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Chapter 18: Problem 21

Balance the following oxidation鈥搑eduction reactions that occur in acidic solution using the half-reaction method. a. \(\mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{NO}(g)\) b. \(\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Ce}^{4+}(a q) \rightarrow \mathrm{CO}_{2}(a q)+\mathrm{Ce}^{3+}(a q)\) c. \(\mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)\)

Short Answer

Expert verified
The balanced redox reactions for each given reaction are: a. Cr(s) + 3NO3鈦(aq) + 12H鈦(aq) 鉄 Cr鲁鈦(aq) + 3NO(g) + 6H2O(l) b. CH3OH(aq) + 4H鈦(aq) + 6Ce4鈦(aq) 鉄 CO2(aq) + H2O(l) + 6Ce3鈦(aq) c. 5SO3虏鈦(aq) + 10H鈦(aq) + 5MnO4鈦(aq) + 40H鈦(aq) 鉄 5SO4虏鈦(aq) + 10e鈦 + 5Mn虏鈦(aq) + 20H2O(l)

Step by step solution

01

Assign initial oxidation states

Using the rules for oxidation numbers, we find: Cr: 0 N in NO3鈦: +5 Cr鲁鈦: +3 N in NO: +2
02

Identify elements undergoing oxidation and reduction

Cr is being oxidized, from 0 to +3. N is being reduced, from +5 to +2.
03

Write down the unbalanced half-reactions

Oxidation half-reaction: Cr(s) 鉄 Cr鲁鈦(aq) Reduction half-reaction: NO3鈦(aq) 鉄 NO(g)
04

Balance the atoms in each half-reaction

Oxidation half-reaction is already balanced in terms of atoms. Reduction half-reaction should be balanced for oxygen atoms: NO3鈦(aq) 鉄 NO(g) + 2H2O(l) Now balance the hydrogen atoms: NO3鈦(aq) + 4H鈦(aq) 鉄 NO(g) + 2H2O(l)
05

Balance the charges in each half-reaction

Oxidation half-reaction: Cr(s) 鉄 Cr鲁鈦(aq) + 3e鈦 (add 3e鈦 to the right side) Reduction half-reaction is already balanced in terms of charge.
06

Multiply and add the half-reactions

The oxidation half-reaction has 3e鈦, and the reduction half-reaction has 1e鈦, so we'll multiply the reduction half-reaction by 3 to get equal electrons transferred. Oxidation half-reaction: Cr(s) 鉄 Cr鲁鈦(aq) + 3e鈦 Reduction half-reaction x 3: 3NO3鈦(aq) + 12H鈦(aq) 鉄 3NO(g) + 6H2O(l) Now add the half-reactions: Cr(s) + 3NO3鈦(aq) + 12H鈦(aq) 鉄 Cr鲁鈦(aq) + 3NO(g) + 6H2O(l)
07

Confirm that atoms and charges are balanced

We find that the number of atoms (Cr, N, O, and H) and charges are balanced in the combined equation. b. $\mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Ce}^{4+}(a q) \rightarrow \mathrm{CO}_{2}(a q)+\mathrm{Ce}^{3+}(a q)$ Repeat the steps from part a: 1. Assign initial oxidation states: C in CH3OH: -2 O in CH3OH: -2 C in CO2: +4 Ce4鈦: +4 Ce3鈦: +3 2. Identify elements undergoing oxidation and reduction: C is being oxidized, from -2 to +4. Ce is being reduced, from +4 to +3. 3. Write down the unbalanced half-reactions: Oxidation half-reaction: CH3OH(aq) 鉄 CO2(aq) Reduction half-reaction: Ce4鈦(aq) 鉄 Ce3鈦(aq) 4. Balance the atoms in each half-reaction: Oxidation half-reaction: CH3OH(aq) 鉄 CO2(aq) + H2O(l) (balance O) Reduction half-reaction: already balanced 5. Balance the charges in each half-reaction: Oxidation half-reaction: CH3OH(aq) + 4H鈦(aq) 鉄 CO2(aq) + H2O(l) + 6e鈦 (balance charges) Reduction half-reaction: Ce4鈦(aq) + e鈦 鉄 Ce3鈦(aq) 6. Multiply and add the half-reactions: Oxidation half-reaction: CH3OH(aq) + 4H鈦(aq) 鉄 CO2(aq) + H2O(l) + 6e鈦 Reduction half-reaction x 6: 6Ce4鈦(aq) + 6e鈦 鉄 6Ce3鈦(aq) Now add the half-reactions: CH3OH(aq) + 4H鈦(aq) + 6Ce4鈦(aq) 鉄 CO2(aq) + H2O(l) + 6Ce3鈦(aq) 7. Confirm that atoms and charges are balanced. c. $\mathrm{SO}_{3}^{2-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{SO}_{4}^{2-}(a q)+\mathrm{Mn}^{2+}(a q)$ Repeat the steps from part a: 1. Assign initial oxidation states: S in SO3虏鈦: +4 O in SO3虏鈦: -2 S in SO4虏鈦: +6 Mn in MnO4鈦: +7 Mn虏鈦: +2 2. Identify elements undergoing oxidation and reduction: S is being oxidized, from +4 to +6. Mn is being reduced, from +7 to +2. 3. Write down the unbalanced half-reactions: Oxidation half-reaction: SO3虏鈦(aq) 鉄 SO4虏鈦(aq) Reduction half-reaction: MnO4鈦(aq) 鉄 Mn虏鈦(aq) 4. Balance the atoms in each half-reaction: Oxidation half-reaction is balanced. Reduction half-reaction: MnO4鈦(aq) 鉄 Mn虏鈦(aq) + 4H2O(l) (balance O) Add 8H鈦 on left side: MnO4鈦(aq) + 8H鈦(aq) 鉄 Mn虏鈦(aq) + 4H2O(l) (balance H) 5. Balance the charges in each half-reaction: Oxidation half-reaction: SO3虏鈦(aq) + 2H鈦(aq) 鉄 SO4虏鈦(aq) + 2e鈦 (balance charges) Reduction half-reaction is balanced. 6. Multiply and add the half-reactions: Oxidation half-reaction x 5: 5SO3虏鈦(aq) + 10H鈦(aq) 鉄 5SO4虏鈦(aq) + 10e鈦 Reduction half-reaction x 5: 5MnO4鈦(aq) + 40H鈦(aq) 鉄 5Mn虏鈦(aq) + 20H2O(l) Now add the half-reactions: 5SO3虏鈦(aq) + 10H鈦(aq) + 5MnO4鈦(aq) + 40H鈦(aq)鉄 5SO4虏鈦(aq) + 10e鈦 + 5Mn虏鈦(aq) + 20H2O(l) 7. Confirm that atoms and charges are balanced.

