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Chapter 21: Problem 88

In the electrolysis of molten \(\mathrm{BaI}_{2}\) (a) What product forms at the negative electrode? (b) What product forms at the positive electrode?

Short Answer

Expert verified
Barium metal forms at the negative electrode and iodine gas forms at the positive electrode.

Step by step solution

01

Understanding Electrolysis

Electrolysis is a chemical process where electrical energy is used to drive a non-spontaneous reaction. In this case, molten \(\mathrm{BaI}_2\) is subjected to electrolysis.
02

Identify the Ions

\(\mathrm{BaI}_2\) dissociates into its ions: \(\mathrm{Ba}^{2+}\) and \(\mathrm{I}^{-}\). These ions will move towards the electrodes based on their charges.
03

Determine the Cathode (Negative Electrode) Reaction

At the negative electrode (cathode), reduction occurs. The \(\mathrm{Ba}^{2+}\) ions gain electrons. The reaction is: \[\mathrm{Ba}^{2+} + 2e^{-} \rightarrow \mathrm{Ba} (s)\]. Thus, barium metal is formed at the cathode.
04

Determine the Anode (Positive Electrode) Reaction

At the positive electrode (anode), oxidation occurs. The \(\mathrm{I}^{-}\) ions lose electrons. The reaction is: \[2\mathrm{I}^{-} \rightarrow \mathrm{I}_2 (g) + 2e^{-}\]. Thus, iodine gas is formed at the anode.

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Key Concepts

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

electrode reactions
During the electrolysis of molten \(\text{BaI}_2\), the ions move towards their respective electrodes based on their charge. The system has a negative electrode (cathode) and a positive electrode (anode). The nature of the reactions occurring at these electrodes depends on the movement of ions.

At the cathode, reduction takes place. Reduction is the gain of electrons. In this process, \(\text{Ba}^{2+}\) ions migrate to the cathode where they gain electrons to form barium metal (\(\text{Ba (s)}\)).

At the anode, oxidation occurs. Oxidation is the loss of electrons. Here, \(\text{I}^{-}\) ions move towards the anode where they lose electrons to form iodine gas (\(\text{I}_2 (g)\)).

This movement of ions ensures that each ion goes to the appropriate electrode to undergo the correct electrochemical reaction.
oxidation and reduction
The principles of oxidation and reduction are essential in understanding electrolysis.

Oxidation: This involves the loss of electrons. During the electrolysis of molten \(\text{BaI}_2\), \(\text{I}^{-}\) ions at the anode (positive electrode) lose electrons to form iodine gas.
The reaction at the anode is:
\[2\text{I}^{-} \rightarrow \text{I}_2 (g) + 2e^{-}\]

Reduction: This is the gain of electrons. In the same scenario, \(\text{Ba}^{2+}\) ions at the cathode (negative electrode) gain electrons to form solid barium.
The reaction at the cathode is:
\[\text{Ba}^{2+} + 2e^{-} \rightarrow \text{Ba (s)}\]

In simple terms, oxidation always involves losing electrons and reduction always involves gaining electrons. The mnemonic 'OIL RIG' can help you remember: Oxidation Is Loss, Reduction Is Gain.
ion movement in electrolysis
In electrolysis, the movement of ions is key to the reaction occurring.

When molten \(\text{BaI}_2\) is subjected to electrolysis, it dissociates into \(\text{Ba}^{2+}\) and \(\text{I}^{-}\) ions.
  • Positive ions (cations) like \(\text{Ba}^{2+}\) move towards the negative electrode (cathode).
  • Negative ions (anions) like \(\text{I}^{-}\) move towards the positive electrode (anode).
This directional movement is driven by the attraction between opposite charges.

At the cathode, the positive \(\text{Ba}^{2+}\) ions gain electrons (reduction) to form barium metal. At the anode, the negative \(\text{I}^{-}\) ions lose electrons (oxidation) to form iodine gas.

By understanding this movement and the resulting electrode reactions, one can predict the products of electrolysis effectively.

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Most popular questions from this chapter

Use the following half-reactions to write three spontaneous reactions, calculate \(E_{\text {cell }}^{\circ}\) for each reaction, and rank the strengths of the oxidizing and reducing agents: (1) \(2 \mathrm{HClO}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(I)\) \(E^{\circ}=1.63 \mathrm{~V}\) (2) \(\mathrm{Pt}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pt}(s) \quad E^{\circ}=1.20 \mathrm{~V}\) (3) \(\mathrm{PbSO}_{4}(s)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Pb}(s)+\mathrm{SO}_{4}^{2-}(a q) \quad E^{\circ}=-0.31 \mathrm{~V}\)

During reconstruction of the Statue of Liberty, Teflon spacers were placed between the iron skeleton and the copper plates that cover the statue. What purpose do these spacers serve?

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?

A voltaic cell consists of \(\mathrm{A} / \mathrm{A}^{+}\) and \(\mathrm{B} / \mathrm{B}^{+}\) half-cells, where A and \(B\) are metals and the A electrode is negative. The initial \(\left[\mathrm{A}^{+}\right] /\left[\mathrm{B}^{+}\right]\) is such that \(E_{\text {cell }}>E_{\text {cell }}^{\circ}\) (a) How do \(\left[\mathrm{A}^{+}\right]\) and \(\left[\mathrm{B}^{+}\right]\) change as the cell operates? (b) How does \(E_{\text {cell }}\) change as the cell operates? (c) What is \(\left[\mathrm{A}^{+}\right] /\left[\mathrm{B}^{+}\right]\) when \(E_{\text {cell }}=E_{\text {cell }}^{\circ} ?\) Explain. (d) Is it possible for \(E_{\text {cell to be less than } E_{\text {cell }}^{\circ} \text { ? Explain. }}\)

Both a D-sized and an AAA-sized alkaline battery have an output of \(1.5 \mathrm{~V}\). What property of the cell potential allows this to occur? What is different about these two batteries?
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