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Chapter 4: Problem 92

Balance the following oxidation鈥搑eduction 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. \(\mathrm{H}_{3} \mathrm{AsO}_{4}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{AsH}_{3}(g)+\mathrm{Zn}^{2+}(a q)\)

Short Answer

Expert verified
The balanced reactions are: a. \(3Cu(s) + 2NO_3^-(aq) + 12H^+(aq) \rightarrow 3Cu^{2+}(aq) + 2NO(g) + 8H_2O(l)\) b. \(7Cl^-(aq) + 2Cr_2O_7^{2-}(aq) + 28H^+(aq) \rightarrow 7Cl_2(g) + 4Cr^{3+}(aq) + 14H_2O(l)\) c. \(Pb(s) + PbO_2(s) + H_2SO_4(aq) \rightarrow PbSO_4(s) + 2H_2O(l)\) d. \(3Mn^{2+}(aq) + 15NaBiO_3(s) + 6H^+(aq) \rightarrow 3MnO_4^-(aq) + 5Bi^{3+}(aq) + 8H_2O(l) + 15Na^+(aq)\) e. \(3Zn(s) + 2H_3AsO_4(aq) + 6H^+(aq) \rightarrow 3Zn^{2+}(aq) + 2AsH_3(g) + 6H_2O(l)\)

Step by step solution

01

a. Balancing Cu(s) + NO鈧冣伝(aq) 鈫 Cu虏鈦(aq) + NO(g) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Cu 鈫 Cu虏鈦 + 2e鈦 Reduction half-reaction: NO鈧冣伝 + e鈦 鈫 NO Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 2H鈧侽 to balance oxygen and 6H鈦 to balance hydrogen: NO鈧冣伝 + 2H鈧侽 + 6H鈦 + 3e鈦 鈫 NO + 6H鈧侽 Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2: 3Cu 鈫 3Cu虏鈦 + 6e鈦 2(NO鈧冣伝 + 2H鈧侽 + 6H鈦 + 3e鈦) 鈫 2(NO + 6H鈧侽) Step 4: Add the half-reactions back together and cancel electrons: 3Cu + 2NO鈧冣伝 + 4H鈧侽 + 12H鈦 鈫 3Cu虏鈦 + 2NO + 12H鈧侽 Step 5: Simplify the reaction, if possible: 3Cu(s) + 2NO鈧冣伝(aq) + 4H鈧侽(l) + 12H鈦(aq) 鈫 3Cu虏鈦(aq) + 2NO(g) + 12H鈧侽(l) Removing 4H鈧侽 from both sides: 3Cu(s) + 2NO鈧冣伝(aq) + 12H鈦(aq) 鈫 3Cu虏鈦(aq) + 2NO(g) + 8H鈧侽(l) The balanced reaction is: 3Cu(s) + 2NO鈧冣伝(aq) + 12H鈦(aq) 鈫 3Cu虏鈦(aq) + 2NO(g) + 8H鈧侽(l)
02

b. Balancing Cr鈧侽鈧嚶测伝(aq) + Cl鈦(aq) 鈫 Cr鲁鈦(aq) + Cl鈧(g) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Cl鈦 鈫 Cl鈧 + 2e鈦 Reduction half-reaction: Cr鈧侽鈧嚶测伝 鈫 2Cr鲁鈦 Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 14H鈦 to balance the change and 7H鈧侽 to balance oxygen: Cr鈧侽鈧嚶测伝 + 14H鈦 鈫 2Cr鲁鈦 + 7H鈧侽 Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 7 and the reduction half-reaction by 2: 7(Cl鈦 鈫 Cl鈧 + 2e鈦) 2(Cr鈧侽鈧嚶测伝 + 14H鈦) 鈫 2(2Cr鲁鈦 + 7H鈧侽) Step 4: Add the half-reactions back together and cancel electrons: 7Cl鈦 + 2Cr鈧侽鈧嚶测伝 + 28H鈦 鈫 7Cl鈧 + 4Cr鲁鈦 + 14H鈧侽 The balanced reaction is: 7Cl鈦(aq) + 2Cr鈧侽鈧嚶测伝(aq) + 28H鈦(aq) 鈫 7Cl鈧(g) + 4Cr鲁鈦(aq) + 14H鈧侽(l)
03

