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Use the method of half reactions to balance the following equations: (a) \(\begin{aligned} \operatorname{CrI}_{3}(s) &+\mathrm{Cl}_{2}(g) \rightarrow \\ & \mathrm{CrO}_{4}^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \end{aligned} \quad\) (basic) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+\mathrm{I}_{3}^{-}(a q) \rightarrow\) \(\mathrm{CO}_{2}(g)+\mathrm{CHO}_{2}^{-}(a q)+\mathrm{CHI}_{3}(a q)+\mathrm{I}^{-}(a q) \quad\) (acidic)

Short Answer

Expert verified
Use half-reactions: balance atoms, add H鈧侽 for O, H鈦/OH鈦 for H, e鈦 for charge; equalize electrons; combine.

Step by step solution

01

Separate Unbalanced Reactions into Half-Reactions

For reaction (a), split into chromium and iodine half-reactions: 1. CrI鈧(s) 鈫 CrO鈧劼测伝(aq) 2. I鈦(from CrI鈧) 鈫 IO鈧勨伝 For reaction (b), split into ethanol and triiodide half-reactions: 1. C鈧侶鈧匫H 鈫 CO鈧 and CHO鈧傗伝 2. I鈧冣伝 鈫 I鈦 and CHI鈧
02

Balance Atoms Other than H and O in Each Half-Reaction

For (a): 1. Cr: already balanced. 2. I: I in CrI鈧 鈫 IO鈧勨伝. For (b): 1. C: C鈧侶鈧匫H 鈫 CO鈧 and CHO鈧傗伝 2. I: I鈧冣伝 鈫 I鈦 and CHI鈧
03

Balance Oxygen Atoms Using Water (H鈧侽)

For (a): 1. Add 4 H鈧侽 to the right side of the chromium half-reaction: CrI鈧 鈫 CrO鈧劼测伝 + 4H鈧侽 2. Add 4 H鈧侽 to the left side of the iodine half-reaction: I鈦 + 4H鈧侽 鈫 IO鈧勨伝 For (b): 1. Add H鈧侽 molecules as necessary to balance O in both carbon and iodine half-reactions.
04

Balance Hydrogen Atoms Using H鈦 (for Acidic) or OH鈦 (for Basic)

For (a): 1. Add 8 OH鈦 to the right of the chromium half-reaction to balance with 4H鈧侽. 2. Add 8 OH鈦 to the left of the iodine half-reaction to react with 4H鈧侽. For (b): 1. Add H鈦 to balance H from ethanol half-reaction since it is acidic.
05

Balance Charges by Adding Electrons (e鈦)

For (a): 1. Add electrons to balance both half-reactions. 2. Cr: CrI鈧 + 8 OH鈦 鈫 CrO鈧劼测伝 + 4 H鈧侽 + 3 e鈦 3. I: I鈦 + 4 H鈧侽 鈫 IO鈧勨伝 + 8 e鈦 + 8 OH鈦 For (b): 1. Add electrons to balance each half-reaction.
06

Equalize the Number of Electrons in Both Half-Reactions and Combine

For (a): 1. Multiply both half-reactions to equalize electrons. 2. Combine the balanced half-reactions and simplify. For (b): 1. Ensure both half-reactions have the same electrons and combine.
07

Double-Check Balancing and Conditions

For both reactions, ensure the total mass and charge balance. Check if the reactions meet the conditions (basic or acidic environment).

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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 powerful technique used to balance redox reactions. This approach breaks a complex reaction into two simpler parts based on changes in oxidation numbers. Each part is called a half-reaction: one for oxidation and another for reduction.
These half-reactions make it easier to ensure that the electron transfer is balanced. By focusing on separate paths for reduction (gain of electrons) and oxidation (loss of electrons), complexities in reactions become manageable.
  • Identify the redox pairs.
  • Write separate half-equations for each process.
  • Balance elements other than O and H first.
  • Adjust oxygen using water and hydrogen using either hydroxide or hydronium depending on the conditions.
  • Finally, balance the charge with electrons.

