Question

Use the half-reaction method to balance the following redox equations. (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}+\mathrm{Fe}+\mathrm{H}^{+} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}^{+} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}+\mathrm{Cr}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}_{2}+\mathrm{MnO}_{4}^{-}+\mathrm{H}_{2} \mathrm{O} \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{OH}+\mathrm{MnO}_{2}+\mathrm{OH}^{-}\).

Short answer

The balanced equations using half-reaction method are:(a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}+6\mathrm{Fe}+18\mathrm{H}^{+}\rightarrow \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^{+}+3\mathrm{Fe}^{3+}+8\mathrm{H}_{2}O\)(b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}+2\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+16\mathrm{H}^{+}\rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\mathrm{H}+4\mathrm{Cr}^{3+}+9\mathrm{H}_{2}O\)(c) \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}_{2}+\mathrm{MnO}_{4}^{-}+3\mathrm{H}_{2}O\rightarrow\mathrm{CH}_{3} \mathrm{CHOHCH}_{2}\mathrm{OH}+\mathrm{MnO}_{2}+5\mathrm{OH}^{-}\)

Step-by-step solution

Balance First Equation

Separate every species into its half-reaction: Oxidation: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}\rightarrow \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^{+}\)Reduction: \(\mathrm{Fe}\rightarrow\mathrm{Fe}^{3+}\)Then balance each half-reaction first by atoms other than H and O, then by O atoms, and then by H atoms. Add electrons to the appropriate side to balance charge. Combine and simplify.Oxidation: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}+6\mathrm{H}^{+}\rightarrow \mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{3}^{+}+2\mathrm{H}_{2}O\)Reduction: \(3\mathrm{Fe}+12\mathrm{H}^{+}+3\mathrm{e}^{-}\rightarrow3\mathrm{Fe}^{3+}+6\mathrm{H}_{2}O\)

Balance Second Equation

Separate into half-reactions and balance as above:Oxidation: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\mathrm{H}\)Reduction: \(\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}\rightarrow\mathrm{Cr}^{3+}\)Oxidation: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}+2\mathrm{H}^{+}+2\mathrm{e}^{-}\rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\mathrm{H}+2\mathrm{H}_{2}O\)Reduction: \(2\mathrm{Cr}_{2}\mathrm{O}_{7}^{2-}+14\mathrm{H}^{+}+6\mathrm{e}^{-}\rightarrow4\mathrm{Cr}^{3+}+7\mathrm{H}_{2}O\)

Balance Third Equation

Similarly, separate and balance as the previous steps:Oxidation: \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}_{2}\rightarrow\mathrm{CH}_{3} \mathrm{CHOHCH}_{2}\mathrm{OH}\)Reduction: \(\mathrm{MnO}_{4}^{-}\rightarrow\mathrm{MnO}_{2}\)Oxidation: \(\mathrm{CH}_{3}\mathrm{CH}=\mathrm{CH}_{2}+2\mathrm{H}_{2}O\rightarrow\mathrm{CH}_{3} \mathrm{CHOHCH}_{2}\mathrm{OH}+2\mathrm{OH}^{-}+\mathrm{e}^{-}\)Reduction: \(\mathrm{MnO}_{4}^{-}+\mathrm{e}^{-}+2\mathrm{H}_{2}O\rightarrow\mathrm{MnO}_{2}+\4\mathrm{OH}^{-} \)

Key concepts

Half-Reaction Method

The half-reaction method is a systematic process for balancing redox reactions, where the overall reaction is separated into two parts – the oxidation half-reaction and the reduction half-reaction. Each half-reaction represents either the loss or gain of electrons by species undergoing redox change.

To apply this method effectively, start by identifying which species are oxidized and which are reduced. Next, write out the separate half-reactions, ensuring each half-reaction has balanced mass and charge by adding H₂O, H⁺, and electrons where necessary. Once both half-reactions are balanced individually, they are then combined with the stoichiometry carefully adjusted to ensure that the electrons lost in the oxidation half equal the electrons gained in the reduction half. This method ensures that the law of conservation of mass is respected, and the overall charge is balanced.

Oxidation and Reduction

Oxidation and reduction are fundamental concepts in chemistry, particularly in the field of electrochemistry. These processes involve the transfer of electrons between chemical species.

Oxidation is the loss of electrons, often accompanied by an increase in oxidation state. It is essential to note that the oxidizing agent, which facilitates the oxidation process, is itself reduced. Conversely, reduction is the gain of electrons and involves a decrease in the oxidation state. The substance that undergoes reduction is the reducing agent, which is oxidized in the process.

Identifying Oxidation and Reduction

In a redox reaction, the substance that loses electrons (is oxidized) is called the reducing agent, while the substance that gains electrons (is reduced) becomes the oxidizing agent. Finding and labeling these agents correctly is a crucial step in balancing redox reactions.

Stoichiometry

Stoichiometry deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It is based on the conservation of mass and the principles of balanced chemical equations.

When balancing redox reactions, stoichiometry becomes essential in ensuring that atoms and charges are balanced on both sides of the reaction equation. These calculations often require the careful addition of coefficients to balance the number of atoms of each element and charges throughout the reaction.

Stoichiometric Coefficients

These coefficients are the numbers placed before compounds and elements in a reaction. They indicate the proportions in which substances react and are formed. By adjusting stoichiometric coefficients appropriately, chemists ensure that the principle of 'matter cannot be created or destroyed in a chemical reaction' holds true for both the atoms and the charge.

Electrochemistry

Electrochemistry is the branch of chemistry that studies the relationship between electrical energy and chemical changes, which includes redox reactions. These reactions are at the heart of processes like battery operation, corrosion, and electrolysis.

Understanding redox reactions is crucial for mastering electrochemistry, as it provides insight into how electrons are transferred in chemical processes and how these transfers can be harnessed to do work or produce energy.

Electrochemical Cells

In an electrochemical cell, a redox reaction is divided into two half-reactions that occur at different electrodes, driving the flow of electrons through an external circuit. Mastery of redox principles, including balancing redox reactions, is not only academically important but also vital for numerous industrial and technological applications.