c. Balancing Pb(s) + PbO鈧(s) + H鈧係O鈧(aq) 鈫 PbSO鈧(s) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Pb 鈫 Pb虏鈦 + 2e鈦 Reduction half-reaction: PbO鈧 + 4H鈦 + 2e鈦 鈫 Pb虏鈦 + 2H鈧侽 Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. Add one SO鈧劼测伝 to the reduction half-reaction: PbO鈧 + 4H鈦 + SO鈧劼测伝 鈫 PbSO鈧 + 2H鈧侽 Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 1 (no need to multiply): Step 4: Add half-reactions back together, and cancel electrons: Pb + PbO鈧 + 4H鈦 + SO鈧劼测伝 鈫 Pb虏鈦 + PbSO鈧 + 2H鈧侽 Step 5: Simplify the reaction, if possible: Pb(s) + PbO鈧(s) + H鈧係O鈧(aq) 鈫 PbSO鈧(s) + 2H鈧侽(l) The balanced reaction is: Pb(s) + PbO鈧(s) + H鈧係O鈧(aq) 鈫 PbSO鈧(s) + 2H鈧侽(l)
04

d. Balancing Mn虏鈦(aq) + NaBiO鈧(s) 鈫 Bi鲁鈦(aq) + MnO鈧勨伝(aq) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Mn虏鈦 鈫 MnO鈧勨伝 Reduction half-reaction: BiO鈧冣伝 鈫 Bi鲁鈦 Step 2: Balance each half-reaction For the oxidation half-reaction, add 4H鈧侽 to balance oxygen and 8H鈦 to balance hydrogen, and then subtract 5e鈦 to balance the charges: Mn虏鈦 + 4H鈧侽 鈫 MnO鈧勨伝 + 8H鈦 + 5e鈦 For the reduction half-reaction, add 2H鈧侽 to balance oxygen and 6H鈦 to balance hydrogen, and then add 3e鈦 to balance the charges: BiO鈧冣伝 + 6H鈦 + 3e鈦 鈫 Bi鲁鈦 + 3H鈧侽 Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 5: 3(Mn虏鈦 + 4H鈧侽 鈫 MnO鈧勨伝 + 8H鈦 + 5e鈦) 5(BiO鈧冣伝 + 6H鈦 + 3e鈦 鈫 Bi鲁鈦 + 3H鈧侽) Step 4: Add the half-reactions back together and cancel electrons: 3Mn虏鈦 + 12H鈧侽 + 15BiO鈧冣伝 + 30H鈦 鈫 3MnO鈧勨伝 + 24H鈦 + 5Bi鲁鈦 + 15H鈧侽 Step 5: Simplify the reaction: 3Mn虏鈦(aq) + 12H鈧侽(l) + 15NaBiO鈧(s) + 6H鈦(aq) 鈫 3MnO鈧勨伝(aq) + 5Bi鲁鈦(aq) + 8H鈧侽(l) + 15Na鈦(aq) The balanced reaction is: 3Mn虏鈦(aq) + 15NaBiO鈧(s) + 6H鈦(aq) 鈫 3MnO鈧勨伝(aq) + 5Bi鲁鈦(aq) + 8H鈧侽(l) + 15Na鈦(aq)
05

