Chlorine gas was first prepared in 1774 by C. W. Scheele by oxidizing sodium chloride with manganese(IV) oxide. The reaction is \(\begin{aligned} \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) \longrightarrow & \mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g) \end{aligned}\) Balance this equation.

Short Answer

Expert verified
The balanced chemical equation is: \(2\text{NaCl}(aq)+\text{H}_{2}\text{SO}_{4}(aq)+\text{MnO}_{2}(s) \longrightarrow \text{MnCl}_{2}(aq)+4\text{H}_{2}\text{O}(l)+\text{Cl}_{2}(g)\)

Step by step solution

01

Identify the species in the given equation

: Reactants: NaCl (aq), H2SO4 (aq), and MnO2 (s) Products: MnCl2 (aq), H2O (l), and Cl2 (g)
02

Balance the chlorine atoms

: The chlorine atom occurs as follows in the equation: Reactant side: NaCl Product side: MnCl2, Cl2 There is only 1 chlorine atom on the left side, while there are 3 chlorine atoms on the right side. To balance the chlorines, we add a coefficient of 2 before NaCl: \(2\text{NaCl}(aq)+\text{H}_{2}\text{SO}_{4}(aq)+\text{MnO}_{2}(s) \longrightarrow \text{MnCl}_{2}(aq)+\text{H}_{2}\text{O}(l)+\text{Cl}_{2}(g)\) Now, we have 2 chlorine atoms on each side of the equation.
03

Balance the hydrogen atoms

: The hydrogen atom occurs as follows in the equation: Reactant side: H2SO4 Product side: H2O There are 2 hydrogen atoms on the left side and 2 hydrogen atoms on the right side. So the hydrogen atoms balanced.
04

Balance the sodium atoms

: The sodium atom occurs as follows in the equation: Reactant side: 2NaCl (due to the coefficient of 2) Product side: none To balance the sodium atoms, there's no need to make any changes as there are 2 sodium atoms on both sides of the equation.
05

Balance the manganese atoms

: The manganese atom occurs as follows in the equation: Reactant side: MnO2 Product side: MnCl2 There is 1 manganese atom on both sides of the equation. The manganese atoms are already balanced.
06

Balance the oxygen atoms

: The oxygen atom occurs as follows in the equation: Reactant side: H2SO4, MnO2 Product side: H2O There are 4 + 2 = 6 oxygen atoms on the left side (4 in H2SO4 and 2 in MnO2) and 1 oxygen atom on the right side (H2O). To balance the oxygen atoms, add a coefficient of 4 before H2O on the product side: \(2\text{NaCl}(aq)+\text{H}_{2}\text{SO}_{4}(aq)+\text{MnO}_{2}(s) \longrightarrow \text{MnCl}_{2}(aq)+4\text{H}_{2}\text{O}(l)+\text{Cl}_{2}(g)\) Now, we have 6 oxygen atoms on both sides of the equation.
07

Check the balanced equation

: Reactants: 2 Na, 6 Cl, 2 H, 1 Mn, 4 O Products: 2 Na, 6 Cl, 2 H, 1 Mn, 4 O The balanced chemical equation is: \(2\text{NaCl}(aq)+\text{H}_{2}\text{SO}_{4}(aq)+\text{MnO}_{2}(s) \longrightarrow \text{MnCl}_{2}(aq)+4\text{H}_{2}\text{O}(l)+\text{Cl}_{2}(g)\)

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Key Concepts

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

Chemical Reaction
A chemical reaction is a process where reactants are transformed into new substances, known as products. During this transformation, chemical bonds are broken and new bonds are formed. The chlorine gas preparation you've encountered in your exercise is a classic example of a chemical reaction, where chlorine gas is produced through the interaction of sodium chloride, sulfuric acid, and manganese(IV) oxide.