This method provides clear visibility into how substances interchange electrons, leading to a more straightforward path to the overall balanced equation.
Balancing Chemical Equations
Balancing chemical equations is crucial as it ensures the conservation of mass in any reaction. In redox reactions, this involves not only balancing atoms but also ensuring the charge is consistent on both sides. This often begins after the equations have been split into half-reactions.
When you balance each half-reaction, you'll focus firstly on non-oxygen and non-hydrogen elements. Once these are settled, oxygen is balanced using water (H鈧侽), and hydrogen is balanced using either hydrogen ions (H鈦) in acidic conditions or hydroxide ions (OH鈦) in basic settings.
Ultimately, electrons are added to each half-reaction to ensure that the charges balance appropriately. Only when these constraints are satisfied can the reaction be considered balanced.
Achieving this requires multiplying each half-reaction by appropriate values so that the electrons cancel out, reflecting the true stoichiometric coefficients that guide the reaction's reality.
Oxidation-Reduction
Oxidation-reduction, often abbreviated as redox, reactions involve a transfer of electrons between two species. In the context of the half-reaction method, oxidation refers to the loss of electrons while reduction denotes the gain of electrons.
Understanding which element gets oxidized and which one gets reduced is key. This dictates how the half-reactions are formulated.
When identifying oxidation and reduction:"
  • Track the oxidation states of various elements in the reactants and products.
  • Determine which element's oxidation state increases (oxidation) and which decreases (reduction).
  • The element undergoing oxidation releases electrons, which are then gained by the reduced element.

By comprehending the transfer of electrons, the essence of the redox process becomes clearer and provides insight into how chemical energy is transformed within a system.
Acidic and Basic Conditions in Redox Reactions
The process of balancing redox reactions can differ significantly depending on whether the reaction occurs under acidic or basic conditions. These conditions dictate how hydrogen and oxygen are adjusted in the half-reaction method.

Acidic Conditions

When dealing with acidic conditions, hydrogen ions (H鈦) are used to balance hydrogen atoms in the half-reactions. If there is a deficit of oxygen, it's balanced by adding water (H鈧侽). The use of these ions is crucial in reaching a balanced state under acidic conditions.

Basic Conditions

In basic conditions, hydroxide ions (OH鈦) serve to balance hydrogen atoms. Similarly, depending on the half-reaction, water may be used to balance oxygen. Adding the right mol quantities can sometimes reveal hidden complexities, like needing to add the same ion type on both sides to cancel out charges effectively.
Adaptation to these environments in the balancing procedure gives students a deep understanding of the chemical nature and adaptability required in practical settings. This aspect not only refines problem-solving skill but also allows an appreciation for real-world chemical dynamics.

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Most popular questions from this chapter

Complete and balance the following equations: (a) \(\mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)\) (acidic) (b) \(\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3} \cdot 3 \mathrm{H}_{2} \mathrm{O}(s)\) (basic)

Identify the oxidizing and reducing agents and write the oxidation and reduction half-reaction equations for the following chemical equations: (a) \(\operatorname{In}^{+}(a q)+2 \mathrm{Fe}^{3+}(a q) \rightarrow 2 \mathrm{Fe}^{2+}(a q)+\mathrm{In}^{3+}(a q)\) (b) \(\mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{ClO}^{-}(a q) \rightarrow \underline{\mathrm{S}}(s)+\mathrm{Cl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Sodium sulfide is manufactured by reacting sodium sulfate with carbon in the form of coke accord- ing to the equation $$ \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+4 \mathrm{C}(s) \rightarrow \mathrm{Na}_{2} \mathrm{~S}(s)+4 \mathrm{CO}(g) $$ Identify the oxidizing and reducing agents in this equation.

Use the method of half reactions to balance the following equations: (a) \(\mathrm{N}_{2} \mathrm{H}_{4}(a q)+\mathrm{Cu}(\mathrm{OH})_{2}(s) \rightarrow \mathrm{N}_{2}(g)+\mathrm{Cu}(s) \quad\) (basic) (b) \(\mathrm{H}_{3} \mathrm{AsO}_{3}(a q)+\mathrm{I}_{2}(a q) \rightarrow\) \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{I}^{-}(a q) \quad\) (acidic)

An ore is to be analyzed for its iron content by an oxidation-reduction titration with permanganate ion. A \(4.24\) -gram sample of the ore is dissolved in hydrochloric acid and passed over a reducing agent so that all the iron is in the form \(\mathrm{Fe}^{2+}(a q)\). The \(\mathrm{Fe}^{2+}(a q)\) is completely oxidized by \(31.6 \mathrm{~mL}\) of a \(0.0512-\mathrm{M}\) aqueous solution of \(\mathrm{KMnO}_{4}(a q) .\) The unbalanced equation for the reaction is $$ \begin{aligned} &\mathrm{KMnO}_{4}(a q)+\mathrm{HCl}(a q)+\mathrm{FeCl}_{2}(a q) \rightarrow \\\ &\mathrm{MnCl}_{2}(a q)+\mathrm{FeCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{KCl}(a q) \end{aligned} $$ Calculate the amount of iron in the sample and its mass percentage in the ore.

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