaction is: \(\mathrm{MnO}_{4}^{-} + 8\mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4\mathrm{H}_{2}\mathrm{O}\).

Balance Electrons Transferred

To balance the electrons transferred, multiply the oxidation half reaction by 5 and the reduction half reaction by 2, to equalize the number of electrons in both half equations. This results in: \(5\mathrm{H}_{2}\mathrm{O}_{2} \rightarrow 5\mathrm{O}_{2} + 10\mathrm{H}^{+} + 10e^{-}\) and \(2\mathrm{MnO}_{4}^{-} + 16\mathrm{H}^{+} + 10e^{-} \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_{2}\mathrm{O}\).

Combine Half Reactions and Simplify

Combine the balanced half reactions and simplify to form the balanced overall reaction. Cancel out the electrons and adjust other atoms and charges as necessary:\[2\mathrm{MnO}_{4}^{-} + 16\mathrm{H}^{+} + 5\mathrm{H}_{2}\mathrm{O}_{2} \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O} + 5\mathrm{O}_{2}\], making sure both sides are equal in the numbers of each type of atom and charge.

Key concepts

Balancing Equations

Balancing chemical equations is crucial for ensuring that the number of atoms for each element is the same on both sides of the reaction. In a balanced equation, the mass and charge are conserved, meaning that no atoms are lost or gained in the process. Balancing requires understanding the stoichiometry of the reaction, which involves the coefficients that indicate the number of molecules or moles involved. For redox reactions, which involve the transfer of electrons, balancing is done by splitting the reaction into two half-reactions: oxidation and reduction. Each needs to be balanced separately before combining them into a final balanced equation. In the final balanced form of the reaction, the coefficients indicate the proportion in which reactants are consumed and products are formed.In our example, the coefficient in front of \(\mathrm{H}_{2}\mathrm{O}_{2}\) is 5 when the equation is fully balanced. Consistently using the correct coefficients helps in understanding the stoichiometry and the dynamics of the chemical reaction.

Oxidation States

Oxidation states, or oxidation numbers, are used to keep track of electron transfer in redox reactions. They help determine what is getting oxidized and what is getting reduced during the reaction.An element's oxidation state is an imaginary charge that it would have if all bonds to atoms of different elements were 100% ionic. In simple terms, the increase in an atom's oxidation state signifies oxidation (loss of electrons), whereas a decrease indicates reduction (gain of electrons). In our redox reaction, \(\mathrm{H}_{2}\mathrm{O}_{2}\) goes from an oxidation state of -1 for each oxygen to 0 in \(\mathrm{O}_{2}\), meaning it loses electrons. \(\mathrm{MnO}_{4}^{-}\) goes from an oxidation state of +7 for manganese to +2 in \(\mathrm{Mn}^{2+}\), signifying that it gains electrons. Accurately identifying these changes is fundamental in balancing redox equations.

Half-Reaction Method

The half-reaction method is a systematic approach for balancing redox reactions. It focuses on separately managing the oxidation and reduction processes before combining them into a balanced overall equation. 1. **Identify the Half-Reactions**: First, determine the oxidation and reduction half-reactions. In our case, \(\mathrm{H}_{2}\mathrm{O}_{2}\) is oxidized to \(\mathrm{O}_{2}\)\,and \(\mathrm{MnO}_{4}^{-}\) is reduced to \(\mathrm{Mn}^{2+}\). 2. **Balance Each Half-Reaction**: Make sure each half-reaction is balanced in terms of atoms and charges. Add electrons to one side to neutralize the charge. 3. **Balance Electrons Transfer**: Ensure that there are an equal number of electrons on both sides by multiplying the half-reactions by appropriate coefficients. This ensures that electrons lost equal the electrons gained. 4. **Combine Half-Reactions**: Add the balanced half-reactions together, canceling out the electrons and any other species that appear on both sides. This gives the balanced redox reaction.The half-reaction method is valuable as it simplifies the balancing of complicated redox reactions and provides a clear framework for addressing both the atom and charge conservation principles.

Electrons Transfer

Understanding electron transfer is key in grasping redox reactions. Redox is short for reduction-oxidation, describing the process where electrons move between substances. - **Oxidation** is the loss of electrons, leading to an increase in oxidation state.- **Reduction** is the gain of electrons, resulting in a decrease in oxidation state. In the reaction, electrons transfer from \(\mathrm{H}_{2}\mathrm{O}_{2}\), which loses electrons (oxidized), to \(\mathrm{MnO}_{4}^{-}\), which gains electrons (reduced). Accurate accounting of the electrons in each half-reaction is necessary to balance the overall reaction. By balancing the transfer of electrons, the reaction demonstrates the law of conservation of charge. This ensures the reaction's legitimacy on a chemical basis, allowing us to predict the behavior and products of the reaction accurately. Electrons transfer is essentially the heart of the redox process, driving the transformation of the reactants.