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Magazine Chapter 9: Problem 88
In the redox reaction \(\mathrm{xKMnO}_{4}+\mathrm{yNH}_{3} \longrightarrow \mathrm{KNO}_{3}+\mathrm{MnO}_{2}+\mathrm{KOH}\) \(+\mathrm{H}_{2} \mathrm{O}\) (a) \(x=3, y=8\) (b) \(x=6, y=3\) (c) \(x=5, y=10\) (d) \(x=8, y=3\)
Short Answer
Expert verified
Option (a) x=3, y=8 is the balanced scenario for this reaction.
Step by step solution
01
Analyze the Reaction Types
First, identify the type of reaction, which is a redox reaction. In redox reactions, oxidation and reduction occur simultaneously. Potassium permanganate (KMnO鈧�) is reduced, and ammonia (NH鈧�) is oxidized.
02
Assign Oxidation Numbers
Assign oxidation numbers to each element in the reaction to track electron transfer. For KMnO鈧�, K is +1, Mn is +7, and O is -2. In NH鈧�, N is -3, and H is +1. In the products, MnO鈧� has Mn at +4, and in KNO鈧�, N is +5.
03
Determine Change in Oxidation States
Determine how the oxidation numbers change. Mn in KMnO鈧� changes from +7 to +4, indicating reduction. N in NH鈧� changes from -3 to +5, indicating oxidation.
04
Balance Oxidation and Reduction Half-Reactions
Write and balance the two half-reactions: the reduction of MnO鈧勨伝 to MnO鈧� and the oxidation of NH鈧� to NO鈧冣伝. Ensure electrons lost in oxidation equal electrons gained in reduction.
05
Balance the Electrons
Balance the electrons transferred in the half-reactions. Add appropriate coefficients to the chemicals in order to equate electron gain and loss.
06
Combine the Balanced Half-Reactions
Combine the balanced half-reactions into a single balanced equation, verifying that the number of atoms and charge are conserved on both sides.
07
Check Coefficient Options
Compare the balanced equation to the given options (a) to (d). Calculate expected values for coefficients x and y to determine which option is correctly balanced based on stoichiometry.
08
Choose and Validate the Correct Answer
Choose the option where the coefficients x and y lead to a balanced chemical equation. Calculate based on stoichiometry which scenario balances all elements.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Oxidation States
Understanding oxidation states is key to mastering redox reactions. An oxidation state tells us whether an element has lost or gained electrons during a chemical process. It helps us identify which species are oxidized and which are reduced. In the given reaction, let's break down the process:
Quantifying these changes allows us to see how electrons are transferred during the reaction, a critical first step in balancing it.
- Potassium permanganate (KMnO鈧�) contains Mn with an oxidation state of +7. As it turns into MnO鈧�, the Mn oxidation state changes to +4. This change signifies that Mn gains electrons, indicating reduction.
- Ammonia (NH鈧�) has nitrogen with an oxidation state of -3. As it transforms into KNO鈧�, nitrogen's oxidation state becomes +5. Here, nitrogen loses electrons, a sign of oxidation.
Quantifying these changes allows us to see how electrons are transferred during the reaction, a critical first step in balancing it.
Balancing Chemical Equations
Balancing a chemical equation ensures that the same number of each type of atom appears on both sides of the reaction. In redox reactions, this balancing also needs to account for electron transfer. Here's how you tackle it:
Balancing is a dynamic process, often requiring you to iterate your solution until all parts fit into place snugly, reflecting conservation of mass and charge.
- Write the unbalanced chemical equation first. For our problem, the equation involves KMnO鈧� and NH鈧� as reactants, and they must form KNO鈧�, MnO鈧�, KOH, and H鈧侽.
- Next, utilizing the oxidation states, determine which elements have changed oxidation numbers, thus identifying the ones that need balancing.
- Introduce coefficients to the equation so that atoms and charges are equivalent on both sides.
Balancing is a dynamic process, often requiring you to iterate your solution until all parts fit into place snugly, reflecting conservation of mass and charge.
Half-Reactions
Half-reactions are a convenient way to separately consider the oxidation and reduction processes within a redox reaction. By doing this, we can determine how electrons are exchanged:
This division simplifies tracking and balancing electron flow, essential for correctly finalizing the chemical equation.
- The oxidation half-reaction describes how NH鈧� is transformed, with nitrogen losing electrons to become NO鈧冣伝.
- The reduction half-reaction handles the transformation of MnO鈧勨伝 in KMnO鈧� to MnO鈧�, with manganese gaining electrons.
- By separately considering these half-reactions, you can balance the electron transfer before bringing them back together.
This division simplifies tracking and balancing electron flow, essential for correctly finalizing the chemical equation.
Electron Transfer
Electron transfer is at the heart of redox reactions, defining their very nature. In each reaction, one reactant gives up electrons (oxidation), while another gains them (reduction). Here's how to focus on electron flow:
By following the electrons' journey through a reaction, you gain a clearer understanding of the fundamental exchanges that drive chemical transformations.
- Identify the reducing agent, which is the species that donates electrons. In our case, NH鈧� acts as a reducing agent by losing electrons.
- The oxidizing agent, here KMnO鈧�, accepts these electrons, thus becoming reduced. Mn in KMnO鈧� accepts electrons and is reduced to MnO鈧�.
- The total number of electrons gained must equal the total lost to ensure the overall reaction is balanced.
By following the electrons' journey through a reaction, you gain a clearer understanding of the fundamental exchanges that drive chemical transformations.
