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Balance each of the following oxidation-reduction reactions by using the oxidation states method. a. \(\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{Mg}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{H}_{2}(g)\) c. \(\mathrm{Co}^{3+}(a q)+\mathrm{Ni}(s) \rightarrow \mathrm{Co}^{2+}(a q)+\mathrm{Ni}^{2+}(a q)\) d. \(\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)\)

Short Answer

Expert verified
The balanced reactions are: a. \[\mathrm{C}_{2}\mathrm{H}_{6}(g) + \frac{7}{2}\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g) + 3\mathrm{H}_{2}\mathrm{O}(g)\] b. \[\mathrm{Mg}(s) + 2\mathrm{HCl}(a q) \rightarrow \mathrm{Mg}^{2+}(a q) + 2\mathrm{Cl}^{-}(a q) + \mathrm{H}_{2}(g)\] c. \[2\mathrm{Co}^{3+}(a q) + 2\mathrm{Ni}(s) \rightarrow 2\mathrm{Co}^{2+}(a q) + 2\mathrm{Ni}^{2+}(a q)\] d. \[\mathrm{Zn}(s) + \mathrm{H}_{2}\mathrm{SO}_{4}(a q) \rightarrow \mathrm{ZnSO}_{4}(a q) + \mathrm{H}_{2}(g)\]

Step by step solution

01

1. Identify oxidation and reduction processes

In this reaction, carbon in \(\mathrm{C}_{2}\mathrm{H}_{6}\) gets oxidized to carbon dioxide \(\mathrm{CO}_{2}\), and oxygen in \(\mathrm{O}_{2}\) gets reduced to water \(\mathrm{H}_{2}\mathrm{O}\).
02

2. Determine the oxidation states of elements

The oxidation state of hydrogen is +1, carbon in ethane is -3, carbon in carbon dioxide is +4, and oxygen in oxygen gas and water is -2.
03

3. Balance the electrons gained and lost

Carbon goes from -3 to +4, losing 7 electrons in the process. Oxygen goes from 0 to -2, gaining 2 electrons. To balance the electrons, we need 2 carbon atoms and 7 oxygen atoms.
04

4. Balance the elements and overall reaction

The balanced reaction is: \[\mathrm{C}_{2}\mathrm{H}_{6}(g) + \frac{7}{2}\mathrm{O}_{2}(g) \rightarrow 2\mathrm{CO}_{2}(g) + 3\mathrm{H}_{2}\mathrm{O}(g)\] #b. Balancing \(\mathrm{Mg}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{H}_{2}(g)\)#
05

1. Identify oxidation and reduction processes

In this reaction, magnesium is oxidized to \(\mathrm{Mg}^{2+}\) and hydrogen in \(\mathrm{HCl}\) is reduced to \(\mathrm{H}_{2}\).
06

2. Determine the oxidation states of elements

The oxidation state of magnesium is 0, magnesium in \(\mathrm{Mg}^{2+}\) is +2, hydrogen in \(\mathrm{HCl}\) is +1, and hydrogen in \(\mathrm{H}_{2}\) is 0.
07

3. Balance the electrons gained and lost

Magnesium goes from 0 to +2, losing 2 electrons. Hydrogen goes from +1 to 0, gaining 1 electron. To balance the electrons, we need 2 hydrogen atoms.
08

4. Balance the elements and overall reaction

The balanced reaction is: \[\mathrm{Mg}(s) + 2\mathrm{HCl}(a q) \rightarrow \mathrm{Mg}^{2+}(a q) + 2\mathrm{Cl}^{-}(a q) + \mathrm{H}_{2}(g)\] #c. Balancing \(\mathrm{Co}^{3+}(a q)+\mathrm{Ni}(s) \rightarrow \mathrm{Co}^{2+}(a q)+\mathrm{Ni}^{2+}(a q)\)#
09