To understand a chemical reaction, it's crucial to look at the substances involved before and after the reaction. For our exercise, the reactants are NaCl (aqueous), H2SO4 (aqueous), and MnO2 (solid), while the products are MnCl2 (aqueous), H2O (liquid), and Cl2 (gas). The balancing of this equation requires careful accounting of each atom type on both sides of the equation to ensure they are equal, reflecting the conservation of matter.
Stoichiometry
Stoichiometry is the quantitative relationship between the amounts of reactants and products in a chemical reaction. It's based on the principle that matter cannot be created or destroyed in chemical reactions (the law of conservation of mass). When resolving stoichiometric problems, such as balancing the provided chemical equation, one must ensure that the number of atoms for each element involved in the reaction is the same on both sides of the equation.

In our exercise, stoichiometry was used to determine the correct coefficients for reactants and products. This ensures that the amounts of sodium, chlorine, hydrogen, manganese, and oxygen atoms are conserved from reactants to products. The balancing steps you went through systematically adjusted coefficients to achieve this conservation. The step-by-step process exemplifies stoichiometrical calculations that are essential in predicting yields of reactions and understanding the proportions of substances required.
Oxidation-Reduction Reactions
Oxidation-reduction (redox) reactions are a subset of chemical reactions involving the transfer of electrons between two species. Oxidation is the loss of electrons, while reduction is the gain of electrons. These reactions play a significant role in chemistry, including biological processes, energy storage, and corrosion.

In the context of our exercise, the preparation of chlorine gas can be viewed as a redox reaction. Manganese(IV) oxide (MnO2) acts as an oxidizing agent and loses electrons (is reduced) while sodium chloride (NaCl) is the source of chlorine which is oxidized to form chlorine gas (Cl2). The successful balancing of the equation reflects the nuances of redox balancing, where attention must be paid to both mass and charge conservation. Balancing redox reactions involves additional steps compared to other reactions because both the number of atoms and the charges must balance, which adds another layer of complexity to the stoichiometric calculations.

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Most popular questions from this chapter

Sketch the galvanic cells based on the following half-reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are \(1.0 M\) and that all partial pressures are \(1.0 \mathrm{~atm}\). a. \(\mathrm{Cl}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=1.36 \mathrm{~V}\) \(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-} \quad \mathscr{8}^{\circ}=1.09 \mathrm{~V}\) b. \(\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{C}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{IO}_{4}^{-}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{IO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.60 \mathrm{~V}\)

A silver concentration cell is set up at \(25^{\circ} \mathrm{C}\) as shown below:The \(\mathrm{AgCl}(s)\) is in excess in the left compartment. a. Label the anode and cathode, and describe the direction of the electron flow. b. Determine the value of \(K_{\text {sp }}\) for \(\mathrm{AgCl}\) at \(25^{\circ} \mathrm{C}\).

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains \(1.0 M \mathrm{M}^{2+}\). Solution B in the other cell compartment has a volume of \(1.00 \mathrm{~L}\). At the beginning of the experiment \(0.0100\) mole of \(\mathrm{M}\left(\mathrm{NO}_{3}\right)_{2}\) and \(0.0100\) mole of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) are dissolved in solution \(\mathrm{B}\) (ignore volume changes), where the reaction $$ \mathrm{M}^{2+}(a q)+\mathrm{SO}_{4}{ }^{2-}(a q) \rightleftharpoons \mathrm{MSO}_{4}(s) $$ occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be \(0.44 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). Assume that the process $$ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} $$ has a standard reduction potential of \(-0.31 \mathrm{~V}\) and that no other redox process occurs in the cell. Calculate the value of \(K_{\text {sp }}\) for \(\operatorname{MSO}_{4}(s)\) at \(25^{\circ} \mathrm{C}\).

Consider the standard galvanic cell based on the following half-reactions: $$ \begin{aligned} \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\ \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag} \end{aligned} $$

A fuel cell designed to react grain alcohol with oxygen has the following net reaction: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ The maximum work that 1 mole of alcohol can do is \(1.32 \times\) \(10^{3} \mathrm{~kJ}\). What is the theoretical maximum voltage this cell can achieve at \(25^{\circ} \mathrm{C} ?\)

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