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The chemical reaction, \(2 \mathrm{AgCl}(\mathrm{s})+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HCl}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})\) taking place in a galvanic cell is represented by the notation (a) \(\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_{2}(\mathrm{~g}), 1\) bar \(\mid 1 \mathrm{M} \mathrm{KCl}\) (aq) \(|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}\) (s) (b) \(\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_{2}(\mathrm{~g}), 1\) bar \(\mid 1 \mathrm{M} \mathrm{KCl}\) (aq) \(\mid 1 \mathrm{M} \mathrm{Ag}^{+}\)(aq) \(\mathrm{Ag}(\mathrm{s})\) (c) \(\mathrm{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{~g}), 1 \mathrm{bar}\right| 1 \mathrm{M} \mathrm{KCl}\) (aq) \(|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}\) (s) (d) \(\mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_{2}(\mathrm{~g}), 1\) bar \(\mid 1 \mathrm{M} \mathrm{KCl}\) (aq) \(|\mathrm{Ag}(\mathrm{s})| \mathrm{AgCl}\)

Short Answer

Expert verified
The correct cell notation is option (a).

Step by step solution

01

Identify the Reaction Components

The reaction is given as \(2 \mathrm{AgCl} (\mathrm{s}) + \mathrm{H}_{2} (\mathrm{g}) \rightarrow 2 \mathrm{HCl} (\mathrm{aq}) + 2 \mathrm{Ag} (\mathrm{s})\). This chemical equation involves two half-reactions: oxidation and reduction.
02

Determine Oxidation and Reduction

In this reaction, \( \mathrm{AgCl} \) is reduced to \( \mathrm{Ag} \), while \( \mathrm{H}_2 \) is oxidized to \( \mathrm{HCl} \). The respective half-reactions are: \( \mathrm{AgCl} + e^- \rightarrow \mathrm{Ag} + \mathrm{Cl}^- \) (reduction) and \( \mathrm{H}_2 \rightarrow 2\mathrm{H}^+ + 2e^- \) (oxidation).
03

Apply Galvanic Cell Notation for the Reaction

In galvanic cells, a particular notation is used to represent the cell. The representation starts from the anode (where oxidation occurs) and proceeds to the cathode (where reduction occurs). For this cell, the anode reaction \(\mathrm{H}_2\) turning into \(\mathrm{H}^+\) occurs at a \(\mathrm{Pt}\) electrode, and the cathode reaction \( \mathrm{AgCl} \) to \( \mathrm{Ag} \) occurs at a \(\mathrm{Ag}\) electrode.
04

Write the Correct Cell Representation

Using the standard notation for galvanic cells, each side is separated by a double vertical line indicating the salt bridge or the phase boundary. The correct notation should include \(\mathrm{Pt}(\mathrm{s})\) as the electrode for the \( \mathrm{H}_2 \) reaction, \(\mathrm{H}_2\) at 1 bar, and \(\mathrm{Ag}(\mathrm{s})\) with \(\mathrm{AgCl}(\mathrm{s})\): \( \mathrm{Pt}(\mathrm{s}) \mid \mathrm{H}_2(\mathrm{g}), 1 \text{ bar} \mid 1 \mathrm{M} \mathrm{Cl}^- \mid \mathrm{AgCl}(\mathrm{s}) \mid \mathrm{Ag}(\mathrm{s})\).
05

Identify the Matching Option

Compare the cell notation with the options provided. The correct cell representation using this notation is option (a), which is the correct representation of the reaction in a galvanic cell.

