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Chapter 18: Problem 24

Balance the following reactions, and specify which species is oxidized and which is reduced. (a) \(\mathrm{Na}+\mathrm{Hg}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{NaCl}+\mathrm{Hg}\) (b) \(\mathrm{HCl}+\mathrm{Zn} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2}\) (c) \(\mathrm{H}_{2}+\mathrm{CO}_{2} \rightarrow \mathrm{CO}+\mathrm{H}_{2} \mathrm{O}\)

Short Answer

Expert verified
(a) Na oxidized, Hg reduced; (b) Zn oxidized, HCl reduced; (c) H2 oxidized, CO2 reduced.

Step by step solution

01

Understanding Oxidation and Reduction

Oxidation refers to the loss of electrons by a species, while reduction refers to the gain of electrons. Identifying changes in oxidation states helps us determine which species are oxidized and which are reduced.
02

Balance Reaction (a)

The unbalanced reaction is \( \mathrm{Na} + \mathrm{Hg}_{2} \mathrm{Cl}_{2} \rightarrow \mathrm{NaCl} + \mathrm{Hg} \). Balance the elements on both sides:- Sodium: \( \mathrm{Na} \rightarrow \mathrm{NaCl} \).- Mercury: \( \mathrm{Hg}_{2}\rightarrow 2 \mathrm{Hg} \).- Chlorine: \( \mathrm{Cl}_{2} \rightarrow 2 \mathrm{Cl} \).Balanced equation: \( 2\,\mathrm{Na} + \mathrm{Hg}_{2} \mathrm{Cl}_{2} \rightarrow 2\,\mathrm{NaCl} + 2\,\mathrm{Hg} \).In this reaction, sodium (Na) is oxidized and mercury (Hg) in \(\mathrm{Hg}_2\mathrm{Cl}_2\) is reduced.
03

Balance Reaction (b)

The unbalanced reaction is \( \mathrm{HCl} + \mathrm{Zn} \rightarrow \mathrm{ZnCl}_{2} + \mathrm{H}_{2} \). Balance the elements:- Zinc: \( \mathrm{Zn} \rightarrow \mathrm{ZnCl}_2 \).- Chlorine: \( 2\,\mathrm{HCl} \rightarrow \mathrm{ZnCl}_2 \).- Hydrogen: \( 2\,\mathrm{HCl} \rightarrow \mathrm{H}_2 \).Balanced equation: \( 2\,\mathrm{HCl} + \mathrm{Zn} \rightarrow \mathrm{ZnCl}_2 + \mathrm{H}_2 \).Here, \(\mathrm{Zn}\) is oxidized and hydrogen in \(\mathrm{HCl}\) is reduced.
04

Balance Reaction (c)

The unbalanced reaction is \( \mathrm{H}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{CO} + \mathrm{H}_{2}\mathrm{O} \). Balance the elements:- Carbon: already balanced \( \mathrm{CO}_{2} \rightarrow \mathrm{CO} \).- Hydrogen: \( \mathrm{H}_{2} \rightarrow \mathrm{H}_{2}\mathrm{O} \).Balanced equation (already balanced): \( \mathrm{H}_{2} + \mathrm{CO}_{2} \rightarrow \mathrm{CO} + \mathrm{H}_{2}\mathrm{O} \).Here, hydrogen in \(\mathrm{H}_2\) is oxidized, and carbon in \(\mathrm{CO}_2 \) is reduced.
05

Conclusion

In (a), \(\mathrm{Na}\) is oxidized and \(\mathrm{Hg}_2\mathrm{Cl}_2\)'s \(\mathrm{Hg}\) is reduced. In (b), \(\mathrm{Zn}\) is oxidized and \(\mathrm{HCl}\)'s \(\mathrm{H}\) is reduced. In (c), \(\mathrm{H}_2\) is oxidized and \(\mathrm{CO}_2\)'s \(\mathrm{C}\) is reduced.

