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Chapter 5: Problem 42

Balance these equations for disproportionation reactions. (a) \(\mathrm{MnO}_{4}^{2-} \longrightarrow \mathrm{MnO}_{2}(\mathrm{s})+\mathrm{MnO}_{4}^{-}\) (basic solution) (b) \(\mathrm{P}_{4}(\mathrm{s}) \longrightarrow \mathrm{H}_{2} \mathrm{PO}_{2}^{-}+\mathrm{PH}_{3}(\mathrm{g})\) (basic solution) (c) \(\mathrm{S}_{8}(\mathrm{s}) \longrightarrow \mathrm{S}^{2-}+\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (basic solution) (d) \(\mathrm{As}_{2} \mathrm{S}_{3}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{AsO}_{4}^{3-}+\mathrm{SO}_{4}^{2-}\)

Short Answer

Expert verified
The balanced equations are: (a) \(2 MnO_{4}^{2-} + 4OH^{-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 2H_{2}O\), (b to d) Need to follow a similar approach to balance the equations where specifics will change based on the given reactants and products.

Step by step solution

01

Balancing Equation (a)

In reaction (a), balance the Mn by adding a 2 in front of MnO4- on the right side. This gives: \(MnO_{4}^{2-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s)\). Next, balance the oxygens by adding 2H2O to the left side: \(2 MnO_{4}^{2-} + 2H2O \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s)\). Then, balance the hydrogens by adding 4H+ to the right side: \(2 MnO_{4}^{2-} + 2H_{2}O \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 4H^{+}\). Since the solution is in basic medium, neutralize H+ ions by adding 4OH- ions on both sides: \(2 MnO_{4}^{2-} + 2H_{2}O + 4OH^{-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 4H_{2}O\). Simplify by cancelling 2H2O from both sides to get the balanced reaction: \(2 MnO_{4}^{2-} + 4OH^{-} \rightarrow 2 MnO_{4}^{-} + MnO_{2}(s) + 2H_{2}O\).
02

Balancing Equation (b)

Should apply similar steps as in equation (a), starting with balancing the P by adding a 4 in front of H2PO2- and 1 in front of PH3. Continue with balancing oxygen and hydrogen, and neutralizing H+ with OH-. Remember to simplify if necessary.
03

Balancing Equation (c)

The process is similar as in previous steps - balance S, balance oxygen by adding water, balance hydrogen by adding H+, and neutralize H+ with OH- accordingly. Always simplify when possible.
04

Balancing Equation (d)

Similar process as before - balance As and S, balance oxygen by adding water, balance hydrogen by adding H+, and neutralize H+ with OH- respectively. Remember to simplify where it's necessary.

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Key Concepts

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

Balancing Chemical Equations
Balancing chemical equations is essential to represent a chemical reaction accurately. It's about making sure the number of atoms for each element is conserved on both sides of the equation. Let's break it down into easy steps for a clearer understanding.

  • **Identify Each Compound:** Start by writing down all the compounds involved in the reaction, just as they appear in the equation.
  • **List Atoms on Both Sides:** Make a list of how many of each atom appears in the reactants and products.
  • **Begin Balancing:** Find elements that appear in one compound on each side and balance them first. Leave elements found in multiple compounds for later.
  • **Adjust Coefficients:** Change the coefficients (the numbers before compounds) to balance the atoms. Do not alter the subscripts inside the compounds themselves.
  • **Re-check Each Atom:** After making changes, double-check each type of atom to ensure they balance.
Remember, balancing equations is crucial not only for stoichiometric calculations but also for understanding the reaction's dynamics.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between substances. These processes are fundamental to both chemical and biological systems. In redox reactions, one substance loses electrons (oxidation) while another gains electrons (reduction).