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Key Concepts

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

Oxidation and Reduction
Oxidation and reduction reactions are essential processes in chemistry, particularly in redox reactions where they occur simultaneously. In any redox reaction, we have two half-reactions: one for oxidation and one for reduction. Oxidation involves the loss of electrons from a substance, while reduction involves the gain of electrons.
For example, in the chemical reaction given:
  • Oxidation: The reaction of hydrogen gas (\( \mathrm{H}_2 \)) to produce hydrochloric acid (\( \mathrm{HCl} \)), which results in the loss of electrons.
  • Reduction: The conversion of silver chloride (\( \mathrm{AgCl} \)) to silver (\( \mathrm{Ag} \)), which involves gaining electrons.
These redox processes are indispensable in galvanic cells, powering the cell through the flow of electrons from the oxidizing agent to the reducing agent. Understanding oxidation and reduction is fundamental to grasping how electrochemical cells function.
Half-Reactions
In electrochemistry, any redox process can be broken down into two simpler components called half-reactions. This is crucial as it helps in understanding the electron flow within a galvanic cell. The half-reactions illustrate separately what happens to each reactant in terms of electron transfer.
For the given reaction, the half-reactions can be expressed as follows:
  • Oxidation Half-Reaction: \( \mathrm{H}_2 ightarrow 2 \mathrm{H}^+ + 2e^- \). This accounts for the loss of electrons by hydrogen gas leading to the formation of hydrogen ions.
  • Reduction Half-Reaction: \( \mathrm{AgCl} + e^- ightarrow \mathrm{Ag} + \mathrm{Cl}^- \). Here, silver chloride gains electrons to form silver and chloride ions.
Recognizing and writing these half-reactions is vital for constructing and understanding the operations of galvanic cells.
Anode and Cathode
The concepts of anode and cathode are central to understanding galvanic cells. In simple terms, the anode is where oxidation occurs, and the cathode is where reduction takes place. This distinction is key in mapping out the direction of electron flow in a cell.
  • Anode: In a galvanic cell, the anode is the site of oxidation. For the discussed reaction, hydrogen gas (\( \mathrm{H}_2 \)) oxidizes at the anode, releasing electrons.
  • Cathode: The cathode is where reduction occurs. In this case, silver chloride (\( \mathrm{AgCl} \)) at the cathode undergoes reduction to form silver (\( \mathrm{Ag} \)).
This electron movement from anode to cathode is pivotal in generating the electrical energy of a galvanic cell. Correctly identifying these sites is essential for accurately setting up and analyzing galvanic cells.
Electrochemistry Concepts
Electrochemistry is the branch of chemistry that studies the relationship between electricity and chemical reactions. It forms the foundation of understanding how galvanic cells, also known as voltaic cells, operate. In a galvanic cell, chemical energy is transformed into electrical energy through spontaneous redox reactions.
Key principles:
  • Galvanic Cell: It consists of two half-cells connected through a salt bridge, facilitating ion exchange to maintain electrical neutrality.
  • Cell Notation: A symbolic representation used to describe the structure and setup of a galvanic cell. It starts with the anode (oxidation side) and proceeds to the cathode (reduction side).
  • Electron Flow: Electrons flow from anode to cathode, driven by the potential difference between the two electrodes.
  • Salt Bridge: Helps complete the electrical circuit and allows ions to flow, preventing charge buildup that would stop the reaction.
Understanding these concepts is essential to designing, interpreting, and utilizing galvanic cells effectively in both educational and practical scenarios.

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Most popular questions from this chapter

The standard reduction potentials at \(298 \mathrm{~K}\) for the following half- reactions are given against each \(\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Zn}(\mathrm{s})-0.762\) \(\mathrm{Cr}^{3+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Cr}(\mathrm{s}) \quad-0.740\) \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g}) \quad 0.000\) \(\mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Fe}^{2+}\) (aq) \(0.770\) Which is the strongest reducing agent? (a) \(\mathrm{H}_{2}(\mathrm{~g})\) (b) \(\mathrm{Cr}(\mathrm{s})\) (c) \(\mathrm{Zn}(\mathrm{s})\) (d) \(\mathrm{Fe}^{2+}(\mathrm{aq})\)

The specific conductance of \(0.1 \mathrm{~N} \mathrm{KCl}\) solution at \(23^{\circ} \mathrm{C}\) is \(0.012 \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}\). The resistance of cell containing the solution at the same temperature was found to be 55 ohm. The cell constant will be (a) \(0.142 \mathrm{~cm}^{-1}\) (b) \(0.616 \mathrm{~cm}^{-1}\) (c) \(6.16 \mathrm{~cm}^{-1}\) (d) \(616 \mathrm{~cm}^{-1}\)

Identify the compounds in which the sulphur atoms are in different oxidation states? (a) \(\mathrm{K}_{2} \mathrm{~S}_{2} \mathrm{O}_{7}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) (c) \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\) (d) \(\mathrm{K}_{2} \mathrm{~S}_{2} \overline{\mathrm{O}}_{\mathrm{s}}\)

In the ionic equation \(\mathrm{xBrO}_{3}^{-}+\mathrm{yCr}^{3+}+\mathrm{zH}_{2} \mathrm{O} \longrightarrow \mathrm{Br}_{2}+\mathrm{HCrO}_{4}^{-}+\mathrm{H}^{+}\) (a) \(\mathrm{x}=6, \mathrm{y}=8, \mathrm{z}=10\) (b) \(x=10, y=6, z=22\) (c) \(x=6, y=8, z=10\) (d) \(x=6, y=10, z=22\)

The standard emf of a cell, involving one electron change is found to be \(0.591 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction is \(\left(F=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}\right.\) \(\left.=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)\) (a) \(1.0 \times 10^{30}\) (b) \(1.0 \times 10^{1}\) (c) \(1.0 \times 10^{5}\) (d) \(1.0 \times 10^{10}\)

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