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What information can you obtain from the following cell diagram? $$\begin{aligned} \mathrm{Pt}\left[\mathrm{H}_{2}(\mathrm{g}, 1 \mathrm{bar})\right] | 2 \mathrm{H}^{+}\left(\mathrm{aq}, 1.0 \mathrm{mol} \mathrm{dm}^{-3}\right) & \\\ & \vdots \mathrm{Ce}^{4+}(\mathrm{aq}), \mathrm{Ce}^{3+}(\mathrm{aq}) | \mathrm{Pt} \end{aligned}$$

Short Answer

Expert verified
This cell diagram represents an electrochemical cell with hydrogen as the anode and cerium ions as the cathode. The net cell reaction is \(\text{H}_2(g) + 2\text{Ce}^{4+}(aq) \rightarrow 2\text{H}^+(aq) + 2\text{Ce}^{3+}(aq)\).

Step by step solution

01

Understanding the Cell Diagram

The provided cell diagram represents an electrochemical cell, often used to describe the components and reactions occurring in a galvanic or voltaic cell. In this cell, each half-cell is separated by a phase boundary, indicated by the ‘|’ symbol. Here, Pt is a solid electrode that serves as the site for both oxidation and reduction.
02

Identifying the Anode and Cathode

The left side of the cell diagram (\( ext{Pt}[ ext{H}_2(g, 1 ext{ bar})] | 2 ext{H}^+(aq, 1.0 ext{ mol dm}^{-3}) \)) represents the anode where the oxidation reaction occurs, which involves hydrogen gas. The right side (\( ext{Ce}^{4+}(aq), ext{Ce}^{3+}(aq) | ext{Pt} \)) represents the cathode where the reduction reaction occurs, involving cerium ions.
03

Determining Anode Reaction

At the anode, hydrogen gas is oxidized. The reaction can be written as: \( ext{H}_2(g) ightarrow 2 ext{H}^+(aq) + 2e^- \). This means that hydrogen molecules lose electrons and form hydrogen ions, releasing electrons into the external circuit.
04

Determining Cathode Reaction

At the cathode, cerium ions are reduced. The reaction can be expressed as: \( ext{Ce}^{4+}(aq) + e^- ightarrow ext{Ce}^{3+}(aq) \). Here, cerium ions gain electrons from the external circuit, reducing from \( ext{Ce}^{4+} \) to \( ext{Ce}^{3+} \).
05

Net Cell Reaction

The overall cell reaction is the sum of the oxidation and reduction reactions. Combining these, we have: \( ext{H}_2(g) + 2 ext{Ce}^{4+}(aq) ightarrow 2 ext{H}^+(aq) + 2 ext{Ce}^{3+}(aq) \). This shows the conversion taking place throughout the cell.
06

Potential Differences and Equilibrium

This setup can help determine the cell potential, but specific values require standard electrode potentials. However, qualitatively, such descriptions can identify cell efficiencies and reaction tendencies as hydrogen is often involved in standard electrode potential calculations.

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Key Concepts

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

Cell Diagram
A cell diagram is a shorthand notation to represent electrochemical cells, which includes galvanic or voltaic cells. It provides a structured way to display the cell's components and the direction of electron flow. In an electrochemical cell, the cell diagram lists the components of each half-cell, starting from the anode on the left side and ending with the cathode on the right side. It uses symbols like '|', which indicates a phase boundary, and sometimes '||' or a salt bridge if included.
Furthermore, the diagram includes the electrode materials and the concentrations or pressures of substances in the reaction. For example, in the provided cell diagram, Pt is used as an inert electrode, and you can observe phase boundaries between gaseous, solid, and aqueous states. Understanding these representations are crucial for analyzing and predicting the electrochemical behavior of the system.
  • The left side involves oxidation and is called the anode half-cell.
  • The right side involves reduction and is called the cathode half-cell.
  • Concentration and pressure details are included to give full reaction context.
Anode Reaction
The anode reaction occurs at the part of the electrochemical cell where oxidation happens. In our specific example of the cell diagram, the anode involves hydrogen. Hydrogen gas (\(H_2(g)\)) reacts here. During this process, it is oxidized to form hydrogen ions (\(2H^+(aq)\)), releasing electrons.
The reaction at the anode can be described by the equation:\[H_2(g) \rightarrow 2H^+(aq) + 2e^-\]This depicts hydrogen molecules losing electrons, making the electrons available for the external circuit to transport. The free electrons travel from the anode through the external circuit to reach the cathode, thus making a continuous flow of electric current.
  • Oxidation involves loss of electrons.
  • This process contributes electrons to the circuit.
  • It transforms molecules from their elemental state to ionic form.
Cathode Reaction
In electrochemical cells, the cathode is where reduction takes place. This means substances gain electrons at the cathode. In the examined cell diagram, the reduction reaction involves cerium ions. At the cathode, cerium ions (\(Ce^{4+}(aq)\)) gain electrons to be reduced to (\(Ce^{3+}(aq)\)).
The reaction occurring at the cathode is represented as:\[Ce^{4+}(aq) + e^- \rightarrow Ce^{3+}(aq)\]During this reaction, cerium ions accept an electron from the external circuit, reducing its oxidation state by one unit. The flow of electrons means that the electric current is driving the chemical change at the cathode.
  • Reduction involves gaining electrons.
  • This reaction balances the electrons lost at the anode.
  • It alters the oxidation state of ions.
Galvanic Cell
A galvanic cell is a type of electrochemical cell that generates electrical energy from spontaneous chemical reactions occurring within it. It is an excellent educator's tool for demonstrating how electrochemical processes work. The characteristic feature of such cells is their ability to convert chemical energy into electrical energy efficiently.
In our example, the galvanic cell involves two separate half-cells: one being the hydrogen-gas half-cell and the other containing cerium ions. These half-cells are connected by a wire and often a salt bridge which completes the circuit, allowing ions to flow and maintain charge balance.
The spontaneous reactions occurring in each half-cell result in a net flow of electrons from anode to cathode through the external circuit, providing useful electrical work. The cell's arrangement indicates the spontaneity and is an important concept in thermodynamics and electrochemistry.
  • Galvanic cells consist of two half-cells.
  • They spontaneously convert chemical to electrical energy.
  • Useful in batteries and corrosion prevention.

