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Chapter 4: Problem 16

In balancing oxidation–reduction reactions, why is it permissible to add water to either side of the equation?

Short Answer

Expert verified
It is permissible to add water to either side of the equation when balancing oxidation-reduction reactions because water is a common species in redox reactions and does not interfere with the original reaction. Adding water helps balance the hydrogen and oxygen atoms and maintains charge neutrality when balancing the half-reactions separately. This ensures that the final balanced redox reaction contains an equal amount of oxidizing and reducing agents, and the number of electrons lost during oxidation equals the number of electrons gained during reduction.

Step by step solution

01

Write the given redox reaction and identify oxidation and reduction

Write the given redox reaction, and identify which substances are undergoing oxidation (loss of electrons) and reduction (gain of electrons).
02

Split the reaction into half-reactions

Divide the redox reaction into two separate half-reactions for oxidation and reduction. Write each half-reaction with all the reactants and products involved in oxidation and reduction processes.
03

Balance the half-reactions separately

Balance each half-reaction, first by balancing the atoms other than hydrogen and oxygen, and then balancing hydrogen and oxygen atoms using water molecules. In other words, we add water molecules to either side of the half-reactions, adjusting the number of H and O atoms accordingly. Adding water in this step is permissible because water is a common species in redox reactions and forms naturally during the transfer of electrons. The presence of water does not interfere with the original reaction, as it often functions as a solvent or a combination of H⁺ and OH⁻ ions that ensure charge neutrality.
04

Balance the charges in each half-reaction

Balance the charges in each half-reaction by adding appropriate numbers of electrons to the side that needs them. This ensures that the net charges on each side of the half-reaction are equal.
05

Combine the balanced half-reactions

Combine the balanced half-reactions by multiplying each half-reaction by an appropriate integer to ensure that the number of electrons lost during oxidation equals the number of electrons gained during reduction. Add the half-reactions together and simplify by canceling out any ions or molecules that appear on both sides of the equation. The balanced redox reaction will now contain an equal amount of oxidizing and reducing agents, along with any added water molecules required for balancing the atoms and charges in the equation. Since the water molecules only serve to balance the equation and maintain charge neutrality, it is permissible to add them during the balancing process.

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Most popular questions from this chapter

Balance the following oxidation–reduction reactions that occur in acidic solution using the half-reaction method. a. $\mathrm{Cu}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cu}^{2+}(a q)+\mathrm{NO}(g)$ b. $\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cl}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Cl}_{2}(g)$ c. $\mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{PbSO}_{4}(s)$ d. $\mathrm{Mn}^{2+}(a q)+\mathrm{NaBiO}_{3}(s) \rightarrow \mathrm{Bi}^{3+}(a q)+\mathrm{MnO}_{4}-(a q)$ e. $\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)$

Gold metal will not dissolve in either concentrated nitric acid or concentrated hydrochloric acid. It will dissolve, however, in aqua regia, a mixture of the two concentrated acids. The products of the reaction are the AuCl, \(^{-}\) ion and gaseous NO. Write a balanced equation for the dissolution of gold in aqua regia.

Zinc and magnesium metal each react with hydrochloric acid according to the following equations: $$\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ $$\mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{MgCl}_{2}(a q)+\mathrm{H}_{2}(g)$$ A 10.00-g mixture of zinc and magnesium is reacted with the stoichiometric amount of hydrochloric acid. The reaction mixture is then reacted with 156 mL of 3.00 M silver nitrate to produce the maximum possible amount of silver chloride. a. Determine the percent magnesium by mass in the original mixture. b. If 78.0 mL of HCl was added, what was the concentration of the HCl?

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) level can be determined by tirrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The unbalanced redox equation is $$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{CO}_{2}(g)$$ If 31.05 \(\mathrm{mL}\) of 0.0600\(M\) potassium dichromate solution is required to titrate 30.0 \(\mathrm{g}\) of blood plasma, determine the mass percent of alcohol in the blood.

When organic compounds containing sulfur are bumed, sulfur dioxide is produced. The amount of \(\mathrm{SO}_{2}\) formed can be determined by the reaction with hydrogen peroxide: $$\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q)$$ The resulting sulfuric acid is then titrated with a standard NaOH solution. A 1.302 -g sample of coal is burned and the \(\mathrm{SO}_{2}\) is collected in a solution of hydrogen peroxide. It took 28.44 \(\mathrm{mL}\) of a $0.1000-M \mathrm{NaOH}$ solution to titrate the resulting sulfuric acid. Calculate the mass percent of sulfur in the coal sample. Sulfuric acid has two acidic hydrogens.
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