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Aqueous solutions of ammonia \(\left(\mathrm{NH}_{3}\right)\) and bleach (active ingredient \(\mathrm{NaOCl}\) ) are sold as cleaning fluids, but bottles of both of them warn: "Never mix ammonia and bleach, as toxic gases may be produced." One of the toxic gases that can be produced is chloroamine, \(\mathrm{NH}_{2} \mathrm{Cl}\). (a) What is the oxidation number of chlorine in bleach? (b) What is the oxidation number of chlorine in chloramine? (c) Is Cl oxidized, reduced, or neither, upon the conversion of bleach to chloramine? (d) Another toxic gas that can be produced is nitrogen trichloride, \(\mathrm{NCl}_{3}\). What is the oxidation number of \(\mathrm{N}\) in nitrogen trichloride? (e) Is N oxidized, reduced, or neither, upon the conversion of ammonia to nitrogen trichloride?

Short Answer

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(a) +1 (b) 0 (c) Reduced (d) +3 (e) Oxidized

Step by step solution

01

Identify Oxidation Number in Bleach

The active ingredient in bleach is sodium hypochlorite, NaOCl. To find the oxidation number of chlorine (\(\text{Cl}\)), recognize that sodium (\(\text{Na}\)) has an oxidation number of \(+1\) and oxygen (\(\text{O}\)) typically has \(-2\). The sum of the oxidation numbers in a compound equals zero. \[\text{Na: +1,}\ \text{O: -2,}\ \text{Cl: } x \] \[+1 + x - 2 = 0\] \[x = +1\] Thus, the oxidation number of chlorine in bleach (NaOCl) is \(+1\).
02

Identify Oxidation Number in Chloramine

Chloramine is \(\text{NH}_2\text{Cl}\). Here, hydrogen has an oxidation number of \(+1\) and nitrogen usually has \(-3\), as it carries a -3 charge in the ammonia ion. Chlorine's oxidation number we need to calculate in \(\text{NH}_2\text{Cl}\). \[2(+1) + (-3) + x = 0\] \[2 - 3 + x = 0\] \[x = 0\] Therefore, the oxidation number of chlorine in chloramine is \(0\).
03

Determine Change in Chlorine Oxidation State

The oxidation number of chlorine changes from \(+1\) in NaOCl to \(0\) in \(\text{NH}_2\text{Cl}\). A decrease in oxidation number indicates a reduction. Thus, chlorine is reduced when forming chloramine from bleach.
04

Identify Oxidation Number in Nitrogen Trichloride

Consider the compound \(\text{NCl}_3\). Chlorine usually has an oxidation number of \(-1\). We need to find the oxidation number of nitrogen. \[x + 3(-1) = 0\] \[x - 3 = 0\] \[x = +3\] Hence, the oxidation number of nitrogen in nitrogen trichloride is \(+3\).
05

Determine Change in Nitrogen Oxidation State

Nitrogen in ammonia \((\text{NH}_3)\) has an oxidation number of \(-3\). In nitrogen trichloride \((\text{NCl}_3)\), nitrogen's oxidation number changes to \(+3\). An increase in oxidation number indicates oxidation. Therefore, nitrogen is oxidized when ammonia converts to nitrogen trichloride.