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Most popular questions from this chapter

Suggest what products are discharged at the anode and cathode during the electrolysis of (a) molten \(\mathrm{KBr} ;(\mathrm{b})\) fused \(\mathrm{CaCl}_{2} ;(\mathrm{c})\) dilute aqueous \(\mathrm{NaCl}\) (d) concentrated aqueous NaCl (brine); (e) aqueous \(\mathrm{CuSO}_{4}\) using Cu electrodes; (f) dilute \(\mathrm{H}_{2} \mathrm{SO}_{4}\) using Pt electrodes.

The manufacture of \(\mathrm{Al}\) by electrolysis of fused \(\mathrm{Al}_{2} \mathrm{O}_{3}\) and \(\mathrm{K}_{3}\left[\mathrm{AlF}_{6}\right]\) demands especially large amounts of electricity, and aluminium refineries are often associated with hydroelectric schemes. Rationalize why this is so.

Balance the following half-equations by finding the number, \(z,\) of electrons: (a) \(\mathrm{Cr}^{2+}+z \mathrm{e}^{-}=\mathrm{Cr}\) (b) \(\left[\mathrm{Cr}_{2} \mathrm{O}_{7}\right]^{2-}+14 \mathrm{H}^{+}+\mathrm{ze}^{-} \rightleftharpoons 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{S}_{4} \mathrm{O}_{6}\right]^{2-}+\mathrm{ze}^{-}=2\left[\mathrm{S}_{2} \mathrm{O}_{3}\right]^{2-}\) (d) \(\mathrm{O}_{2}+4 \mathrm{H}^{+}+\mathrm{ze}^{-}=2 \mathrm{H}_{2} \mathrm{O}\) (e) \(\left[\mathrm{BrO}_{3}\right]^{-}+3 \mathrm{H}_{2} \mathrm{O}+\mathrm{ze}^{-} \rightleftharpoons \mathrm{Br}^{-}+6[\mathrm{OH}]^{-}\) \((f) P b O_{2}+4 H^{+}+\left[S O_{4}\right]^{2-}+z e^{-}=P b S O_{4}+2 H_{2} O\)

For the reaction:\\[ 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \rightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{Zn}^{2+}(\mathrm{aq}) \\] \(E_{\text {cell }}^{\circ}=1.56 \mathrm{V} .\) Calculate \(\Delta G^{\circ}\) per mole of \(\mathrm{Ag}^{+}\)

Discuss the following statements and data. (a) In dilute aqueous solution, \(\left[\mathrm{ClO}_{4}\right]^{-}\) is very difficult to reduce despite the values of \(E^{\circ}\) for the following half- cells at pH 0 : $$\begin{aligned} \left[\mathrm{ClO}_{4}\right]^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} &=\\\ \left[\mathrm{ClO}_{3}\right]^{-}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & E^{0}=+1.19 \mathrm{V} \\ \left[\mathrm{ClO}_{4}\right]^{-}(\mathrm{aq})+8 \mathrm{H}^{+}(\mathrm{aq})+8 \mathrm{e}^{-} \rightleftharpoons &=\\\\\mathrm{Cl}^{-}(\mathrm{aq})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & E^{\circ}=+1.24 \mathrm{V}\end{aligned}$$ (b) Lead-acid batteries involve the half-cells: $$\begin{aligned} \mathrm{PbSO}_{4}(\mathrm{s})+2 \mathrm{e}^{-} &=\\\ \mathrm{Pb}(\mathrm{s})+\left[\mathrm{SO}_{4}\right]^{2-}(\mathrm{aq}) & E^{0}=-0.36 \mathrm{V} \\ \mathrm{PbO}_{2}(\mathrm{s})+4 \mathrm{H}^{+}(\mathrm{aq})+\left[\mathrm{SO}_{4}\right]^{2-}(\mathrm{aq})+2 \mathrm{e}^{-} &=\\\ \mathrm{PbSO}_{4}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(1) & E^{0}=+1.69 \mathrm{V} \end{aligned}$$ Batteries can be recharged by reversing the cell reaction.

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