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Chapter 4: Problem 66

Write balanced equations for the following reactions in basic solution. (a) \(\mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{N}_{2}(g)\) (b) \(\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}(s)+\mathrm{CrO}_{4}{ }^{2-}(a q)\) (c) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)+\mathrm{BrO}_{4}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{IO}_{4}^{-}(a q) \longrightarrow \mathrm{IO}_{2}^{-}(a q)+\mathrm{O}_{2}(g)\)

Short Answer

Expert verified
Question: Balance the following redox equations in basic solutions: (a) Ni(OH)2(s) + N2H4(aq) 鈫� Ni(s) + N2(g) (b) Fe(OH)3(s) + Cr3+(aq) 鈫� Fe(OH)2(s) + CrO4^2-(aq) (c) MnO4^-(aq) + BrO3^-(aq) 鈫� MnO2(s) + BrO4^-(aq) (d) H2O2(aq) + IO4^-(aq) 鈫� IO2^-(aq) + O2(g) Answer: (a) 4Ni(OH)2 + N2H4 鈫� 4Ni + N2 + 16OH- (b) 3Fe(OH)3 + Cr3+ + 7OH- 鈫� 3Fe(OH)2 + CrO4^2- + 2H2O (c) 3BrO3^- + MnO4^- + 2H2O 鈫� 3BrO4^- + MnO2 (d) H2O2 + IO4^- + 4OH- 鈫� IO2^- + O2 + 2H2O

Step by step solution

01

Write down the half-reactions

We start by separating the reaction into two half-reactions: one for oxidation and one for reduction. Oxidation: Ni(OH)2 鈫� Ni + 2OH- Reduction: N2H4 + 4H+ + 4e- 鈫� N2 + 4H2O
02

Balance atoms and charges

The atoms in the half-reactions are already balanced, so now we need to balance the charges by adding electrons. Oxidation: Ni(OH)2 鈫� Ni + 2OH- + 2e- Reduction: N2H4 + 4H2O 鈫� N2 + 8e- + 8OH-
03

Multiply by a common factor

In order to cancel the electrons, we multiply the oxidation half-reaction by 4 and the reduction half-reaction by 1. 4[Ni(OH)2 鈫� Ni + 2OH- + 2e-] 1[N2H4 + 4H2O 鈫� N2 + 8e- + 8OH-]
04

Add the half-reactions

Now, we add the two half-reactions and cancel the electrons: 4Ni(OH)2 + N2H4 鈫� 4Ni + 8OH- + 8e- + N2 + 8e- + 8OH- The final balanced equation is: 4Ni(OH)2 + N2H4 鈫� 4Ni + N2 + 16OH- (b) Fe(OH)3(s) + Cr3+(aq) 鈫� Fe(OH)2(s) + CrO4^2-(aq)
05

Write down the half-reactions

The oxidation and reduction half-reactions are: Oxidation: Fe(OH)3 鈫� Fe(OH)2 + OH- Reduction: Cr3+ + 4OH- 鈫� CrO4^2-
06

Balance atoms and charges

Add electrons to balance charges: Oxidation: Fe(OH)3 鈫� Fe(OH)2 + OH- + e- Reduction: Cr3+ + 4OH- 鈫� CrO4^2- + 2H2O + 3e-
07

Multiply by a common factor

To cancel the electrons, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1: 3[Fe(OH)3 鈫� Fe(OH)2 + OH- + e-] 1[Cr3+ + 4OH- 鈫� CrO4^2- + 2H2O + 3e-]
08

Add the half-reactions

Now add the two half-reactions and cancel the electrons: 3Fe(OH)3 + Cr3+ + 4OH- 鈫� 3Fe(OH)2 + 3OH- + 3e- + CrO4^2- + 2H2O + 3e- The final balanced equation is: 3Fe(OH)3 + Cr3+ + 7OH- 鈫� 3Fe(OH)2 + CrO4^2- + 2H2O (c) MnO4^-(aq) + BrO3^-(aq) 鈫� MnO2(s) + BrO4^-(aq)
09

Write down the half-reactions

The oxidation and reduction half-reactions are: Oxidation: BrO3^- 鈫� BrO4^- Reduction: MnO4^- 鈫� MnO2
10

Balance atoms and charges

Add electrons and water molecules to balance charges and atoms: Oxidation: BrO3^- 鈫� BrO4^- + e- Reduction: MnO4^- + 2H2O 鈫� MnO2 + 4H+ + 3e-
11

Multiply by a common factor

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 1 to cancel the electrons: 3[BrO3^- 鈫� BrO4^- + e-] 1[MnO4^- + 2H2O 鈫� MnO2 + 4H+ + 3e-]
12

Add the half-reactions

Add the two half-reactions and cancel the electrons and protons with hydroxide ions: 3BrO3^- + MnO4^- + 2H2O 鈫� 3BrO4^- + 3e- + MnO2 + 4H+ + 3e- The final balanced equation is: 3BrO3^- + MnO4^- + 2H2O 鈫� 3BrO4^- + MnO2 (d) H2O2(aq) + IO4^-(aq) 鈫� IO2^-(aq) + O2(g)
13