e. Balancing H鈧傾sO鈧(aq) + Zn(s) 鈫 AsH鈧(g) + Zn虏鈦(aq) reaction

Step 1: Separate the redox reaction into half-reactions Oxidation half-reaction: Zn 鈫 Zn虏鈦 + 2e鈦 Reduction half-reaction: H鈧傾sO鈧 鈫 AsH鈧 Step 2: Balance each half-reaction Oxidation half-reaction is already balanced. For the reduction half-reaction, add 3H鈦 + 3e鈦 to balance hydrogen and charges: H鈧傾sO鈧 + 3H鈦 + 3e鈦 鈫 AsH鈧 + 3H鈧侽 Step 3: Equalize the number of electrons in both half-reactions by multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 2: 3(Zn 鈫 Zn虏鈦 + 2e鈦) 2(H鈧傾sO鈧 + 3H鈦 + 3e鈦 鈫 AsH鈧 + 3H鈧侽) Step 4: Add the half-reactions back together and cancel electrons: 3Zn + 2H鈧傾sO鈧 + 6H鈦 鈫 3Zn虏鈦 + 2AsH鈧 + 6H鈧侽 The balanced reaction is: 3Zn(s) + 2H鈧傾sO鈧(aq) + 6H鈦(aq) 鈫 3Zn虏鈦(aq) + 2AsH鈧(g) + 6H鈧侽(l)

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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 balance oxidation-reduction reactions, especially in an acidic solution. It breaks down the redox reaction into two separate equations: the oxidation half-reaction and the reduction half-reaction.
The oxidation half-reaction describes the loss of electrons from a substance while the reduction half-reaction describes the gain of electrons by another substance.
Each half-reaction is balanced individually for mass and charge, and then combined to form the overall balanced equation. This method ensures that electrons lost in oxidation equal those gained in reduction, maintaining charge neutrality.
Acidic Solution
In chemistry, an acidic solution has a higher concentration of H鈦 ions. Balancing redox reactions in such solutions requires careful handling of hydrogen atoms and charges.
To achieve balance of redox reactions in an acidic solution using the half-reaction method, additional H鈦 ions are often introduced to the equation to balance hydrogen atoms. Water (H鈧侽) may also be added to balance oxygen atoms.
The presence of H鈦 ions affects the electron counts needed for balancing, as they help track and balance charges across the redox process. Understanding the role of these ions is crucial in ensuring the accurate balancing of reactions in these conditions.
Balancing Redox Reactions
Balancing redox reactions involves equalizing the number of electrons transferred during the oxidation and reduction processes. The process typically includes:
  • Simplification of original reaction into two half-reactions.
  • Balancing the atoms in each half-reaction.
  • Ensuring the same number of electrons are present in both half-reactions by multiplying them appropriately.
  • Combining the two half-reactions to cancel out common terms like electrons.
Only through precise balancing of both atoms and charges can the full redox reaction be accurately represented, demonstrating the actual chemical changes occurring during the reaction.
Electron Transfer
Electron transfer is the fundamental process in any redox reaction. It involves the movement of electrons from one chemical species (the reducing agent) to another (the oxidizing agent).
This electron movement is what characterizes a reaction as a redox process, driving the transformation of substances involved. The count of electrons lost in oxidation must directly match the electrons gained in reduction to preserve electron balance.
Monitoring and equalizing this electron transfer is the key objective of using the half-reaction method, ensuring minimal energy loss and correct chemical conversion according to the reaction's stoichiometry.

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Most popular questions from this chapter

You made 100.0 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 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) nitrate solution?

What mass of silver chloride can be prepared by the reaction of 100.0 mL of 0.20 M silver nitrate with 100.0 mL of 0.15 M calcium chloride? Calculate the concentrations of each ion remaining in solution after precipitation is complete

What mass of NaOH is contained in 250.0 mL of a 0.400 M sodium hydroxide solution?

Many over-the-counter antacid tablets are now formulated using calcium carbonate as the active ingredient, which enables such tablets to also be used as dietary calcium supplements. As an antacid for gastric hyperacidity, calcium carbonate reacts by combining with hydrochloric acid found in the stomach, producing a solution of calcium chloride, converting the stomach acid to water, and releasing carbon dioxide gas (which the person suffering from stomach problems may feel as a 鈥渂urp鈥). Write the balanced chemical equation for this process.

The concentration of a certain sodium hydroxide solution was determined by using the solution to titrate a sample of potassium hydrogen phthalate (abbreviated as KHP). KHP is an acid with one acidic hydrogen and a molar mass of 204.22 g/ mol. In the titration, 34.67 mL of the sodium hydroxide solution was required to react with 0.1082 g KHP. Calculate the molarity of the sodium hydroxide.
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