1. Identify oxidation and reduction processes

In this reaction, nickel gets oxidized to \(\mathrm{Ni}^{2+}\), and cobalt in \(\mathrm{Co}^{3+}\) gets reduced to \(\mathrm{Co}^{2+}\).
10

2. Determine the oxidation states of elements

The oxidation state of nickel is 0, nickel in \(\mathrm{Ni}^{2+}\) is +2, cobalt in \(\mathrm{Co}^{3+}\) is +3, and cobalt in \(\mathrm{Co}^{2+}\) is +2.
11

3. Balance the electrons gained and lost

Nickel goes from 0 to +2, losing 2 electrons. Cobalt goes from +3 to +2, gaining 1 electron. To balance the electrons, we need 2 cobalt atoms.
12

4. Balance the elements and overall reaction

The balanced reaction is: \[2\mathrm{Co}^{3+}(a q) + 2\mathrm{Ni}(s) \rightarrow 2\mathrm{Co}^{2+}(a q) + 2\mathrm{Ni}^{2+}(a q)\] #d. Balancing \(\mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g)\)#
13

1. Identify oxidation and reduction processes

In this reaction, zinc gets oxidized to \(\mathrm{ZnSO}_{4}\) and hydrogen in \(\mathrm{H}_{2}\mathrm{SO}_{4}\) gets reduced to \(\mathrm{H}_{2}\).
14

2. Determine the oxidation states of elements

The oxidation state of zinc is 0, zinc in \(\mathrm{ZnSO}_{4}\) is +2, hydrogen in \(\mathrm{H}_{2}\mathrm{SO}_{4}\) is +1, and hydrogen in \(\mathrm{H}_{2}\) is 0.
15

3. Balance the electrons gained and lost

Zinc goes from 0 to +2, losing 2 electrons. Hydrogen goes from +1 to 0, gaining 1 electron. To balance the electrons, we need 2 hydrogen atoms.
16

4. Balance the elements and overall reaction

The balanced reaction is: \[\mathrm{Zn}(s) + \mathrm{H}_{2}\mathrm{SO}_{4}(a q) \rightarrow \mathrm{ZnSO}_{4}(a q) + \mathrm{H}_{2}(g)\]

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Key Concepts

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

Oxidation States Method
Understanding the oxidation states method is essential for balancing redox reactions. This technique helps identify which elements undergo oxidation and reduction during a chemical reaction. Let's break it down:- **Oxidation** refers to the loss of electrons, resulting in an increase in oxidation state.- **Reduction** is the gain of electrons, leading to a decrease in oxidation state.Oxidation states provide a way to keep track of electron transfer. For instance, in the scenario involving ethane (\(\mathrm{C}_{2}\mathrm{H}_{6}\)), the carbon undergoes oxidation, moving from an oxidation state of -3 to +4. Meanwhile, oxygen goes from an oxidation state of 0 to -2 during its reduction, when forming \(\mathrm{H}_{2}\mathrm{O}\). By analyzing these changes, we can identify how many electrons are lost by one element and gained by another, paving the way to balance the reaction effectively.
Balancing Chemical Equations
Balancing chemical equations is crucial to maintain the law of conservation of mass, ensuring that atoms are neither created nor destroyed. In the case of redox reactions, this balance is not just about atoms, but also electrons. The steps include:1. **Identify each element's oxidation states** to understand which are oxidized and reduced.2. **Ensure that the number of electrons lost equals the number of electrons gained.** This means that if X electrons are transferred in oxidation, Y electrons must be transferred in reduction, and X must equal Y for the equation to balance.For example, in the reaction of magnesium (\(\mathrm{Mg}(s)\)) with hydrochloric acid (\(\mathrm{HCl}(aq)\)), one magnesium atom loses two electrons, while each hydrogen ion requires one electron to be reduced. Therefore, we need two \(\mathrm{HCl}\) molecules to accept the two electrons from magnesium, resulting in the balanced equation.
Electron Transfer in Reactions
Electron transfer is an integral part of redox reactions and is essentially the heart of these processes. When electrons are transferred from one substance to another, the substance that loses electrons is oxidized and the one that gains electrons is reduced. Here's how it manifests:- **Oxidizing agent**: The substance that gains electrons (and is reduced).- **Reducing agent**: The substance that loses electrons (and is oxidized).In the case with nickel (\(\mathrm{Ni}(s)\)) and cobalt ions (\(\mathrm{Co}^{3+}(aq)\)), nickel acts as the reducing agent by losing electrons to become \(\mathrm{Ni}^{2+}(aq)\), while cobalt serves as the oxidizing agent by gaining those electrons to change from \(\mathrm{Co}^{3+}(aq)\) to \(\mathrm{Co}^{2+}(aq)\). Recognizing these roles helps in understanding the electron flow and balancing reactions at a molecular level.
Reduction Processes
Reduction processes are vital for balancing redox reactions, focusing on the gain of electrons by a particular substance. This is the counterpart to oxidation, and they occur simultaneously in reactions.A reduction process can be identified by:- A decrease in the oxidation state of an element.- The addition of electrons to an atom or molecule.Take, for example, the reduction of hydrogen in sulfuric acid during the reaction with zinc (\(\mathrm{Zn}(s)\)). Here, hydrogen goes from an oxidation state of +1 in \(\mathrm{H}_{2}\mathrm{SO}_{4}(aq)\) to 0 in \(\mathrm{H}_{2}(g)\), indicating a gain of electrons. These electrons are provided by zinc, which is simultaneously oxidized. In every redox reaction, recognizing the reduction process helps balance both the charge and the atoms involved, ensuring the entire reaction equation is balanced correctly.

