Log out Go to App
Go to App

Learning Materials

Features

Discover

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) → Ni(s) + N2(g) (b) Fe(OH)3(s) + Cr3+(aq) → Fe(OH)2(s) + CrO4^2-(aq) (c) MnO4^-(aq) + BrO3^-(aq) → MnO2(s) + BrO4^-(aq) (d) H2O2(aq) + IO4^-(aq) → IO2^-(aq) + O2(g) Answer: (a) 4Ni(OH)2 + N2H4 → 4Ni + N2 + 16OH- (b) 3Fe(OH)3 + Cr3+ + 7OH- → 3Fe(OH)2 + CrO4^2- + 2H2O (c) 3BrO3^- + MnO4^- + 2H2O → 3BrO4^- + MnO2 (d) H2O2 + IO4^- + 4OH- → 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 → Ni + 2OH- Reduction: N2H4 + 4H+ + 4e- → 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 → Ni + 2OH- + 2e- Reduction: N2H4 + 4H2O → 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 → Ni + 2OH- + 2e-] 1[N2H4 + 4H2O → N2 + 8e- + 8OH-]
04

Add the half-reactions

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

Write down the half-reactions

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

Balance atoms and charges

Add electrons to balance charges: Oxidation: Fe(OH)3 → Fe(OH)2 + OH- + e- Reduction: Cr3+ + 4OH- → 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 → Fe(OH)2 + OH- + e-] 1[Cr3+ + 4OH- → CrO4^2- + 2H2O + 3e-]
08

Add the half-reactions

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

Write down the half-reactions

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

Balance atoms and charges

Add electrons and water molecules to balance charges and atoms: Oxidation: BrO3^- → BrO4^- + e- Reduction: MnO4^- + 2H2O → 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^- → BrO4^- + e-] 1[MnO4^- + 2H2O → 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 → 3BrO4^- + 3e- + MnO2 + 4H+ + 3e- The final balanced equation is: 3BrO3^- + MnO4^- + 2H2O → 3BrO4^- + MnO2 (d) H2O2(aq) + IO4^-(aq) → IO2^-(aq) + O2(g)
13

Write down the half-reactions

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

Balance atoms and charges

Add electrons and water molecules to balance charges and atoms: Oxidation: H2O2 → O2 + 2H+ + 2e- Reduction: IO4^- + 2H2O → 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 → O2 + 2H+ + 2e- + IO2^- + 4H+ + 2e- The final balanced equation is: H2O2 + IO4^- + 4OH- → IO2^- + O2 + 2H2O

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Starting with the solid and adding water, how would you prepare \(2.00 \mathrm{~L}\) of \(0.685 \mathrm{M}\) (a) \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2} ?\) (b) \(\mathrm{CuCl}_{2}\) ? (c) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{6}(\) vitamin \(\mathrm{C})\) ?

What is the molarity of each ion present in aqueous solutions prepared by dissolving \(20.00 \mathrm{~g}\) of the following compounds in water to make 4.50 L of solution? (a) cobalt(III) chloride (b) nickel(III) sulfate (c) sodium permanganate (d) iron(II) bromide

III Classify each of the following half-reactions as oxidation or reduction. (a) \(\mathrm{O}_{2}(g) \longrightarrow \mathrm{O}^{2-}(a q)\) (b) \(\mathrm{MnO}_{4}^{-}(a q) \longrightarrow \mathrm{MnO}_{2}(s)\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)\) (d) \(\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{Cl}_{2}(g)\)

Write net ionic equations to explain the formation of (a) a white precipitate when solutions of calcium sulfate and sodium carbonate are mixed. (b) two different precipitates formed when solutions of iron(III) sulfate and barium hydroxide are mixed.

Write balanced equations for the following reactions in acid solution. (a) \(\mathrm{P}_{4}(s)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{PH}_{3}(g)+\mathrm{Cl}_{2}(g)\) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{NO}_{2}^{-}(a q) \longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{NO}_{3}^{-}(a q)\) (c) \(\mathrm{HBrO}_{3}(a q)+\mathrm{Bi}(s) \longrightarrow \mathrm{HBrO}_{2}(a q)+\mathrm{Bi}_{2} \mathrm{O}_{3}(s)\) (d) \(\mathrm{CrO}_{4}^{2-}(a q)+\mathrm{SO}_{3}^{2-}(a q) \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free