/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 The products formed when an aque... [FREE SOLUTION] | 91影视

91影视

The products formed when an aqueous solution of \(\mathrm{NaBr}\) is electrolyzed in a cell having inert electrodes are a. \(\mathrm{Na}\) and \(\mathrm{Br}_{2}\) b. \(\mathrm{Na}\) and \(\mathrm{O}_{2}\) c. \(\mathrm{H}_{2}, \mathrm{Br}_{2}\) and \(\mathrm{NaOH}\) d. \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\)

Short Answer

Expert verified
The products are H鈧, Br鈧, and NaOH.

Step by step solution

01

Identify the Electrolysis Reaction

In the electrolysis of an aqueous solution of NaBr, we must consider the dissociation of NaBr into Na鈦 and Br鈦 ions, along with the water dissociation into H鈦 and OH鈦 ions.
02

Determine Possible Reactions at the Cathode

At the cathode, reduction reactions occur. Possible reductions include: Na鈦 + e鈦 鈫 Na (not favored in water), and 2H鈧侽 + 2e鈦 鈫 H鈧 + 2OH鈦. In aqueous solutions, the reduction of water to form H鈧 gas is favored.
03

Determine Possible Reactions at the Anode

At the anode, oxidation reactions occur. Br鈦 can be oxidized: 2Br鈦 鈫 Br鈧 + 2e鈦. Water might also be oxidized: 2H鈧侽 鈫 O鈧 + 4H鈦 + 4e鈦, but Br鈦 oxidation is favored over water.
04

Summarize the Electrolysis Products

In electrolysis of aqueous NaBr, the cathode reaction forms H鈧 gas, and the anode reaction forms Br鈧. The solution also becomes basic due to the formation of OH鈦 ions, creating NaOH in solution.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cathode Reaction
When you're dealing with the electrolysis of an aqueous solution of NaBr, the cathode is where the magic of reduction happens. In electrochemical terms, reduction simply means gaining electrons. But what exactly happens at the cathode in our solution?

In the case of aqueous NaBr, two key sets of ions are floating around: from the NaBr itself, we have Na鈦 and Br鈦, and from the water, there are H鈦 and OH鈦. While it might seem logical for the Na鈦 ions to grab electrons (2Na鈦 + 2e鈦 鈫 2Na), the reduction of water is actually favored under these aqueous conditions. Water molecules (2H鈧侽) also participate by accepting electrons, leading to the formation of hydrogen gas (H鈧) and hydroxide ions (2OH鈦).
  • Reduction favored: 2H鈧侽 + 2e鈦 鈫 H鈧 + 2OH鈦
This reaction is important because it determines the gaseous product at the cathode, which is hydrogen (H鈧). Thus, hydrogen gas bubbles up as a product here!
Anode Reaction
On the flip side, the anode is the site for oxidation, which means losing electrons. In our electrochemical cell with an aqueous NaBr solution, this again involves deciding between the possible players: Br鈦 ions from NaBr and water molecules.

For aqueous NaBr, the Br鈦 ions are more readily oxidized than the water molecules. This means they'll lose electrons to form bromine gas (\(2Br^- 鈫 Br鈧 + 2e^-\)). The preference for bromide ion oxidation over water oxidation is crucial because it dictates that bromine (Br鈧) is released at the anode.
  • Bromide oxidation favored: 2Br鈦 鈫 Br鈧 + 2e鈦
Therefore, the primary product at the anode is bromine gas, which you'll observe as bubbles or a brown color at the anode.
Reduction and Oxidation Reactions
Electrolysis encompasses two principal types of reactions: reduction at the cathode and oxidation at the anode. Reduction involves the gain of electrons, whereas oxidation involves the loss.

In the electrolysis of NaBr, reduction occurs at the cathode, where water gains electrons, forming hydrogen gas and hydroxide ions:
  • Reduction: 2H鈧侽 + 2e鈦 鈫 H鈧 + 2OH鈦

At the anode, oxidation occurs as bromide ions shed electrons, forming bromine gas:
  • Oxidation: 2Br鈦 鈫 Br鈧 + 2e鈦
These reactions are crucial in the electrochemical process, where energy is converted into chemical changes, producing gases, and changing the solution's composition.
Formation of NaOH
When NaBr solution undergoes electrolysis, not only gases are produced but also a change in the solution's make-up occurs.

