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Chapter 17: Problem 63

Identify the products of the electrolysis of a 1 -M aqueous solution of NaBr. What species are present in the solution? What is formed at the cathode? What is formed at the anode?

Short Answer

Expert verified
At the cathode, H鈧� is formed; at the anode, Br鈧� is formed.

Step by step solution

01

Identify the species present in the solution

An aqueous solution of NaBr dissociates into its ions: Na鈦� and Br鈦�. Additionally, the solution contains water molecules, H鈧侽, which can also undergo electrolysis.
02

Determine the reactions at the cathode

At the cathode, reduction occurs. In an aqueous solution, two possible reductions can occur:1. Reduction of Na鈦�: \( \text{Na}^+ + e^- \rightarrow \text{Na} \)2. Reduction of water: \( \text{2H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^- \)The reduction potential for water is higher compared to Na鈦�, so water is reduced, producing hydrogen gas (H鈧�) and hydroxide ions (OH鈦�).
03

Determine the reactions at the anode

At the anode, oxidation occurs. In an aqueous solution of Br鈦�, the possible oxidations are:1. Oxidation of Br鈦�: \( 2\text{Br}^- \rightarrow \text{Br}_2 + 2e^- \)2. Oxidation of water: \( 2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \)However, bromide ions (Br鈦�) have a lower oxidation potential compared to water, so they are oxidized, producing bromine gas (Br鈧�).
04

Summarize the products at each electrode

Based on the reactions: - At the cathode, hydrogen gas (H鈧�) is formed. - At the anode, bromine gas (Br鈧�) is formed.

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Key Concepts

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

Aqueous solution
An aqueous solution is simply a mixture where a substance, such as NaBr, dissolves in water. This happens due to water's polar nature, which allows it to break apart the molecules of the solute into ions.
When NaBr is dissolved in water, it separates into sodium ions (Na鈦�) and bromide ions (Br鈦�).
These ions freely move around within the solution and can participate in electrochemical reactions when electricity is applied. The presence of water molecules also means that it can participate in reactions, potentially undergoing both oxidation and reduction.
  • NaBr splits into Na鈦� and Br鈦� ions.
  • Water (H鈧侽) remains present and active in the solution.
Understanding what species are in solution is crucial to predicting the outcomes of electrolysis.
Cathode reactions
At the cathode of an electrochemical cell, reduction reactions take place. In simpler terms, this means that specific ions gain electrons.
Within an aqueous solution of NaBr, both Na鈦� ions and water molecules can potentially undergo reduction.
However, the reduction potential is an important factor to consider here, as it dictates which reaction is more favorable.
For NaBr, water reduction is more favorable, as it has a higher reduction potential compared to Na鈦�.
This means that instead of Na being reduced, the water molecules are reduced to form hydrogen gas (H鈧�) and hydroxide ions (OH鈦�).
  • Electrons are gained at the cathode.
  • Water reduction is the preferred reaction.
  • Hydrogen gas (H鈧�) and hydroxide ions (OH鈦�) form.
Reduction potentials guide which species is reduced at the cathode.
Anode reactions
The anode is where oxidation happens 鈥� this is where electrons are lost by the ions. In the electrolysis of NaBr aqueous solutions, the main players for oxidation are Br鈦� ions and water molecules.
Similar to the cathode, the potential decides which reaction occurs more readily. In this case, the oxidation potential of bromide ions (Br鈦�) is lower than that of water, leading to their oxidation into bromine gas (Br鈧�).
This process results in the liberation of electrons.
  • Oxidation leads to the loss of electrons.
  • Bromide ions are oxidized over water.
  • Produces bromine gas (Br鈧�).
The liberation of electrons at the anode ensures the continued flow of electric current in the solution.
Reduction potential
Reduction potential is a concept referring to the tendency of a substance to gain electrons.
In electrochemistry, these potentials are used to predict which species are likely to undergo reduction. Under standard conditions, each ion or molecule has a tabulated potential value indicating this tendency.
Comparing the reduction potentials of sodium ions (Na鈦�) and water in aqueous solution shows that water has a higher potential.
Hence, water is more likely to be reduced even if Na鈦� ions are present.
  • High reduction potential suggests a stronger tendency to gain electrons.
  • Guides which reactions occur at the cathode.
  • Determines formation of products like H鈧� gas in aqueous NaBr electrolysis.
Knowing reduction potentials is essential in predicting cathodic reactions effectively.
Oxidation potential
Oxidation potential is a measure of how likely a species is to lose electrons and be oxidized.
Like reduction potential, oxidation potential helps us to identify which species are more favorable for oxidation reactions during electrolysis.
For the case of NaBr in water, bromide ions (Br鈦�) have a lower oxidation potential meaning they're more prone to oxidation as opposed to water. As a result, they lose electrons and form bromine gas (Br鈧�).
Understanding these potentials allows chemists to predict and control the outcomes of electrochemical processes accurately.
  • Low oxidation potential leads to a higher likelihood of oxidation.
  • Determines anode reactions during electrolysis.
  • Informs the production of substances like Br鈧� gas from NaBr solutions.
Mastery of oxidation potentials is crucial in designing effective electrochemical cells.

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Most popular questions from this chapter

A hydrogen-oxygen fuel cell operates on the simple reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell) $$ If the cell is designed to produce \(1.5 \mathrm{~A}\) of current, determine how long it can operate if there is an excess of oxygen and only sufficient hydrogen to fill a \(1.0-\mathrm{L}\) tank at 200. bar pressure at \(25^{\circ} \mathrm{C}\).

Write chemical equations for the electrolysis of molten salts of three different alkali halides to produce the corresponding halogens and alkali metals.

Make a drawing showing the principal parts of (a) a voltaic cell: show the anode, the cathode, the direction of electron movement outside the cell, and the direction of ion movement inside the cell. (b) a standard hydrogen electrode: describe the components of the electrode and explain how it works.

Consider these half-reactions: $$ \begin{array}{lr} \hline \text { Half-reaction } & E^{\circ}(\mathrm{V}) \\ \hline \mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(\mathrm{s}) & 1.52 \\ \mathrm{Pt}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}(\mathrm{s}) & 1.118 \\ \mathrm{Co}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}(\mathrm{s}) & -0.277 \\ \mathrm{Mn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mn}(\mathrm{s}) & -1.18 \\ \hline \end{array} $$ (a) Which is the weakest oxidizing agent? (b) Which is the strongest oxidizing agent? (c) Which is the strongest reducing agent? (d) Which is the weakest reducing agent? (e) Will \(\mathrm{Co}(\mathrm{s})\) reduce \(\mathrm{Pt}^{2+}(\mathrm{aq})\) to \(\mathrm{Pt}(\mathrm{s})\) ? (f) Will \(\mathrm{Pt}(\mathrm{s})\) reduce \(\mathrm{Co}^{2+}(\mathrm{aq})\) to \(\mathrm{Co}(\mathrm{s})\) ? (g) Which ions can be reduced by Co(s)?

Explain how reactant-favored electrochemical reactions can be induced to make products.
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