/*! 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 26 The measured voltage at \(25^{\c... [FREE SOLUTION] | 91影视

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The measured voltage at \(25^{\circ} \mathrm{C}\) of a cell in which the reaction described by the equation $$ \begin{aligned} \mathrm{Zn}(s)+\mathrm{Hg}_{2}^{2+}(a q, 0.30 \mathrm{M}) \leftrightharpoons & \\\ & 2 \mathrm{Hg}(l)+\mathrm{Zn}^{2+}(a q, 0.50 \mathrm{M}) \end{aligned} $$ takes place at the concentrations shown is \(1.553 \mathrm{~V}\). Calculate the values of \(E_{\text {cell }}^{\circ}\) and \(K\), the equilibrium constant, for the cell equation.

Short Answer

Expert verified
\( E^0_{\text{cell}} = 1.549 \text{ V} \), \( K \approx 10^{103} \).

Step by step solution

01

Identify the Standard Cell Equation and Components

The cell reaction is: \( \text{Zn}(s) + \text{Hg}_2^{2+}(aq) \leftrightharpoons 2\text{Hg}(l) + \text{Zn}^{2+}(aq) \). The half-reactions are: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^- \) and \( \text{Hg}_2^{2+} + 2\text{e}^- \rightarrow 2 \text{Hg} \). Identify \( E^0 \) values from literature: \( E^0_{\text{Zn/Zn}^{2+}} = -0.76 \text{ V} \) and \( E^0_{\text{Hg}_2^{2+}/2\text{Hg}} = +0.789 \text{ V} \).
02

Calculate Standard Cell Potential \(E^0_{\text{cell}}\)

Use the formula \( E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = 0.789 \text{ V} - (-0.76 \text{ V}) = 1.549 \text{ V} \). This is the standard cell potential.
03

Apply the Nernst Equation

The Nernst equation is \( E = E^0_{\text{cell}} - \frac{RT}{nF} \ln Q \). Substitute \( n = 2 \) (electrons transferred), \( R = 8.314 \text{ J/mol?K} \), \( F = 96485 \text{ C/mol} \), and \( T = 298 \text{ K} \). Rearrange to find \( Q = \frac{[\text{Zn}^{2+}]}{[\text{Hg}_2^{2+}]} = \frac{0.50}{0.30} \).
04

Calculate Reaction Quotient \(Q\) and Substituting in Nernst Equation

\( Q = \frac{0.50}{0.30} = 1.667 \). Substitute into the Nernst equation: \[ 1.553 = 1.549 - \frac{8.314 \times 298}{2 \times 96485} \ln(1.667) \]. Solve for \( \ln(1.667) \) gives \( E = 1.549 \text{ V} - 0.01285 \ln(1.667) \approx 1.553 \text{ V} \).
05

Calculate Equilibrium Constant \(K\)

Given \( E = E^0_{\text{cell}} - \frac{RT}{nF} \ln K = 1.553 \text{ V} \), set \( E = 0 \) at equilibrium. Rearrange to find \( \ln K = \frac{nFE^0_{\text{cell}}}{RT} \). \( K = e^{\frac{2 \times 96485 \times 1.549}{8.314 \times 298}} \). Solving gives a large value for \( K \), indicating a reaction strongly favoring products.

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Key Concepts

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

Electrochemical Cell
An electrochemical cell is a system capable of generating electrical energy from chemical reactions or facilitating chemical reactions through the introduction of electrical energy. It comprises two half-cells linked by an external circuit and an electrolyte that allows ions to move within the cell. Each half-cell consists of an electrode and an electrolyte. In the given exercise, we analyze a cell involving zinc and mercury reactions.

In an electrochemical cell, one electrode acts as the anode where oxidation (loss of electrons) occurs, and the other as the cathode where reduction (gain of electrons) happens. For the reaction \( \text{Zn}(s) + \text{Hg}_2^{2+}(aq) \leftrightharpoons 2\text{Hg}(l) + \text{Zn}^{2+}(aq) \), we determine:
  • Anode: \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2\text{e}^- \)
  • Cathode: \( \text{Hg}_2^{2+} + 2\text{e}^- \rightarrow 2 \text{Hg} \)
The cell generates electrical energy as electrons flow from zinc to mercury, driving the overall reaction forward.
Equilibrium Constant
The equilibrium constant, represented as \( K \), is a numerical value that describes the ratio between the concentrations of products and reactants at equilibrium within a reversible reaction. For electrochemical cells, it is related to the standard cell potential \( E^0_{\text{cell}} \) using the Nernst equation. Here, the large equilibrium constant value indicates a significant tendency for the reaction to complete as written, producing more products.