The key lies in the hydroxide ions generated at the cathode. These 2OH鈦 ions remain in the solution, combining with the Na鈦 ions that had not been reduced. This amalgamation results in the formation of sodium hydroxide (NaOH).
  • Net Result:
    Na鈦 + OH鈦 鈫 NaOH
This means that alongside hydrogen and bromine gases, the resulting solution now contains the base \(NaOH\).

This property contributes to making the solution more alkaline and creates an essential by-product widely used in industry.
Electrochemical Cell
The scene of action for electrolysis is the electrochemical cell. Think of it as a structured laboratory where electricity leads to chemical transformation.

Within our cell for aqueous NaBr, inert electrodes serve as conduits for electrons. The cathode receives electrons where reduction of water occurs, while the anode expels electrons, allowing bromide to oxidize.
  • Cathode (reduction site): \(2H鈧侽 + 2e鈦 鈫 H鈧 + 2OH鈦籠)
  • Anode (oxidation site): \(2Br鈦 鈫 Br鈧 + 2e鈦籠)
This dual-action transforms simple ionic movements into a structured and predictable sequence, producing hydrogen, bromine, and NaOH. Using inert electrodes ensures that they're simply facilitators of electron flow, without participating in the chemical transformations. It's the mixture of science and engineering that makes electrochemical cells powerful tools in chemistry.

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

Based on the following information, \(\mathrm{F}_{2}(\mathrm{~g})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{~F}^{-}(\mathrm{aq}) \quad \mathrm{E}^{\circ}=+2.87 \mathrm{~V}\) \(\mathrm{Mg}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Mg}(\mathrm{s}) \mathrm{E}^{\circ}=-2.37 \mathrm{~V}\) Which of the following chemical species is the strongest reducing agent? a. \(\mathrm{F}^{-}(\mathrm{aq})\) b. \(\mathrm{F}_{2}(\mathrm{~g})\) c. \(\mathrm{Mg}(\mathrm{s})\) d. \(\mathrm{Mg}^{2+}\) (aq)

Two electrochemical cells, \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) and \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) are connected in series. What will be the net e.m.f. of the cell at \(25^{\circ} \mathrm{C}\) ? Given: \(\mathrm{Zn}^{2+} \mid \mathrm{Zn}=-0.76 \mathrm{~V}\) \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}=+0.34 \mathrm{~V}\) \(\mathrm{Fe}^{2+} \mid \mathrm{Fe}=-0.41 \mathrm{~V}\) a. \(-1.85 \mathrm{~V}\) b. \(-0.83 \mathrm{~V}\) c. \(+0.83 \mathrm{~V}\) d. \(+1.85 \mathrm{~V}\)

Match the following: \(\begin{aligned}&\text { Column I } & \text { Column II }\end{aligned}\) A. electrolytic cell (p) Nernst equation B. Ecell = (q) 96500 coulombs \(\frac{0.059}{\mathrm{n}} \log \frac{\text { Cathode }}{\text { Canode }}\) C. connect two half cells (r) EMF of cell D. 1 faraday (s) salt bridge (t) device converting electrical energy into chemical energy

The oxidation number of sulphur in \(\mathrm{S}_{8}, \mathrm{~S}_{2} \mathrm{~F}_{2}, \mathrm{H}_{2} \mathrm{~S}\) respectively, are a. \(0,+1\) and \(-2\) b. \(+2,+1\) and \(-2\) c. \(0,+1\) and \(+2\) d. \(-2,+1\) and \(-2\).

Which of the following is/are not redox reaction? a. \(\mathrm{Zn}+2 \mathrm{AgCN} \rightarrow 2 \mathrm{Ag}+\mathrm{Zn}(\mathrm{CN})_{2}\) b. \(\mathrm{Mg}(\mathrm{OH})_{2}+2 \mathrm{NH}_{4} \mathrm{Cl} \rightarrow\) \(\mathrm{MgCl}_{2}+2 \mathrm{NH}_{4} \mathrm{OH}\) c. \(\mathrm{NaCl}+\mathrm{KNO}_{3} \rightarrow \mathrm{NaNO}_{3}+\mathrm{KCl}\) d. \(\mathrm{CaC}_{2} \mathrm{O}_{4}+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_{2}+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.