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Key Concepts

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

Half-Reaction Method
The half-reaction method is a systematic approach to balancing oxidation-reduction reactions, or redox reactions. In this method, we split the overall reaction into two separate parts: one for oxidation and one for reduction.
Each half-reaction is treated independently and balanced for both mass and charge. By doing so, we can clearly see how electrons are transferred between species. This approach simplifies complex reactions and helps ensure accuracy in balancing.
  • Oxidation Half-Reaction: Involves loss of electrons.
  • Reduction Half-Reaction: Involves gain of electrons.
By balancing each half separately and combining them, the overall chemical equation is ensured to obey the law of conservation of mass and charge.
Acidic Solution Balancing
Balancing redox reactions in acidic solutions requires special attention to hydrogen and oxygen atoms. In such conditions, we achieve balance by adding water (H鈧侽) for oxygen and hydrogen ions (H鈦) for hydrogen.
This often involves these steps:
  • Balance Atoms: Oxygen atoms are balanced by adding H鈧侽.
  • Balance Hydrogen With H鈦: After balancing oxygen, add H鈦 ions to balance hydrogen atoms.
Finally, balance the overall charge in each half-reaction, typically using electrons, and then combine the half-reactions to form the final balanced equation.
Oxidation States
Understanding oxidation states is crucial for identifying how electrons are transferred in redox reactions. An oxidation state indicates the degree of oxidation of an atom within a molecule.
Here's how you can find oxidation states:
  • Assign Initial Values: Some elements follow common rules, like oxygen is usually -2, and hydrogen is +1.
  • Change in State: Recognize which elements undergo a change in oxidation state. This helps in identifying oxidation and reduction processes.
In a redox reaction, the element that increases in oxidation state is oxidized, while the one that decreases in oxidation state is reduced.
Chemical Equations
Chemical equations represent reactions where reactants are transformed into products. In oxidation-reduction reactions, these equations indicate the loss and gain of electrons.
Here's a simplified approach to understanding them:
  • Identify Reactants and Products: Write down the initial and final species.
  • Write Half-Reactions: Separate the equation into oxidation and reduction parts.
  • Balance and Combine: Ensure both mass and charge are equal on both sides.
Balanced chemical equations provide a way to quantify the relationships between molecules involved in the reaction.

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Most popular questions from this chapter

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}$$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ} :\) $$\quad\quad\quad \mathrm{H}_{2} \mathrm{O}(l)=-237 \mathrm{kJ} / \mathrm{mol}$$ $$\mathrm{H}_{2}(g)=0.0$$ $$\quad\quad\quad \mathrm{OH}^{-}(a q)=-157 \mathrm{kJ} / \mathrm{mol}$$ $$\quad \mathrm{e}^{-}=0.0$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 18.1

A solution containing \(\mathrm{Pt}^{4+}\) is electrolyzed with a current of 4.00 \(\mathrm{A} .\) How long will it take to plate out 99\(\%\) of the platinum in 0.50 \(\mathrm{L}\) of a \(0.010-M\) solution of \(\mathrm{Pt}^{4+} ?\)

An electrochemical cell consists of a nickel metal electrode immersed in a solution with \(\left[\mathrm{Ni}^{2+}\right]=1.0 M\) separated by a porous disk from an aluminum metal electrode. a. What is the potential of this cell at \(25^{\circ} \mathrm{C}\) if the aluminum electrode is placed in a solution in which \(\left[\mathrm{Al}^{3+}\right]=7.2 \times 10^{-3} M?\) b. When the aluminum electrode is placed in a certain solution in which \(\left[\mathrm{Al}^{3+}\right]\) is unknown, the measured cell potential at \(25^{\circ} \mathrm{C}\) is 1.62 \(\mathrm{V}\) . Calculate \(\left[\mathrm{Al}^{3+}\right]\) in the unknown solution. (Assume Al is oxidized.)

Is the following statement true or false? Concentration cells work because standard reduction potentials are dependent on concentration. Explain.

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of \(\mathrm{NaCl}\) . b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of \(\mathrm{Nal}\). c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) at \(25^{\circ} \mathrm{C}\)
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