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Most popular questions from this chapter

A \(2.20-g\) sample of an unknown acid (empirical formula = \(\left.\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{3}\right)\) is dissolved in \(1.0 \mathrm{L}\) of water. A titration required \(25.0 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{NaOH}\) to react completely with all the acid present. Assuming the unknown acid has one acidic proton per molecule, what is the molecular formula of the unknown acid?

You made \(100.0 \mathrm{mL}\) of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.0 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take \(2.00 \mathrm{mL}\) of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 3.407 g. What was the concentration of the original lead(II)

How would you prepare 1.00 L of a \(0.50-M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" \((18 \mathrm{M})\) sulfuric acid b. HCl from "concentrated" (12 M) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO_ from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

What mass of \(\mathrm{Na}_{2} \mathrm{CrO}_{4}\) is required to precipitate all of the silver ions from \(75.0 \mathrm{mL}\) of a \(0.100-M\) solution of \(\mathrm{AgNO}_{3} ?\)

The units of parts per million (ppm) and parts per billion (ppb) are commonly used by environmental chemists. In general, I ppm means 1 part of solute for every \(10^{6}\) parts of solution. Mathematically, by mass: $$ \mathrm{ppm}=\frac{\mu \mathrm{g} \text { solute }}{\mathrm{g} \text { solution }}=\frac{\mathrm{mg} \text { solute }}{\mathrm{kg} \text { solution }} $$ In the case of very dilute aqueous solutions, a concentration of \(1.0 \mathrm{ppm}\) is equal to \(1.0 \mu \mathrm{g}\) of solute per \(1.0 \mathrm{mL},\) which equals 1.0 g solution. Parts per billion is defined in a similar fashion. Calculate the molarity of each of the following aqueous solutions. a. \(5.0 \mathrm{ppb} \mathrm{Hg}\) in \(\mathrm{H}_{2} \mathrm{O}\) b. \(1.0 \mathrm{ppb} \mathrm{CHCl}_{3}\) in \(\mathrm{H}_{2} \mathrm{O}\) c. \(10.0 \mathrm{ppm}\) As in \(\mathrm{H}_{2} \mathrm{O}\) d. \(0.10 \mathrm{ppm} \mathrm{DDT}\left(\mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}\right)\) in \(\mathrm{H}_{2} \mathrm{O}\)

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