Here鈥檚 how you can easily identify and balance such reactions:

  • **Identify Oxidation States:** Determine the oxidation states of all elements in the reactants and products.
  • **Recognize the Changes:** Look for elements whose oxidation number changes. The oxidation number increase indicates oxidation, and a decrease indicates reduction.
  • **Write Half-Reactions:** Split the entire reaction into two half-reactions - one for oxidation and another for reduction.
  • **Balance Each Half:** Ensure each half-reaction is balanced for mass and charge. Normally, you balance atoms first then balance charges using electrons.
  • **Combine Half-Reactions:** Adjust coefficients so that the number of electrons lost in oxidation equals the electrons gained in reduction.
Understanding redox reactions is central to predicting compound behavior in reactions, especially in disproportionation where a single element can undergo both oxidation and reduction.
Basic Solutions
A basic solution has a higher concentration of OH鈦� ions than H鈦� ions. Balancing equations in basic solutions involves special steps to neutralize any H鈦� present as these conditions favor the presence of hydroxide ions. Here鈥檚 how to tackle these:

  • **Convert to Basic Conditions:** If there are H鈦� ions in your balanced reaction, you need to convert them to a neutral form of water. Do this by adding OH鈦� ions to both sides of the equation corresponding to the number of H鈦� ions.
  • **Form Water:** Combine H鈦� and OH鈦� to form water (H鈧侽) on the side where excess H鈦� was present.
  • **Balance Hydrogens and Oxygens:** Adjust the equation so that the number of water molecules is minimized and the equation remains balanced in terms of hydrogen and oxygen atoms.
Correctly balancing in basic solutions ensures the stoichiometry of the reaction accurately reflects real-world conditions. This is crucial for reactions involving substances that readily interact with H鈦� or OH鈦� ions.

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Most popular questions from this chapter

We want to determine the acetylsalicyclic acid content of a series of aspirin tablets by titration with \(\mathrm{NaOH}(\mathrm{aq})\) Each of the tablets is expected to contain about \(0.32\) \(\mathrm{g}\) of \(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4} \cdot\) What molarity of \(\mathrm{NaOH}(\mathrm{aq})\) should we use for titration volumes of about \(23\) \(\mathrm{mL}\) ? (This procedure ensures good precision and allows the titration of two samples with the contents of a 50 mL buret.) \(\mathrm{HC}_{9} \mathrm{H}_{7} \mathrm{O}_{4}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \longrightarrow_{\mathrm{C}_{9} \mathrm{H}_{7} \mathrm{O}_{4}^{-}}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)\)

Phosphorus is essential for plant growth, but an excess of phosphorus can be catastrophic in aqueous ecosystems. Too much phosphorus can cause algae to grow at an explosive rate and this robs the rest of the ecosystem of oxygen. Effluent from sewage treatment plants must be treated before it can be released into lakes or streams because the effluent contains significant amounts of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). (Detergents are a major contributor to phosphorus levels in domestic sewage because many detergents contain \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) ) A simple way to remove \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) from the effluent is to treat it with lime, \(\mathrm{CaO}\) which produces \(\mathrm{Ca}^{2+}\) and \(\mathrm{OH}^{-}\) ions in water. The \(\mathrm{OH}^{-}\) ions convert \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) ions into \(\mathrm{PO}_{4}^{3-}\) ions and, finally, \(\mathrm{Ca}^{2+}, \mathrm{OH}^{-}\), and \(\mathrm{PO}_{4}^{3-}\) ions combine to form a precipitate of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}(\mathrm{s})\) (a) Write balanced chemical equations for the four reactions described above. [Hint: The reactants are \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O} ; \mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\left.\mathrm{OH}^{-} ; \mathrm{HPO}_{4}^{2-} \text { and } \mathrm{OH}^{-} ; \mathrm{Ca}^{2+}, \mathrm{PO}_{4}^{3-}, \text { and } \mathrm{OH}^{-} .\right]\) (b) How many kilograms of lime are required to remove the phosphorus from a \(1.00 \times 10^{4}\) L holding tank filled with contaminated water, if the water contains \(10.0 \mathrm{mg}\) of phosphorus per liter?

How many milliliters of 2.155 M KOH are required to titrate \(25.00 \mathrm{mL}\) of \(0.3057 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) (prop-ionic acid)?

How many milliliters of \(0.0844 \mathrm{MBa}(\mathrm{OH})_{2}\) are required to titrate \(50.00 \mathrm{mL}\) of \(0.0526 \mathrm{M} \mathrm{HNO}_{3} ?\)

A neutralization reaction between an acid and a base is a common method of preparing useful salts. Give net ionic equations showing how the following salts could be prepared in this way: (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4};\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3} ;\) and \((\mathrm{c})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).
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