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Key Concepts

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

Understanding Oxidation Number Determination
Oxidation numbers are like chemical bookkeeping. They help us track how electrons are transferred in a chemical reaction. Determining these numbers involves rules that are quite simple to follow. Consider sodium hypochlorite (NaOCl), where sodium (Na) has an oxidation number of \(+1\), and oxygen (O) is typically \(-2\). To find chlorine's oxidation number, you sum the known oxidation numbers and set the total equal to zero because compounds are neutral. In NaOCl, chlorine ends up with an oxidation number of \(+1\). Similarly, in chloramine (NH\(_2\)Cl), recognizing that hydrogen (H) is \(+1\) and nitrogen (N) is \(-3\), allows us to deduce that chlorine must be \(0\) for the sum to remain zero. These systematic rules make oxidation number determination straightforward, promoting better understanding of how elements interact.
The Role of Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are at the heart of many chemical processes. They involve the transfer of electrons between substances, changing their oxidation states. In the transformation from bleach (NaOCl) to chloramine (NH\(_2\)Cl), chlorine undergoes a change in oxidation state from \(+1\) to \(0\). This reduction occurs because chlorine gains electrons. Conversely, when ammonia (NH\(_3\)) becomes nitrogen trichloride (NCl\(_3\)), nitrogen's oxidation number increases from \(-3\) to \(+3\), signifying oxidation as it loses electrons. Understanding redox reactions is crucial as they explain the chemical principles behind processes like metabolism and the action of batteries.
Ensuring Chemical Safety
Chemical safety is paramount when dealing with reactive substances like ammonia and bleach. Mixing these common household cleaners can generate toxic gases, such as chloroamine (NH\(_2\)Cl) and nitrogen trichloride (NCl\(_3\)). Both can pose serious health risks. It is essential to read labels carefully and follow safety precautions when using chemicals. Always work in well-ventilated areas, and avoid direct inhalation of fumes. By understanding the potential hazards and the chemistry behind them, you can prevent accidental exposure to harmful substances. Prioritizing safety ensures that chemistry remains a useful and beneficial pursuit rather than a hazardous endeavor.
Fundamentals of Oxidation-Reduction Concepts
Oxidation-reduction, or redox concepts, form the backbone of understanding chemical reactions involving electron transfer. During oxidation, a species loses electrons, resulting in an increase in oxidation number. Oxygen plays a central role in these processes, although it is not always involved. For example, when ammonia becomes nitrogen trichloride, nitrogen is oxidized because its oxidation state rises from \(-3\) to \(+3\). Reduction, the opposite, involves a gain of electrons, as observed with chlorine reducing from an oxidation state of \(+1\) in NaOCl to \(0\) in NH\(_2\)Cl. These concepts help us predict and interpret how chemicals change and interact, providing a deeper understanding of molecular transformations essential for any scientific study involving chemistry.

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Most popular questions from this chapter

Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. (b) For the reaction \(\mathrm{Fe}^{3+}(a q)+\mathrm{Co}^{2+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\) \(\mathrm{Co}^{3+}(a q), \mathrm{Fe}^{3+}(a q)\) is the reducing agent and \(\mathrm{Co}^{2+}(a q)\) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction.

Copper corrodes to cuprous oxide, \(\mathrm{Cu}_{2} \mathrm{O},\) or cupric oxide, CuO, depending on environmental conditions. (a) What is the oxidation state of copper in cuprous oxide? (b) What is the oxidation state of copper in cupric oxide? (c) Copper peroxide is another oxidation product of elemental copper. Suggest a formula for copper peroxide based on its name. (d) Copper(III) oxide is another unusual oxidation product of elemental copper. Suggest a chemical formula for copper(III) oxide.

Metallic gold is collected from below the anode when a mixture of copper and gold metals is refined by electrolysis. Explain this behavior.

(a) Write the reactions for the discharge and charge of a nickel-cadmium (nicad) rechargeable hattery. (b) Given the following reduction potentials, calculate the standard emf of the cell: $$ \begin{array}{r} \mathrm{Cd}(\mathrm{OH})_{2}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cd}(s)+2 \mathrm{OH}^{-}(a q) \\ \qquad \begin{aligned} E_{\mathrm{red}}^{\circ} &=-0.76 \mathrm{~V} \end{aligned} \\ \mathrm{NiO}(\mathrm{OH})(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{e}^{-} \longrightarrow \mathrm{Ni}(\mathrm{OH})_{2}(s)+\mathrm{OH}^{-}(a q) \\ E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{~V} \end{array} $$ (c) A typical nicad voltaic cell generates an emf of \(+1.30 \mathrm{~V}\). Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}, P_{\mathrm{H}_{2}}=\) \(96.3 \mathrm{kPa},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M}\), and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

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