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(a) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell.(b) Write the half-reaction that occurs at a hydrogen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard hydrogen electrode?

Short Answer

Expert verified
The half-reaction for a hydrogen electrode when it acts as a cathode in acidic aqueous solution is: \[ 2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g) \] The half-reaction for a hydrogen electrode when it acts as an anode in acidic aqueous solution is: \[ H_{2}(g) \rightarrow 2H^{+}(aq) + 2e^{-} \] The standard hydrogen electrode (SHE) serves as a reference electrode and is maintained under these specific conditions: 1 M concentration of H+ ions, 1 atm pressure of H2 gas, and 25°C temperature.

Step by step solution

01

Understanding Redox Reactions

A redox reaction involves two half-reactions: one for the reduction process and one for the oxidation process. The substance that gains electrons is reduced, while the substance that loses electrons is oxidized. Step 2: Writing the half-reaction for a hydrogen electrode when it acts as a cathode
02

Writing the half-reaction for a hydrogen electrode when it acts as a cathode

When the hydrogen electrode is acting as a cathode, it will undergo a reduction reaction. In acidic aqueous solution, hydrogen ions (H+) will gain electrons to form hydrogen gas (H2). The half-reaction can be written as follows: \[ 2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g) \] Step 3: Writing the half-reaction for a hydrogen electrode when it acts as an anode
03

Writing the half-reaction for a hydrogen electrode when it acts as an anode

When the hydrogen electrode is acting as an anode, it will undergo an oxidation reaction. In acidic aqueous solution, hydrogen gas (H2) will lose electrons to form hydrogen ions (H+). The half-reaction can be written as follows: \[ H_{2}(g) \rightarrow 2H^{+}(aq) + 2e^{-} \] Step 4: Understanding the standard hydrogen electrode
04

Understanding the standard hydrogen electrode

The standard hydrogen electrode (SHE) is used as a reference electrode in electrochemical cells. The conditions for the standard hydrogen electrode are defined as follows: 1. The concentration of hydrogen ions (H+) in aqueous solution is 1 M (1 mol/dm³). 2. The pressure of hydrogen gas (H2) is 1 atmosphere (1 atm). 3. The temperature is 25°C (298K). By maintaining these specific conditions, the SHE can be used as a reference to measure the electromotive force (EMF) of other electrodes and redox reactions.

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Key Concepts

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

Redox Reactions
Redox reactions are fundamental to understanding many chemical processes, such as those occurring in electrochemical cells. These reactions involve the transfer of electrons between two substances. In a redox reaction, one substance gets oxidized while another gets reduced.
  • Reduction is the gain of electrons. The electron receiver is known as the oxidizing agent.
  • Oxidation is the loss of electrons. The electron donor is called the reducing agent.
These reactions are crucial in batteries, rusting of metals, and cellular respiration. In the exercise, two half-reactions demonstrate the process of electrons being transferred through the use of a hydrogen electrode.
Half-Reaction
In the context of redox chemistry, a half-reaction represents either the oxidation or reduction phase of the process. It specifically describes the transfer of electrons for one of the reactants.
Half-reactions are written to show the transfer of electrons explicitly:
  • The reduction half-reaction involves a species gaining electrons.
  • The oxidation half-reaction involves a species losing electrons.
When balancing redox equations, these half-reactions are typically used to ensure that the electrons lost in oxidation are equal to those gained in reduction.
Electrochemical Cells
Electrochemical cells are devices capable of generating electrical energy through redox reactions. They consist of two electrodes immersed in an electrolyte solution.
Each electrode acts as either an anode or a cathode, depending on the direction of the redox reaction.
  • Galvanic cells (voltaic cells) transform chemical energy into electrical energy. They do this by harnessing spontaneous redox reactions.
  • Electrolytic cells, on the other hand, drive non-spontaneous reactions using an external electrical source.
These cells find applications in batteries, performing electrolysis, and metal plating.
Hydrogen Electrode
The hydrogen electrode, often referred to as the standard hydrogen electrode (SHE), plays a critical role in determining electrode potentials. It relies on the half-reaction: \[ 2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g) \] The hydrogen electrode serves as a universal standard for measuring the electrode potential of other reactions. It maintains very specific conditions:
  • The concentration of hydrogen ions is standardized at 1 M.
  • The pressure of the hydrogen gas is kept at 1 atm.
  • The temperature is maintained at 25°C (298K).
These conditions enable it to provide a reference potential of 0 volts, offering a benchmark for measuring and comparing other electrode reactions.
Cathode and Anode Reactions
In electrochemical cells, the terms cathode and anode define the roles of the electrodes where redox reactions occur. Their nomenclature is based on the direction of electronic flow:
  • The **Cathode** is the electrode where reduction occurs. In a galvanic cell, it is the positive electrode.
  • The **Anode** is the electrode where oxidation occurs. In a galvanic cell, it is the negative electrode.
In the hydrogen electrode example from the exercise:
- As a cathode, hydrogen ions gain electrons to form hydrogen gas: \[ 2H^{+}(aq) + 2e^{-} \rightarrow H_{2}(g) \] - As an anode, hydrogen gas loses electrons to form hydrogen ions: \[ H_{2}(g) \rightarrow 2H^{+}(aq) + 2e^{-} \] Understanding these reactions helps in comprehending how electrochemical cells function and their applications.

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Most popular questions from this chapter

Given the following half-reactions and associated standard reduction potentials: $$ \begin{array}{c}{\text { AuBr }_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{V}} \\ {\mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q)} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad {E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{V}}\end{array} $$ $$ \begin{array}{r}{\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q)} \\\ {E_{\mathrm{red}}^{\circ}=+0.49 \mathrm{V}}\end{array} $$ (a) Write the equation for the combination of these half-cell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.

A voltaic cell is constructed with all reactants and products in their standard states. Will the concentration of the reactants increase, decrease, or remain the same as the cell operates?

The standard reduction potential of \(\mathrm{Eu}^{2+}(a q)\) is \(-0.43 \mathrm{V}\) . Using Appendix E, which of the following substances is capable of reducing Eu' \((a q)\) to \(\mathrm{Eu}^{2+}(a q)\) under standard conditions: Al, Co, \(\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{N}_{2} \mathrm{H}_{5}^{+}, \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} ?\)

A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: $$ \begin{array}{l}{\text { (a) } \mathrm{Ni}^{+}(a q) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{Ni}(s)(\text { acidic solution })} \\ {\text { (b) } \operatorname{MnO}_{4}^{2-}(a q) \longrightarrow \mathrm{MnO}_{4}^{-}(a q)+\mathrm{MnO}_{2}(s) \text { (acidic }} \\ \quad {\text { solution) }} \\\ {\text { (c) } \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{S}(s)+\mathrm{HSO}_{4}^{-}(a q)(\text { acidic solution })} \\ {\text { (d) } \mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{Cl}^{-}(a q)+\mathrm{ClO}^{-}(a q) \text { (basic solution) }}\end{array} $$

From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent: $$ \begin{array}{l}{\text { (a) } \mathrm{Cl}_{2}(g) \text { or } \mathrm{Br}_{2}(l)} \\ {\text { (b) } \mathrm{Zn}^{2+}(a q) \text { or } \mathrm{Cd}^{2+}(a q)} \\ {\text { (c) } \mathrm{Cl}^{-}(a q) \text { or } \mathrm{ClO}_{3}(a q)} \\ {\text { (d) } \mathrm{H}_{2} \mathrm{O}_{2}(a q) \text { or } \mathrm{O}_{3}(\mathrm{g})}\end{array} $$

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