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In the Breathalyzer test, blood alcohol is determined by reaction of the alcohol with potassium dichromate: \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow\) \(4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l)\) (a) Calculate the oxidation numher of \(\mathrm{Cr}\) in \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) (b) Calculate the oxidation number of \(\mathrm{C}\) in \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) and in \(\mathrm{CO}_{2}\) (c) Identify the oxidizing agent and the reducing agent in this reaction.

Short Answer

Expert verified
(a) Cr in Cr鈧侽鈧嚶测伝 has an oxidation number of +6. (b) C in C鈧侶鈧匫H: -2; in CO鈧: +4. (c) Oxidizing agent: Cr鈧侽鈧嚶测伝, Reducing agent: C鈧侶鈧匫H.

Step by step solution

01

Determine the oxidation number of Cr in Cr鈧侽鈧嚶测伝

To find the oxidation number of Cr in Cr鈧侽鈧嚶测伝, use the rule that the sum of the oxidation numbers in a compound must equal the overall charge. Each oxygen atom typically has an oxidation number of -2. In Cr鈧侽鈧嚶测伝, there are 7 oxygen atoms, so:\[\text{Sum of oxidation numbers of oxygen = } -2 \times 7 = -14.\]Let \(x\) be the oxidation number of Cr. Since there are 2 Cr atoms:\[2x + (-14) = -2.\]\[2x = 12.\]\[x = +6.\] Thus, the oxidation number of Cr in Cr鈧侽鈧嚶测伝 is +6.
02

Determine the oxidation number of C in C鈧侶鈧匫H

In \(C_2H_5OH\) or ethanol, carbon, hydrogen, and oxygen have specific typical oxidation states we can use. Typically, \(H = +1\) and \(O = -2\), so we solve for \(C\). Let \(x\) be the oxidation number of carbon in ethanol. Thus:\[2x + 5(+1) + (-2) + (+1) = 0.\]\[2x + 5 + 1 - 2 = 0.\]\[2x + 4 = 0.\]\[2x = -4.\]\[x = -2.\] The oxidation number of C in C鈧侶鈧匫H is -2.
03

Determine the oxidation number of C in CO鈧

In \(CO_2\), the oxygen has an oxidation state of -2. Therefore, the equation for the oxidation states will be:\[x + 2(-2) = 0.\]\[x - 4 = 0.\]\[x = +4.\] The oxidation number of C in CO鈧 is +4.
04

Identify the oxidizing and reducing agents

The oxidizing agent gains electrons and its oxidation number decreases, while the reducing agent loses electrons and its oxidation number increases. - From ethanol \(C_2H_5OH\) with \(C = -2\) to carbon dioxide \(CO_2\) with \(C = +4\), carbon has lost electrons (oxidation number increased), so ethanol is the reducing agent.- Chromium changes from \(+6\) in \(Cr_2O_7^{2-}\) to \(+3\) in \(Cr^{3+}\), gaining electrons (oxidation number decreased), so \(Cr_2O_7^{2-}\) is the oxidizing agent.

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Key Concepts

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

Breathalyzer Test Chemistry
The breathalyzer test is a fascinating application of chemistry that helps determine blood alcohol content (BAC). When a person breathes into a breathalyzer, any ethanol present in their breath reacts with a chemical substance called potassium dichromate. This reaction is a key component that makes these devices work.

In the chemical equation for the reaction, ethanol \(C_2H_5OH\) is oxidized, while the potassium dichromate contains chromium atoms that get reduced. The transformation of these elements as they react is useful both for detecting the presence of alcohol and for understanding chemical processes involved.
  • The primary reaction involves ethanol being oxidized to carbon dioxide, \(CO_2\), and water, while chromium is reduced from dichromate, \(Cr_2O_7^{2-}\), to \(Cr^{3+}\).
  • This chemical transformation changes the color of the potassium dichromate solution, which provides an indication of the amount of alcohol present.
The innovative use of metal oxidation and reduction in the breathalyzer applications showcases how chemistry is intelligently used in everyday technology.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are processes in which one or more elements change their oxidation states. They are fundamental to a wide range of chemical and biological processes, including the breathalyzer test.

In a redox reaction, one substance loses electrons (oxidation) while another gains them (reduction). This electron transfer results in changes in the oxidation numbers of the involved elements.
  • For the breathalyzer reaction, ethanol \(C_2H_5OH\) is oxidized to carbon dioxide \(CO_2\). This is because the carbon in ethanol increases its oxidation number from -2 to +4 by losing electrons.
  • Conversely, the chromium in the dichromate \(Cr_2O_7^{2-}\) is reduced to \(Cr^{3+}\), as it gains electrons, going from an oxidation state of +6 to +3.
Understanding redox reactions is vital as it opens up a world of chemical behaviors and applications, illustrating how substances interact and alter each other at a molecular level.
Oxidizing and Reducing Agents
In chemical reactions, specifically redox reactions, we often talk about oxidizing and reducing agents. These agents are crucial for the transfer of electrons between substances, which is the essence of redox reactions.

- **Oxidizing Agent:** It is the substance that gains electrons. In doing so, it gets reduced. The oxidizing agent helps another substance lose electrons.- **Reducing Agent:** It is the substance that loses electrons. This agent gets oxidized in the process, facilitating the gain of electrons by another substance.

In the breathalyzer test, which involves a redox reaction:
  • The oxidizing agent is potassium dichromate \(Cr_2O_7^{2-}\). It gains electrons, thus getting reduced to \(Cr^{3+}\). Its ability to accept electrons is why it is used to detect ethanol in the breath.
  • The reducing agent is ethanol \(C_2H_5OH\). As it loses electrons and gets oxidized to carbon dioxide \(CO_2\), it allows the oxidation state of the chromium to decrease.
Recognizing oxidizing and reducing agents helps us understand how elements interact, transfer electrons, and ultimately change their states during a reaction. This insight is essential for grasping complex redox processes like those used in breathalyzer tests.

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Most popular questions from this chapter

Balance the reaction for the synthesis of urea, commonly used as a fertilizer: $$ \mathrm{CO}_{2}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{2} \mathrm{CONH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$

Write balanced ionic equations and net ionic equations for the following reactions: (a) A precipitate of barium sulfate forms when aqueous solutions of barium nitrate and potassium sulfate are mixed. (b) Zinc ion and hydrogen gas form when zinc metal reacts with aqueous sulfuric acid.

Where in the periodic table are the most easily reduced elements found? The most easily oxidized?

White phosphorus \(\left(\mathrm{P}_{4}\right)\) is a highly reactive form of elemental phosphorus that reacts with oxygen to form a variety of molecular compounds, including diphosphorus pentoxide. (a) Write the balanced chemical equation for this reaction. (b) Calculate the oxidation number for \(\mathrm{P}\) and \(\mathrm{O}\) on both sides of the reaction, and identify the oxidizing and reducing agents.

Which of the following equations are balanced? Balance those that need it. (a) \(2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(3 \mathrm{Ca}(\mathrm{OH})_{2}(a q)+2 \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow\) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) (d) \(\mathrm{K}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(a q)+\mathrm{H}_{2}(g)\)

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