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Chapter 21: Problem 99

Write a balanced half-reaction for the product that forms at each electrode in the aqueous electrolysis of the following salts: (a) \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} ;\) (b) \(\mathrm{MnCl}_{2}\).

Short Answer

Expert verified
(a) Reduction: \[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \]; Oxidation: \[2\text{H}_2\text{O} \rightarrow O_2 + 4H^+ + 4e^- \]. (b) Reduction: \[ \text{Mn}^{2+} + 2e^- \rightarrow \text{Mn} \]; Oxidation: \[2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \].

Step by step solution

01

- Identify the ions in solution

Identify the ions present in each solution after dissociation. For \(\text{Cr(NO}_3\text{)}_3\), the ions are \( \text{Cr}^{3+} \) and \( \text{NO}_3^- \). For \( \text{MnCl}_2 \), the ions are \( \text{Mn}^{2+} \) and \( \text{Cl}^- \).
02

- Determine possible reactions at the cathode

At the cathode, reduction occurs. Identify which ion is more likely to be reduced from each solution. \(\text{Cr}^{3+}\) can be reduced to \( \text{Cr} \), and \( \text{Mn}^{2+} \) can be reduced to \( \text{Mn} \).
03

- Write the reduction half-reactions

For \( \text{Cr}^{3+} \): \[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \]. For \( \text{Mn}^{2+} \): \[ \text{Mn}^{2+} + 2e^- \rightarrow \text{Mn} \].
04

- Determine possible reactions at the anode

At the anode, oxidation occurs. Identify which ion is more likely to be oxidized from each solution. \( \text{NO}_3^- \) ion is stable under electrolysis, so water is oxidized to produce oxygen gas and hydrogen ions: \[ 2\text{H}_2\text{O} \rightarrow O_2 + 4H^+ + 4e^- \]. For \(\text{Cl}^-\): Oxidation will form chlorine gas: \[2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \].
05

- Write the oxidation half-reactions

For \( \text{Cr(NO}_3\text{)}_3 \): \[2\text{H}_2\text{O} \rightarrow O_2 + 4H^+ + 4e^- \]. For \(\text{MnCl}_2\): \[2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \].

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Key Concepts

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

reduction half-reaction
The reduction half-reaction is a key concept in electrolysis. Reduction always occurs at the cathode. In this process, ions gain electrons. This gain of electrons reduces the positive charge of the ions.
For example, during the electrolysis of \(\text{Cr(NO}_3\text{)}_3\), chromium ions (\(\text{Cr}^{3+}\)) in solution will undergo reduction. The half-reaction can be written as:
\[ \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \]
Similarly, for \(\text{MnCl}_2\), manganese ions (\(\text{Mn}^{2+}\)) will also undergo reduction:
\[ \text{Mn}^{2+} + 2e^- \rightarrow \text{Mn} \]
Remember, in both cases, the ions are gaining electrons.
  • Chromium ion (\(\text{Cr}^{3+}\)) gains 3 electrons to become chromium metal (\(\text{Cr}\)).
  • Manganese ion (\(\text{Mn}^{2+}\)) gains 2 electrons to become manganese metal (\(\text{Mn}\)).
oxidation half-reaction
In the oxidation half-reaction, oxidation occurs at the anode. This means ions lose electrons. For the aqueous electrolysis of \(\text{Cr(NO}_3\text{)}_3\), nitrate ions (\(\text{NO}_3^-\)) are stable, so water is oxidized instead:
\[ 2\text{H}_2\text{O} \rightarrow O_2 + 4H^+ + 4e^- \]
For \(\text{MnCl}_2\), chloride ions (\(\text{Cl}^-\)) will be oxidized to form chlorine gas:
\[2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \]
In both cases, oxidation results in the loss of electrons:
  • For water: Water molecules lose electrons to form oxygen gas and hydrogen ions.
  • For chloride ions: Chloride ions lose electrons to form chlorine gas.

