/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 For each of the following reacti... [FREE SOLUTION] | 91Ó°ÊÓ

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For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. \((\mathbf{c})\) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\)

Short Answer

Expert verified
The balanced equation for (a) is: \( 2\mathrm{I}^{-} + \mathrm{Hg}_{2}^{2+} \rightarrow \mathrm{I}_{2} + 2\mathrm{Hg}^{+} \). Standard emf is \(+0.26 \, \text{V}\). \(\Delta G^{\circ} = -50172.2 \, \text{J/mol}\). \(K = 9.95 \times 10^8\).

Step by step solution

01

Write the Balanced Equation for Reaction (a)

For the reaction where aqueous iodide ion is oxidized to \( \mathrm{I}_{2}(s) \) by \( \mathrm{Hg}_{2}^{2+}(aq) \), we have: \( 2 \mathrm{I}^{-}(aq) + \mathrm{Hg}_{2}^{2+}(aq) \rightarrow \mathrm{I}_{2}(s) + 2 \mathrm{Hg}^{+}(aq) \). The iodide ions are oxidized, and the mercury(I) ions are reduced.
02

Calculate the Standard EMF for Reaction (a)

The standard electrode potentials are: \( E^{\circ}(\mathrm{I}_{2}/\mathrm{I}^-) = +0.54 \, \text{V} \) and \( E^{\circ}(\mathrm{Hg}_2^{2+}/\mathrm{Hg}^+) = +0.80 \, \text{V} \). The standard emf is calculated by \( E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} \), which equals \(+0.80 \, \text{V} - (+0.54 \, \text{V}) = +0.26 \, \text{V} \).
03

Calculate \(\Delta G^{\circ}\) for Reaction (a)

Use the formula \( \Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} \), where \( n \) is the number of moles of electrons transferred (2 moles) and \( F \) is Faraday's constant (96485 C/mol). Therefore, \( \Delta G^{\circ} = -(2)(96485)(0.26) = -50172.2 \, \text{J/mol} \).
04

Calculate the Equilibrium Constant \(K\) for Reaction (a)

Use the formula \( \Delta G^{\circ} = -RT \ln K \). Solve for \( K \): \( K = e^{\Delta G^{\circ}/-RT} \) where \( R = 8.314 \, \text{J/mol·K} \) and \( T = 298 \, \text{K} \). Substituting \( \Delta G^{\circ} = -50172.2 \) gives \( K = e^{50172.2/(8.314\times298)} \approx 9.95 \times 10^8 \).

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Key Concepts

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

Standard Electrode Potential
The standard electrode potential, denoted as \( E^{\circ} \), is a measure of the intrinsic ability of a species to gain or lose electrons when connected to a standard hydrogen electrode under standard conditions (1 M concentration, 1 atm pressure, and 25°C). It reveals if the substance is a good oxidizing or reducing agent. A higher \( E^{\circ} \) value indicates a stronger oxidizing agent, while a lower value signifies a more potent reducing agent.
In the context of our exercise, we looked at the reaction involving iodide ions being oxidized by mercurous ions to form solid iodine and mercury(I) ions. The given standard electrode potentials were \( E^{\circ}(\mathrm{I}_2/\mathrm{I}^-) = +0.54\, \text{V} \) and \( E^{\circ}(\mathrm{Hg}_2^{2+}/\mathrm{Hg}^+) = +0.80\, \text{V} \).
  • The cell potential \( E^{\circ}_{\text{cell}} \) is calculated by taking the potential of the cathode and subtracting the potential of the anode.
  • This resulted in a cell potential of \(+0.26\, \text{V}\).
A positive \( E^{\circ}_{\text{cell}} \) confirms the feasibility of the reaction under standard conditions. This understanding is foundational for predicting the direction of electron flow and the spontaneity of the reaction.
Gibbs Free Energy
Gibbs free energy change, represented as \( \Delta G^{\circ} \), determines whether a process will proceed spontaneously. A negative \( \Delta G^{\circ} \) indicates a spontaneous process. Its relation to electrochemistry is given by the equation:
\[ \Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} \]
where:
  • \( n \) is the number of moles of electrons transferred.
  • \( F \) is Faraday's constant, approximately 96485 C/mol.
  • \( E^{\circ}_{\text{cell}} \) is the standard electrode potential of the cell.
For our exercise's reaction \( a \), with two moles of electrons transferred and a cell potential of \(+0.26\, \text{V}\), the calculation yields \( \Delta G^{\circ} = -50172.2\, \text{J/mol}\). This negative value confirms the spontaneity.Understanding \( \Delta G^{\circ} \) values in electrochemistry helps determine if a particular redox reaction will occur on its own, making it crucial for practical applications like battery and fuel cell design.
Equilibrium Constant
The equilibrium constant, \( K \), is a measure of the position of equilibrium for a chemical reaction at a given temperature. It relates the concentrations of reactants and products at equilibrium. In terms of electrochemistry, \( K \) can be derived from the Gibbs free energy change using the formula:
\[ \Delta G^{\circ} = -RT \ln K \]
where:
  • \( R \) is the ideal gas constant, 8.314 J/mol·K.
  • \( T \) is the temperature in Kelvin.
By rearranging this to solve for \( K \):
\[ K = e^{\Delta G^{\circ}/-RT} \]
For the exercise's example with \( \Delta G^{\circ} = -50172.2\, \text{J/mol} \), substituting \( R = 8.314\) J/mol·K and \( T = 298\, \text{K} \), we find \( K \approx 9.95 \times 10^8 \). Such a large \( K \) value indicates a reaction that heavily favors product formation at equilibrium.
Understanding how to calculate and interpret \( K \) is vital in predicting the extent of reactions, especially in designing processes and systems where controlled reaction conditions are necessary, such as environmental systems and industrial processes.

