Balance the following redox equations by the halfreaction method: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (in acidic solution) (b) \(\mathrm{Cu}+\mathrm{HNO}_{3} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{NO}+\mathrm{H}_{2} \mathrm{O}\) (in acidic solution) (c) \(\mathrm{CN}^{-}+\mathrm{MnO}_{4}^{-} \longrightarrow \mathrm{CNO}^{-}+\mathrm{MnO}_{2}\) (in basic solution) (d) \(\mathrm{Br}_{2} \longrightarrow \mathrm{BrO}_{3}^{-}+\mathrm{Br}^{-}\) (in basic solution) (e) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{I}_{2} \longrightarrow \mathrm{I}^{-}+\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) (in acidic solution)

Short Answer

Expert verified
(a) \(2\mathrm{Fe}^{2+} + \mathrm{H}_{2} \mathrm{O}_{2} + 2\mathrm{H}^{+} \longrightarrow 2\mathrm{Fe}^{3+} + 2\mathrm{H}_{2} \mathrm{O}\) \n(b) \(3\mathrm{Cu} + 8\mathrm{HNO}_{3} \longrightarrow 3\mathrm{Cu}^{2+} + 2\mathrm{NO} + 4\mathrm{H}_{2} \mathrm{O}\)\n(c) \(2\mathrm{CN}^{-} + 2\mathrm{MnO}_{4}^{-} + 3\mathrm{H}_{2}\mathrm{O} \longrightarrow 2\mathrm{CNO}^{-} + 2\mathrm{MnO}_{2} + 4\mathrm{OH}^{-} + \mathrm{C}\)\n(d) \(\mathrm{Br_{2}} + 6\mathrm{OH^{-}} \longrightarrow 2\mathrm{BrO_{3}^{-}} + 6\mathrm{Br^{-}}\)\n(e) \(2\mathrm{S}_{2} \mathrm{O}_{3}^{2-} + 2\mathrm{I}_{2} \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-} + 4\mathrm{I}^{-}\)

Step by step solution

01

Balance (a)

Divide the equation into half-reactions. Oxidation: \(\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}\); Reduction: \(\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{3+}\). Balance elements other than O and H in all half-reactions. Balancing oxidation half-reaction gives \(\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2\mathrm{H}_{2} \mathrm{O}\). Next, balance the O and H by adding H2O and H+ appropriately. Finally, balance the charge by adding electrons (e-). Do the same for the reduction half-reaction then add the two half-reactions. The final balanced equation is: \(2\mathrm{Fe}^{2+} + \mathrm{H}_{2} \mathrm{O}_{2} + 2\mathrm{H}^{+} \longrightarrow 2\mathrm{Fe}^{3+} + 2\mathrm{H}_{2} \mathrm{O}\)
02

Balance (b)

Follow the same procedure as in Step 1. The balanced half-reactions are: Oxidation: \(\mathrm{Cu} \longrightarrow \mathrm{Cu}^{2+} + 2e^{-}\); Reduction: \(4\mathrm{H}^{+} + \mathrm{NO}_{3}^{-} + 3e^{-} \longrightarrow \mathrm{NO} + 2\mathrm{H}_{2} \mathrm{O}\). Adding these gives: \(3\mathrm{Cu} + 8\mathrm{HNO}_{3} \longrightarrow 3\mathrm{Cu}^{2+} + 2\mathrm{NO} + 4\mathrm{H}_{2} \mathrm{O}\)
03

Balance (c)

Since the solution is basic, OH- is used to balance O. The balanced half-reactions are: Oxidation: \(2\mathrm{CN}^{-} \longrightarrow \mathrm{CNO}^{-} + \mathrm{C} + 2e^{-}\); Reduction: \(2\mathrm{MnO}_{4}^{-} + 3\mathrm{H}_{2}\mathrm{O} + 2e^{-} \longrightarrow 2\mathrm{MnO}_{2} + 4\mathrm{OH}^{-}\). Adding these gives: \(2\mathrm{CN}^{-} + 2\mathrm{MnO}_{4}^{-} + 3\mathrm{H}_{2}\mathrm{O} \longrightarrow 2\mathrm{CNO}^{-} + 2\mathrm{MnO}_{2} + 4\mathrm{OH}^{-} + \mathrm{C}\)
04

Balance (d)

Balancing in basic solution uses OH-. The balanced half-reactions are: Oxidation: \(\mathrm{Br_{2}} + 6\mathrm{OH^{-}} \longrightarrow 2\mathrm{BrO_{3}^{-}} + 3\mathrm{H_{2}O} + 6e^{-}\); Reduction: \(6\mathrm{OH^{-}} + 6e^{-} \longrightarrow 6\mathrm{Br^{-}} + 3\mathrm{H_{2}O}\). Adding these gives: \(\mathrm{Br_{2}} + 6\mathrm{OH^{-}} \longrightarrow 2\mathrm{BrO_{3}^{-}} + 6\mathrm{Br^{-}}\)
05

Balance (e)

Follow the same procedure as in Step 1. The balanced half-reactions are: Oxidation: \(2\mathrm{S}_{2} \mathrm{O}_{3}^{2-} \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-} + 2e^{-}\); Reduction: \(2\mathrm{I}_{2} + 2e^{-} \longrightarrow 4\mathrm{I}^{-}\). Adding these gives: \(2\mathrm{S}_{2} \mathrm{O}_{3}^{2-} + 2\mathrm{I}_{2} \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-} + 4\mathrm{I}^{-}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Half-Reaction Method
The half-reaction method is a systematic approach to balancing redox equations, which are equations that involve electron transfer. This approach focuses on separating the oxidation and reduction processes, often referred to as 'half-reactions'.