Write down the half-reactions

The oxidation and reduction half-reactions are: Oxidation: H2O2 鈫� O2 Reduction: IO4^- 鈫� IO2^-
14

Balance atoms and charges

Add electrons and water molecules to balance charges and atoms: Oxidation: H2O2 鈫� O2 + 2H+ + 2e- Reduction: IO4^- + 2H2O 鈫� IO2^- + 4H+ + 2e-
15

Add the half-reactions

Add the two half-reactions and cancel the electrons and protons with hydroxide ions: H2O2 + IO4^- + 2H2O 鈫� O2 + 2H+ + 2e- + IO2^- + 4H+ + 2e- The final balanced equation is: H2O2 + IO4^- + 4OH- 鈫� IO2^- + O2 + 2H2O

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Key Concepts

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

Redox Reactions
Redox reactions stand for reduction-oxidation reactions, where one substance loses electrons (oxidation) and another gains electrons (reduction). Understanding these reactions involves recognizing changes in oxidation states of chemicals involved.
Oxidation means an increase in oxidation state, which equates to losing electrons. Reduction means a decrease in oxidation state, or a gain of electrons. In a redox reaction, these two processes occur simultaneously.
Key Steps in Redox Reactions:
  • Identify oxidation and reduction half-reactions by recognizing changes in oxidation states.
  • Balance each half-reaction separately for atoms and charge.
  • Add the balanced half-reactions together, making sure electrons lost in oxidation are equal to those gained in reduction.
The ability to balance redox reactions correctly is essential in chemistry since they are found in many different fields, such as biology, physics, and environmental science.
Half-Reactions
Half-reactions simplify the process of balancing redox reactions by breaking the reaction into two parts: oxidation and reduction. Each half-reaction is balanced separately before they are combined again.
To write a half-reaction, follow these steps:
  • First, write the reactants and products for each segment (oxidation and reduction).
  • Balance the atoms aside from hydrogen and oxygen.
  • Balance oxygen atoms by adding water molecules to the appropriate side.
  • Balance hydrogen atoms using hydrogen ions (H+).
  • Finally, balance the charges by adding electrons (e-).
After balancing, the half-reactions are recombined. The crucial step is ensuring the numbers of electrons lost and gained are equal, often achieved by multiplying half-reactions by appropriate factors.
Basic Solution
In basic solutions, the presence of OH- ions instead of H+ (as in acidic solutions) modifies how redox reactions are balanced. It is crucial to adjust half-reactions based on the basic medium.
Here's how to manage redox reactions in a basic solution:
  • Write down the half-reactions and balance as you would normally under acidic conditions.
  • After balancing with H+, add OH- ions to both sides of each half-reaction to neutralize the H+ ions, forming water (H2O).
  • Simplify the number of water molecules on both sides of the half-reaction if possible.
Ensure the overall reaction remains balanced both in terms of mass and charge. Basic solutions are common in natural waters and many industrial processes, making this skill useful in practical applications.

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Most popular questions from this chapter

Write a net ionic equation for any precipitation reaction that occurs when \(0.1 \mathrm{M}\) solutions of the following are mixed. (a) zinc nitrate and nickel(II) chloride (b) potassium phosphate and calcium nitrate (c) sodium hydroxide and zinc nitrate (d) iron(III) nitrate and barium hydroxide

Consider the following balanced redox reaction in basic medium. \(3 \mathrm{Sn}^{2+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+4 \mathrm{H}_{2} \mathrm{O} \longrightarrow\) \(3 \mathrm{Sn}^{4+}(a q)+\mathrm{Cr}_{2} \mathrm{O}_{3}(s)+8 \mathrm{OH}^{-}(a q)\) (a) What is the oxidizing agent? (b) What species has the element that increases its oxidation number? (c) What species contains the element with the highest oxidation number? (d) If the reaction were to take place in acidic medium, what species would not be included in the reaction?

Calcium in blood or urine can be determined by precipitation as call cium oxalate, \(\mathrm{CaC}_{2} \mathrm{O}_{4}\). The precipitate is dissolved in strong acid and titrated with potassium permanganate. The products of the reaction are carbon dioxide and manganese(II) ion. A 24-hour urine sample is collected from an adult patient, reduced to a small volume, and titrated with \(26.2 \mathrm{~mL}\) of \(0.0946 \mathrm{M} \mathrm{KMnO}_{4}\). How many grams of calcium oxalate are in the sample? Normal range for \(\mathrm{Ca}^{2+}\) output for an adult is 100 to \(300 \mathrm{mg}\) per 24 hour. Is the sample within the normal range?

Write net ionic equations for the formation of (a) a precipitate when solutions of magnesium nitrate and potassium hydroxide are mixed. (b) two different precipitates when solutions of silver(I) sulfate and barium chloride are mixed.

The percentage of sodium hydrogen carbonate, \(\mathrm{NaHCO}_{3}\), in a powder for stomach upsets is found by titrating with \(0.275 M\) hydrochloric acid. If \(15.5 \mathrm{~mL}\) of hydrochloric acid is required to react with \(0.500 \mathrm{~g}\) of the sample, what is the percentage of sodium hydrogen carbonate in the sample? The balanced equation for the reaction that takes place is $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$
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