/*! 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 87 In the electrolysis of molten \(... [FREE SOLUTION] | 91Ó°ÊÓ

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In the electrolysis of molten \(\mathrm{NaBr}\) (a) What product forms at the anode? (b) What product forms at the cathode?

Short Answer

Expert verified
The product at the anode is bromine gas (\(\text{Br}_2\)), and the product at the cathode is sodium metal (\(\text{Na}\)).

Step by step solution

01

Understand Electrolysis of Molten Ionic Compounds

Electrolysis is a process that uses electric current to drive a non-spontaneous chemical reaction. When molten \(\text{NaBr}\) is electrolyzed, it dissociates into \(\text{Na}^{+}\) and \(\text{Br}^{-}\) ions.
02

Identify Reactions at the Electrodes

In electrolysis, the anode is positively charged and attracts anions (negative ions), while the cathode is negatively charged and attracts cations (positive ions). The relevant ions here are \(\text{Na}^{+}\) and \(\text{Br}^{-}\).
03

Determine the Product at the Anode

At the anode, \(\text{Br}^{-}\) ions lose electrons (oxidation) and form bromine gas. The half-reaction is: \(\text{2Br}^{-} \rightarrow \text{Br}_2 + 2e^{-}\).
04

Determine the Product at the Cathode

At the cathode, \(\text{Na}^{+}\) ions gain electrons (reduction) and form sodium metal. The half-reaction is: \(\text{Na}^{+} + e^{-} \rightarrow \text{Na}\).

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Key Concepts

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

Oxidation
Oxidation is a key concept in chemistry, especially in electrolysis. When we talk about oxidation, we mean the process where an atom or ion loses electrons. For instance, in the electrolysis of molten \text{NaBr}\, the \text{Br}^{-}\ ions at the anode lose electrons to become bromine gas \text{Br}_2\. This process can be described by the half-reaction:
\[\text{2Br}^{-} \rightarrow \text{Br}_2 + 2e^{-}\]
Here, the bromine ions are undergoing oxidation because they are losing electrons.

In electrochemical terms, oxidation always happens at the anode, which is positively charged in an electrolytic cell. The anode attracts negative ions (anions) and it is at this point where the extraction of electrons occurs. Remember, oxidation is often remembered by the acronym 'OIL' in 'OIL RIG', where OIL stands for Oxidation Is Loss (of electrons).
Reduction
Reduction is the opposite of oxidation. It involves the gain of electrons by an atom or ion. During the electrolysis of molten \text{NaBr}\, the \text{Na}^{+}\ ions at the cathode gain electrons to form sodium metal. The half-reaction for this process is:
\[\text{Na}^{+} + e^{-} \rightarrow \text{Na}\]
Here, the sodium ions are undergoing reduction because they are gaining electrons.

In an electrolytic cell, reduction takes place at the cathode, which is negatively charged. The cathode attracts positive ions (cations) and provides them with the electrons they need to become neutral atoms. Similar to oxidation, reduction can be remembered with the acronym 'RIG' in 'OIL RIG', where RIG stands for Reduction Is Gain (of electrons).
Half-Reaction
Understanding half-reactions is crucial for grasping how electrolysis works. A half-reaction shows either the oxidation or reduction process separately, highlighting the transfer of electrons. Let's break down the half-reactions in the electrolysis of molten \text{NaBr}\.

At the anode, we have the oxidation half-reaction:
\[\text{2Br}^{-} \rightarrow \text{Br}_2 + 2e^{-}\]
This half-reaction shows how each bromide ion loses one electron, which collectively forms bromine gas.

At the cathode, we have the reduction half-reaction:
\[\text{Na}^{+} + e^{-} \rightarrow \text{Na}\]
This half-reaction displays how each sodium ion gains one electron to become sodium metal.

By separately examining each half-reaction, we can better understand the overall process occurring during electrolysis. It also helps to identify the specific reactions happening at the anode and cathode. These half-reactions are essential for balancing the overall redox equations correctly.

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Most popular questions from this chapter

Comparing the standard electrode potentials \(\left(E^{\circ}\right)\) of the Group \(1 \mathrm{~A}(1)\) metals \(\mathrm{Li}, \mathrm{Na},\) and \(\mathrm{K}\) with the negative of their first ionization energies reveals a discrepancy: Ionization process reversed: \(\mathrm{M}^{+}(g)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(g)\) Electrode reaction: $$\mathrm{M}^{+}(a q)+\mathrm{e}^{-} \rightleftharpoons \mathrm{M}(s)$$ $$\begin{array}{lcc}\text { Metal } & -\text { IE (kJ/mol) } & E^{\circ} \text { (V) } \\\\\hline \text { Li } & -520 & -3.05 \\\\\text { Na } & -496 & -2.71 \\\ \text { K } & -419 & -2.93\end{array}$$ Note that the electrode potentials do not decrease smoothly down the group, whereas the ionization energies do. You might expect that, if it is more difficult to remove an electron from an atom to form a gaseous ion (larger IE), then it would be less difficult to add an electron to an aqueous ion to form an atom (smaller \(E^{\circ}\) ), yet \(\mathrm{Li}^{+}(a q)\) is more difficult to reduce than \(\mathrm{Na}^{+}(a q) .\) Applying Hess's law, use an approach similar to a Born-Haber cycle to break down the process occurring at the electrode into three steps, and label the energy involved in each step. How can you account for the discrepancy?

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Why does a voltaic cell not operate unless the two compartments are connected through an external circuit?

Electrodes used in electrocardiography are disposable, and many of them incorporate silver. The metal is deposited in a thin layer on a small plastic "button," and then some is converted to AgCl: $$\operatorname{Ag}(s)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \operatorname{AgCl}(s)+\mathrm{e}^{-}$$ (a) If the surface area of the button is \(2.0 \mathrm{~cm}^{2}\) and the thickness of the silver layer is \(7.5 \times 10^{-6} \mathrm{~m},\) calculate the volume (in \(\mathrm{cm}^{3}\) ) of \(\mathrm{Ag}\) used in one electrode. (b) The density of silver metal is \(10.5 \mathrm{~g} / \mathrm{cm}^{3}\). How many grams of silver are used per electrode? (c) If \(\mathrm{Ag}\) is plated on the button from an \(\mathrm{Ag}^{+}\) solution with a current of \(12.0 \mathrm{~mA}\), how many minutes does the plating take? (d) If bulk silver costs \(\$ 28.93\) per troy ounce \((31.10 \mathrm{~g}),\) what is the cost (in cents) of the silver in one disposable electrode?

21.115 A thin circular-disk earring \(4.00 \mathrm{~cm}\) in diameter is plated with a coating of gold \(0.25 \mathrm{~mm}\) thick from an \(\mathrm{Au}^{3+}\) bath. (a) How many days does it take to deposit the gold on one side of one earring if the current is \(0.013 \mathrm{~A}\left(d\right.\) of gold \(\left.=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\) (b) How many days does it take to deposit the gold on both sides of the pair of earrings? (c) If the price of gold is \(\$ 1595\) per troy ounce \((31.10 \mathrm{~g}),\) what is the total cost of the gold plating?

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