In simple terms, the method involves breaking down the overall chemical reaction into two separate parts: one in which a species is losing electrons (oxidation), and one in which another species is gaining electrons (reduction). Each half-reaction is balanced independently for mass and charge. Then, these balanced half-reactions are added together to form the balanced redox equation.

Through the process, certain rules apply: balance all atoms except hydrogen and oxygen first, then balance the oxygen atoms using water molecules (H2O), next balance hydrogen atoms using hydrogen ions (H+ in acidic solutions or OH- in basic solutions), and finally balance the charges by adding electrons (e-).
Oxidation State
Understanding oxidation states is crucial for the half-reaction method as it tells us about the transfer of electrons during a reaction. An oxidation state, often referred to as oxidation number, essentially indicates the degree of oxidation of an atom in a compound.

An atom's oxidation state is a hypothetical charge it would have if electrons in bonds were distributed according to certain rules, with more electronegative atoms taking the electrons. In the context of redox reactions, when the oxidation state of an element increases, it has undergone oxidation (loss of electrons), and when it decreases, it has undergone reduction (gain of electrons).

By comparing the changes in these states in a reaction, we can identify which species are oxidized and which are reduced, thereby setting the stage for applying the half-reaction method.
Acidic Solution
In the context of redox reactions, the term 'acidic solution' implies that an excess of H+ ions is present in the environment where the reaction is occurring. This has implications on how the half-reactions are balanced.

For half-reactions in acidic solutions, hydrogen ions (H+) or water (H2O) are used to balance the oxygen and hydrogen atoms respectively. To balance oxygen atoms, add H2O to the side lacking oxygen; for hydrogen atoms, add H+ to the side lacking hydrogen. Once balanced for atoms, electrons are added to balance the charge. This makes balancing redox reactions in acid solutions distinct from those in basic solutions, which use OH- instead of H+.
Basic Solution
When dealing with redox reactions in basic solutions, it is essential to recognize that OH- ions are abundant in the reaction environment.

Balancing redox reactions in these solutions requires a different approach from acidic solutions. After balancing for atoms other than oxygen and hydrogen, we balance oxygen by adding water molecules (H2O) to the side needing oxygen and then balance hydrogen by adding OH- ions to the side needing hydrogen. Finally, electrons are added to balance the electrical charge. This distinct procedure accommodates the presence of OH- ions, characteristic of a basic or alkaline environment.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is grounded in the conservation of mass and the consistent ratios called 'mole ratios' derived from the coefficients of a balanced chemical equation.

In redox reactions, stoichiometry not only involves balancing atoms but also managing the electrons that are transferred between species. Once the half-reactions are balanced separately, their stoichiometry must be reconciled by ensuring that the same number of electrons are lost in the oxidation half-reaction as are gained in the reduction half-reaction. This is achieved by multiplying the half-reactions by appropriate factors so that the electrons cancel out when they are combined to give the final balanced equation.

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Most popular questions from this chapter

In a certain electrolysis experiment, \(1.44 \mathrm{~g}\) of \(\mathrm{Ag}\) were deposited in one cell (containing an aqueous \(\mathrm{AgNO}_{3}\) solution), while \(0.120 \mathrm{~g}\) of an unknown metal X was deposited in another cell (containing an aqueous \(\mathrm{XCl}_{3}\) solution) in series with the \(\mathrm{AgNO}_{3}\) cell. Calculate the molar mass of \(\mathrm{X}\).

Industrially, copper is purified by electrolysis. The impure copper acts as the anode, and the cathode is made of pure copper. The electrodes are immersed in a \(\mathrm{CuSO}_{4}\) solution. During electrolysis, copper at the anode enters the solution as \(\mathrm{Cu}^{2+}\) while \(\mathrm{Cu}^{2+}\) ions are reduced at the cathode. (a) Write half-cell reactions and the overall reaction for the electrolytic process. (b) Suppose the anode was contaminated with \(\mathrm{Zn}\) and \(\mathrm{Ag} .\) Explain what happens to these impurities during electrolysis. (c) How many hours will it take to obtain \(1.00 \mathrm{~kg}\) of \(\mathrm{Cu}\) at a current of \(18.9 \mathrm{~A} ?\)

Consider a galvanic cell composed of the SHE and a half-cell using the reaction \(\mathrm{Ag}^{+}(a q)+e^{-} \rightarrow \operatorname{Ag}(s)\). (a) Calculate the standard cell potential. (b) What is the spontaneous cell reaction under standard-state conditions? (c) Calculate the cell potential when \(\left[\mathrm{H}^{+}\right]\) in the hydrogen electrode is changed to (i) \(1.0 \times\) \(10^{-2} M\) and (ii) \(1.0 \times 10^{-5} M,\) all other reagents being held at standard-state conditions. (d) Based on this cell arrangement, suggest a design for a pH meter.

What is a cell diagram? Write the cell diagram for a galvanic cell consisting of an Al electrode placed in a \(1 \mathrm{MAl}\left(\mathrm{NO}_{3}\right)_{3}\) solution and a Ag electrode placed in a \(1 M \mathrm{AgNO}_{3}\) solution.

A sample of iron ore weighing \(0.2792 \mathrm{~g}\) was dissolved in an excess of a dilute acid solution. All the iron was first converted to \(\mathrm{Fe}(\mathrm{II})\) ions. The solution then required \(23.30 \mathrm{~mL}\) of \(0.0194 \mathrm{M} \mathrm{KMnO}_{4}\) for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.

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