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Key Concepts

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

Understanding Oxidation
Oxidation is a fundamental concept in redox reactions that refers to the loss of electrons from a chemical species. When a substance undergoes oxidation, its oxidation state increases because it is losing negatively charged electrons. For example, when sodium (\(\mathrm{Na}\)) reacts in a chemical equation, it often loses an electron to become \(\mathrm{Na}^+\).
  • Oxidation is characterized by an increase in oxidation state.
  • The substance that loses electrons is known as the reducing agent because it "donates" electrons.
During a reaction, identify the element that has lost electrons by looking for an increase in its oxidation state from reactants to products. In our exercises:
  • In reaction (a), sodium (\(\mathrm{Na}\)) is oxidized as it goes from elemental \(\mathrm{Na}\) to \(\mathrm{Na}^+\).
  • In reaction (b), zinc (\(\mathrm{Zn}\)) is oxidized as it changes from \(\mathrm{Zn}\) to \(\mathrm{Zn}^{2+}\).
  • In reaction (c), the hydrogen in \(\mathrm{H}_2\) is oxidized as it is transformed into water (\(\mathrm{H}_2\mathrm{O}\)).
Recognizing oxidation is crucial to balancing redox equations and understanding how chemical reactions occur on a molecular level.
The Nature of Reduction
Reduction is the opposite of oxidation and involves the gain of electrons by a chemical species. During reduction, a substance's oxidation state decreases because it receives extra electrons. For example, mercury in \(\mathrm{Hg}_2\mathrm{Cl}_2\) gains electrons when it is reduced to \(\mathrm{Hg}\).
  • Reduction is marked by a decrease in oxidation state.
  • The substance that gains electrons is considered the oxidizing agent because it "accepts" electrons.
Examining redox reactions helps us identify which species is reduced by observing a decrease in the oxidation state from reactants to products. In the given reactions:
  • In reaction (a), mercury (\(\mathrm{Hg}\)) in \(\mathrm{Hg}_2\mathrm{Cl}_2\) is reduced as it gains electrons.
  • In reaction (b), the hydrogen in hydrochloric acid (\(\mathrm{HCl}\)) is reduced as it forms \(\mathrm{H}_2\) gas.
  • In reaction (c), carbon in \(\mathrm{CO}_2\) is reduced as it loses an oxygen atom to form \(\mathrm{CO}\).
Understanding reduction processes is essential for predicting the outcome of redox reactions and balancing equations accurately.
Mastering Balancing Chemical Equations
Balancing chemical equations is a critical skill in chemistry that ensures that the number of each type of atom is conserved across a chemical reaction. In any chemical equation, the same number of each type of atom must appear on both sides of the equation. This balance reflects the law of conservation of mass.
Balancing redox reactions involves ensuring that both the mass and charge are balanced. When approaching a redox equation:
  • Balance elements that appear in only one reactant and one product first.
  • Balance the remaining atoms, usually starting with oxygen and hydrogen.
  • Ensure that charges are balanced if dealing with ionic species.
In our sample reactions:
  • For reaction (a), merely adjusting coefficients balances \(\mathrm{Na}\) and \(\mathrm{Hg}\).
  • In reaction (b), adding a coefficient of 2 in front of \(\mathrm{HCl}\) balances both \(\mathrm{Cl}\) and \(\mathrm{H}_2\)
  • Reaction (c) is already balanced because each side has the same number of each atom present.
Mastering the balancing of chemical equations facilitates a deeper understanding of chemical reactions, ensuring they adhere to fundamental laws and behave predictably in applications.

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Most popular questions from this chapter

What is the voltage of a concentration cell of \(\mathrm{Cl}^{-}\) ions where the concentrations are 1.045 and \(0.085 \mathrm{M}\) ? What is the spontaneous reaction?

The standard free energy change at \(25^{\circ} \mathrm{C}, \Delta G^{\circ},\) is equal to \(-34.3 \mathrm{~kJ}\) for $$ 2 \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{~s}) $$ Calculate the standard potential for this reaction.

A Consider the standard reduction potentials of cesium and lithium. $$ \begin{array}{ll} \mathrm{Cs}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Cs}(\mathrm{s}) & E^{\circ}=-3.026 \mathrm{~V} \\\ \mathrm{Li}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Li}(\mathrm{s}) & E^{\circ}=-3.095 \mathrm{~V} \end{array} $$ The periodic trends in the properties of the element indicate that fluorine is the most chemically reactive nonmetal, so perhaps it is not surprising that the standard reduction potential of fluorine has the highest positive value for a nonmetallic element. However, periodic properties of the elements also indicate that cesium should be the most reactive metal. Comparison of the voltage of the cesium half-reaction with that of lithium shows that the standard reduction potential of lithium is less negative than that of cesium, indicating that lithium is a better oxidizer than is cesium. (a) Calculate the standard cell voltages of the voltaic cells based on the reaction $$ 2 \mathrm{M}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{M}^{+}(\mathrm{aq})+2 \mathrm{~F}^{-}(\mathrm{aq}) $$ where \(\mathrm{M}\) is \(\mathrm{Cs}\) and \(\mathrm{Li}\). (b) Assuming that the pressure of \(\mathrm{F}_{2}(\mathrm{~g})\) stays at \(1.00 \mathrm{~atm}\), what concentration does \(\mathrm{Li}^{+}(\mathrm{aq})\) have to be for the voltage of the \(\mathrm{Li} / \mathrm{F}_{2}\) voltaic cell to equal the standard voltage of the \(\mathrm{Cs} / \mathrm{F}_{2}\) voltaic cell? (c) Can you suggest a reason why the standard reduction potential of lithium is lower than that of cesium, even though periodic trends indicate that cesium is the more reactive metal? (d) Calculate \(\Delta G^{o}\) for both the \(\mathrm{Li} / \mathrm{F}_{2}\) and the \(\mathrm{Cs} / \mathrm{F}_{2}\) voltaic cells from their \(E^{\circ} \mathrm{s}\). Compare this with the Gibbs' free energies of formation of \(2 \mathrm{~mol} \mathrm{LiF}\) and CsF. Can you explain the difference? (e) Given the fact that alkali metals react rather violently with water, it would be unlikely that any voltaic cell can be constructed using Li(s) or \(\mathrm{Cs}(\mathrm{s})\) in the presence of water. A more likely scenario is that the voltaic cell would have no solvent, so that the voltaic cell reaction would be $$ 2 \mathrm{M}(\mathrm{s})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{MF}(\mathrm{xtal}) $$ where \(\mathrm{M}\) is \(\mathrm{Cs}\) or \(\mathrm{Li}\). What would be the \(E^{\circ}\) s of the two different voltaic cells if this were the reaction? (Hint: See your answer to part d.)

Complete and balance each half-reaction in acid solution, and identify it as an oxidation or a reduction. (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightarrow \mathrm{NO}_{3}^{-}(\mathrm{aq})\) (b) \(\mathrm{Mn}^{3+}(\mathrm{aq}) \rightarrow \mathrm{MnO}_{4}^{-}(\mathrm{aq})\) (c) \(\mathrm{HOCl}(\mathrm{aq}) \rightarrow \mathrm{ClO}_{3}^{-}(\mathrm{aq})\)

Like zinc, sodium is a rather active metal. Would it be possible to use metallic sodium for cathodic protection of the iron hull of an ocean vessel? Explain.
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