Thus, half-reactions at the anode involve the removal of electrons.
aqueous electrolysis
Aqueous electrolysis involves passing an electric current through an aqueous solution to cause a chemical change.
In aqueous solutions, water itself can also participate in oxidation or reduction reactions.
It’s important to consider both the ions from the salt as well as the water in the solution.
  • Reduction Potential: Water can be reduced to form hydrogen gas and hydroxide ions.
  • Oxidation Potential: Water can be oxidized to form oxygen gas and hydrogen ions.
For example, in the aqueous electrolysis of \(\text{Cr(NO}_3\text{)}_3\), water is oxidized because nitrate ions are stable.
Whereas for \(\text{MnCl}_2\), chloride ions are oxidized to form chlorine gas.
ion identification
Identifying ions in solution is the first step in solving electrolysis problems. When salts dissolve in water, they dissociate into their constituent ions.
For instance:
  • \(\text{Cr(NO}_3\text{)}_3\) dissociates into chromium ions (\(\text{Cr}^{3+}\)) and nitrate ions (\(\text{NO}_3^-\)).
  • \(\text{MnCl}_2\) dissociates into manganese ions (\(\text{Mn}^{2+}\)) and chloride ions (\(\text{Cl}^-\)).
Recognizing these ions is crucial for predicting the possible reactions at the electrodes.
Once you know the ions, you'll determine which ones are likely to be reduced or oxidized:
  • At the cathode, positive ions (cations) will be reduced.
  • At the anode, negative ions (anions) will be oxidized.
This fundamental step ensures accurate half-reaction predictions in electrolysis problems.

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Most popular questions from this chapter

A chemist designs an ion-specific probe for measuring \(\left[\mathrm{Ag}^{+}\right]\) in an \(\mathrm{NaCl}\) solution saturated with \(\mathrm{AgCl}\). One half-cell has an Ag wire electrode immersed in the unknown AgCl-saturated \(\mathrm{NaCl}\) solution. It is connected through a salt bridge to the other half-cell, which has a calomel reference electrode [a platinum wire immersed in a paste of mercury and calomel (Hg \(_{2} \mathrm{Cl}_{2}\) ) ] in a saturated KCl solution. The measured \(E_{\text {cell }}\) is \(0.060 \mathrm{~V}\). (a) Given the following standard half-reactions, calculate \(\left[\mathrm{Ag}^{+}\right]\). Calomel: \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Hg}(I)+2 \mathrm{Cl}^{-}(a q) \quad E^{\circ}=0.24 \mathrm{~V}\) Silver: \(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{Ag}(s)\) $$E^{\circ}=0.80 \mathrm{~V}$$ (Hint: Assume that [Cl \(^{-}\) ] is so high that it is essentially constant.) (b) A mining engineer wants an ore sample analyzed with the \(\mathrm{Ag}^{+}-\) selective probe. After pretreating the ore sample, the chemist measures the cell voltage as \(0.53 \mathrm{~V}\). What is \(\left[\mathrm{Ag}^{+}\right] ?\)

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

A voltaic cell using \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) and \(\mathrm{Sn} / \mathrm{Sn}^{2+}\) half-cells is set up at standard conditions, and each compartment has a volume of \(345 \mathrm{~mL}\). The cell delivers 0.17 A for \(48.0 \mathrm{~h}\). (a) How many grams of \(\mathrm{Cu}(s)\) are deposited? (b) What is the \(\left[\mathrm{Cu}^{2+}\right]\) remaining?

A voltaic cell consists of two \(\mathrm{H}_{2} / \mathrm{H}^{+}\) half-cells. Half-cell \(\mathrm{A}\) has \(\mathrm{H}_{2}\) at 0.95 atm bubbling into \(0.10 \mathrm{M} \mathrm{HCl}\). Half-cell \(\mathrm{B}\) has \(\mathrm{H}_{2}\) at 0.60 atm bubbling into \(2.0 \mathrm{M} \mathrm{HCl}\). Which half-cell houses the anode? What is the voltage of the cell?

To examine the effect of ion removal on cell voltage, a chemist constructs two voltaic cells, each with a standard hydrogen electrode in one compartment. One cell also contains a \(\mathrm{Pb} / \mathrm{Pb}^{2+}\) half-cell; the other contains a \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell. (a) What is \(E^{\circ}\) of each cell at \(298 \mathrm{~K} ?\) (b) Which electrode in each cell is negative? (c) When \(\mathrm{Na}_{2} \mathrm{~S}\) solution is added to the \(\mathrm{Pb}^{2+}\) electrolyte, solid \(\mathrm{PbS}\) forms. What happens to the cell voltage? (d) When sufficient \(\mathrm{Na}_{2} \mathrm{~S}\) is added to the \(\mathrm{Cu}^{2+}\) electrolyte, CuS forms and \(\left[\mathrm{Cu}^{2+}\right]\) drops to \(1 \times 10^{-16} \mathrm{M} .\) Find the cell voltage.
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