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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) → 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

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Most popular questions from this chapter

A sample of limestone weighing \(1.005 \mathrm{~g}\) is dissolved in \(75.00 \mathrm{~mL}\) of \(0.2500 \mathrm{M}\) hydrochloric acid. The following reaction occurs: $$ \mathrm{CaCO}_{3}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$ It is found that \(19.26 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{NaOH}\) is required to titrate the excess \(\mathrm{HCl}\) left after reaction with the limestone. What is the mass percent of \(\mathrm{CaCO}_{3}\) in the limestone?

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 $$ \mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O} $$

Write the formulas of the following compounds and decide which are soluble in water. (a) sodium sulfate (b) iron(III) nitrate (c) silver chloride (d) chromium(III) hydroxide

The average adult has about \(16 \mathrm{~g}\) of sodium ions in her blood. Assuming a total blood volume of \(5.0 \mathrm{~L}\), what is the molarity of \(\mathrm{Na}^{+}\) ions in blood?

Assign oxidation numbers to each element in (a) \(\mathrm{P}_{2} \mathrm{O}_{5}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{CO}_{3}{ }^{2-}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) (e) \(\mathrm{N}_{2} \mathrm{H}_{4}\)
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