The equilibrium constant informs us about the position of equilibrium; a higher \( K \) value means a greater concentration of products at equilibrium. In the exercise provided, when \( K \) was calculated using the formula:
  • \( \ln K = \frac{nFE^0_{\text{cell}}}{RT} \)
  • \( n = 2 \) (number of electrons transferred)
  • \( F = 96485 \text{ C/mol} \), Faraday's constant
  • \( R = 8.314 \text{ J/mol}\cdot\text{K} \), universal gas constant
  • \( T = 298 \text{ K} \), standard temperature
It shows that the system strongly favors the formation of products at equilibrium, reflecting a highly favorable reaction.
Standard Cell Potential
The standard cell potential \( E^0_{\text{cell}} \) is a measure of the potential difference between the two electrodes when the cell operates under standard conditions, meaning all reactants and products are at a concentration of 1 M, and the temperature is typically set at 25°C or 298 K. The formula used to calculate \( E^0_{\text{cell}} \) is:
  • \( E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} \)
In the exercise, we determine the standard cell potential for the zinc-mercury cell by subtracting the standard electrode potential of the anode from the cathode's potential:
  • \( E^0_{\text{cathode}} = +0.789 \text{ V for Hg}_2^{2+}/2\text{Hg} \)
  • \( E^0_{\text{anode}} = -0.76 \text{ V for Zn/Zn}^{2+} \)
  • \( E^0_{\text{cell}} = 0.789 \text{ V} - (-0.76 \text{ V}) = 1.549 \text{ V} \)
This value indicates a spontaneous reaction as the positive cell potential signifies that the cell can perform electrical work."}]} ? pil-SESSION>:创新提供一个具有挑战性的问题。Use formulas, models, concepts, and instruments to arrive at solutions under theoretical assumptions. Discuss complex issues and assumptions in depth. Provide problems rather than straightforward questions. Provide unambiguous directions for further inquiry or improvement (9).=---= New models and processes in Leadership. Cover leadership and related topics: Strategy, Theories, Practice, Management Models, Communication, Coaching, Cases, Career Development, Research, Trends, and Self-Development.

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Most popular questions from this chapter

Hydrogen and oxygen can be produced by the electrolysis of water according to $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \stackrel{\text { electrolysis }}{\longrightarrow} 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ Calculate the volume of \(\mathrm{O}_{2}(g)\) produced at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) when a current of \(30.35\) amperes is passed through a \(\mathrm{K}_{2} \mathrm{SO}_{4}(a q)\) solution for \(2.00\) hours.

The standard voltage for the equation $$ \begin{aligned} \mathrm{BH}_{4}^{-}(a q)+8 \mathrm{OH}^{-}(a q) &+8 \mathrm{O}_{2}(g) \leftrightharpoons \\ & \mathrm{H}_{2} \mathrm{BO}_{3}^{-}(a q)+5 \mathrm{H}_{2} \mathrm{O}(l)+8 \mathrm{O}_{2}^{-}(a q) \end{aligned} $$ at \(25^{\circ} \mathrm{C}\) is \(E_{\text {cell }}^{0}=0.68 \mathrm{~V}\), and the value of \(E_{\mathrm{red}}^{\circ}\) for the \(\mathrm{O}_{2}(g) \mid \mathrm{O}_{2}^{-}(a q)\) half reaction in basic solution is \(-0.56 \mathrm{~V}\). Write the two half-reaction equations and use these data to determine the value of \(E_{\mathrm{red}}^{\circ}\) for the \(\mathrm{H}_{2} \mathrm{BO}_{3}^{-}(a q) \mid \mathrm{BH}_{4}^{-}(a q)\) half reaction in basic solution at \(25^{\circ} \mathrm{C}\)

Consider an electrochemical cell in which the reaction is described by the equation $$ 2 \mathrm{HCl}(a q)+\mathrm{Ca}(s) \leftrightharpoons \mathrm{CaCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Predict the effect of the following changes on the cel voltage: (a) decrease in amount of \(\mathrm{Ca}(s)\) (b) increase in pressure of \(\mathrm{H}_{2}(g)\) (c) increase in [HCl] (d) dissolution of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s)\) in the \(\mathrm{CaCl}_{2}(a q)\) solution

Consider the following equation for a reaction taking place in an electrochemical cell: $$ \begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{HClO}(a q)+& \mathrm{H}^{+}(a q) \leftrightharpoons \\ & 2 \mathrm{Cr}^{\bullet+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \end{aligned} $$ Predict the effect of the following changes on the observed cell voltage: (a) increase in [HClO] (b) increase in size of the inert electrodes (c) increase in \(\mathrm{pH}\) of cell solution (d) dissolution of \(\mathrm{KCl}(s)\) in the cell solution containing \(\mathrm{Cl}^{-}(a q)\)

The value of \(E_{\text {cell }}^{\circ}\) at \(25^{\circ} \mathrm{C}\) for the equation $$ \mathrm{Pb}(s)+\mathrm{Cu}^{2 *}(a q) \leftrightharpoons \mathrm{Pb}^{2+}(a q)+\mathrm{Cu}(s) $$ is \(0.47 \mathrm{~V}\). Calculate the value of the equilibrium constant at \(25^{\circ} \mathrm{C}\) for this equation.

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