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Chapter 18: Problem 65

Hydrazine is somewhat toxic. Use the half-reactions shown below to explain why household bleach (a highly alkaline solution of sodium hypochlorite) should not be mixed with household ammonia or glass cleansers that contain ammonia. \(\mathrm{ClO}^{-}+\mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=0.90 \mathrm{~V}\) \(\mathrm{N}_{2} \mathrm{H}_{4}+2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{NH}_{3}+2 \mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 \mathrm{~V}\)

Short Answer

Expert verified
In short, mixing household bleach containing sodium hypochlorite (ClO⁻) with ammonia (NH₃) results in a spontaneous redox reaction with a positive potential of 1.00 V, forming the highly toxic and volatile compound nitrogen trichloride (NCl₃). Thus, they should not be mixed due to the potential hazards to human health posed by the toxic product.

Step by step solution

01

Identify the Redox Reaction

To determine the reaction that occurs when household bleach and ammonia are mixed, we must first combine the given half-reactions: 1. \(\mathrm{ClO}^{-}+\mathrm{H}_{2}\mathrm{O}+2e^{-}\rightarrow 2\mathrm{OH}^{-}+\mathrm{Cl}^{-} \quad \mathscr{E}^{\circ}=0.90 V \) 2. \(\mathrm{N}_{2}\mathrm{H}_{4}+2\mathrm{H}_{2}\mathrm{O}+2e^{-}\rightarrow 2\mathrm{NH}_{3}+2\mathrm{OH}^{-} \quad \mathscr{E}^{\circ}=-0.10 V \)
02

Balance the Electrons

To combine the two half-reactions, we need an equal number of electrons on both reactions. The first reaction has 2 electrons, and the second reaction also has 2 electrons. Therefore, the half-reactions are already balanced.
03

Add the Half-Reactions

Next, we will add the two half-reactions together: \(\mathrm{ClO}^{-}+\mathrm{H}_{2}\mathrm{O}+2e^{-}\rightarrow 2\mathrm{OH}^{-}+\mathrm{Cl}^{-}\) \(\mathrm{N}_{2}\mathrm{H}_{4}+2\mathrm{H}_{2}\mathrm{O}+2e^{-}\rightarrow 2\mathrm{NH}_{3}+2\mathrm{OH}^{-}\) Add the equations to get the full redox reaction: \(\mathrm{ClO}^{-}+\mathrm{N}_{2}\mathrm{H}_{4}+3\mathrm{H}_{2}\mathrm{O}\rightarrow 2\mathrm{NH}_{3}+4\mathrm{OH}^{-}+\mathrm{Cl}^{-}\)
04

Determine the Reaction's Potential

Since we have the standard potentials \(\mathscr{E}^{\circ}\) for each half-reaction, we can calculate the overall reaction potential: \(\Delta \mathscr{E}^{\circ} = \mathscr{E}^{\circ}_{cathode} - \mathscr{E}^{\circ}_{anode} = 0.90 V - (-0.10 V) = 1.00 V\) Since the overall reaction potential is positive, the reaction is spontaneous, and the toxic product will be formed when bleach and ammonia are mixed.
05

Identify the Toxic Product

From the balanced redox equation, we can identify the toxic product formed as a result of the reaction. When bleach containing sodium hypochlorite (ClO⁻) and ammonia (NH₃) are mixed, nitrogen trichloride (NCl₃) is produced. Nitrogen trichloride is a highly toxic and volatile compound which poses hazards to human health. In conclusion, household bleach containing sodium hypochlorite should not be mixed with ammonia or glass cleansers containing ammonia due to the formation of toxic nitrogen trichloride in the resulting redox reaction.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Ag}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) & \mathscr{E}^{\circ}=0.80 \mathrm{~V} \\ \mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & \mathscr{E}^{\circ}=0.34 \mathrm{~V} \end{aligned} $$ In this cell, the silver compartment contains a silver electrode and excess \(\operatorname{AgCl}(s)\left(K_{\mathrm{sp}}=1.6 \times 10^{-10}\right)\), and the copper compartment contains a copper electrode and \(\left[\mathrm{Cu}^{2+}\right]=2.0 M\). a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\). b. Assuming \(1.0 \mathrm{~L}\) of \(2.0 \mathrm{M} \mathrm{Cu}^{2+}\) in the copper compartment, calculate the moles of \(\mathrm{NH}_{3}\) that would have to be added to give a cell potential of \(0.52 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) (assume no volume change on addition of \(\mathrm{NH}_{3}\) ). $$ \begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons & \\ \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}{ }^{2+}(a q) & K=1.0 \times 10^{13} \end{aligned} $$

Consider the following half-reactions: $$ \begin{aligned} \mathrm{IrCl}_{6}{ }^{3-}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Ir}+6 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.77 \mathrm{~V} \\ \mathrm{PtCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.73 \mathrm{~V} \\ \mathrm{PdCl}_{4}{ }^{2-}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pd}+4 \mathrm{Cl}^{-} & \mathscr{E}^{\circ}=0.62 \mathrm{~V} \end{aligned} $$ A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant \(1.0 M\) in chloride ion and \(0.020 M\) in each complex ion. Is it feasible to separate the three metals from this solution by electrolysis? (Assume that \(99 \%\) of a metal must be plated out before another metal begins to plate out.)

The equation \(\Delta G^{\circ}=-n F \mathscr{E}^{\circ}\) also can be applied to halfreactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and \(\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ}\right.\) for \(\left.\mathrm{e}^{-}=0 .\right)\)

Direct methanol fuel cells (DMFCs) have shown some promise as a viable option for providing "green" energy to small electrical devices. Calculate \(\mathscr{E}^{\circ}\) for the reaction that takes place in DMFCs: $$ \mathrm{CH}_{3} \mathrm{OH}(l)+3 / 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ Use values of \(\Delta G_{\mathrm{f}}^{\circ}\) from Appendix \(4 .\)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.957 \mathrm{~V}\) \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.775 \mathrm{~V}\) a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm? Assume that no other gases are present and that the change in acid concentration can be neglected.
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