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Most popular questions from this chapter

A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \mathrm{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 \mathrm{M},\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cl^{-} ] will increase, decrease, or stay the same as the cell operates.

(a) What is electrolysis? (b) Are electrolysis reactions thermodynamically spontaneous? (c) What process occurs at the anode in the electrolysis of molten NaCl? (d) Why is sodium metal not obtained when an aqueous solution of NaCl undergoes electrolysis?

Consider the voltaic cell illustrated in Figure \(20.5,\) which is based on the cell reaction $$ \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) $$ Under standard conditions, what is the maximum electrical work, in joules, that the cell can accomplish if \(50.0 \mathrm{~g}\) of copper is formed?

Cytochrome, a complicated molecule that we will represent as \(\mathrm{CyFe}^{2+}\), reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions (Section 19.7). At \(\mathrm{pH} 7.0\) the following reduction potentials pertain to this oxidation of \(\mathrm{CyFe}^{2+}\) : $$ \begin{aligned} \mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & E_{\mathrm{red}}^{\circ} &=+0.82 \mathrm{~V} \\\ \mathrm{CyFe}^{3+}(a q)+\mathrm{e}^{-} & \longrightarrow \mathrm{CyFe}^{2+}(a q) & E_{\mathrm{red}}^{\circ} &=+0.22 \mathrm{~V} \end{aligned} $$ (a) What is \(\Delta G\) for the oxidation of \(\mathrm{CyFe}^{2+}\) by air? \((\mathbf{b})\) If the synthesis of \(1.00 \mathrm{~mol}\) of ATP from adenosine diphosphate (ADP) requires a \(\Delta G\) of \(37.7 \mathrm{~kJ},\) how many moles of ATP are synthesized per mole of \(\mathrm{O}_{2} ?\)

Disulfides are compounds that have \(\mathrm{S}-\mathrm{S}\) bonds, like peroxides have \(\mathrm{O}-\mathrm{O}\) bonds. Thiols are organic compounds that have the general formula \(\mathrm{R}-\mathrm{SH},\) where \(\mathrm{R}\) is a generic hydrocarbon. The \(\mathrm{SH}^{-}\) ion is the sulfur counterpart of hydroxide, \(\mathrm{OH}^{-}\). Two thiols can react to make a disulfide, \(\mathrm{R}-\mathrm{S}-\mathrm{S}-\mathrm{R} .\) (a) What is the oxidation state of sulfur in a thiol? (b) What is the oxidation state of sulfur in a disulfide? (c) If you react two thiols to make a disulfide, are you oxidizing or reducing the thiols? (d) If you wanted to convert a disulfide to two thiols, should you add a reducing agent or oxidizing agent to the solution? (e) Suggest what happens to the H's in the thiols when they form disulfides.

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