Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 86

Given that $$ \begin{array}{cl} 2 \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg}_{2}^{2+}(a q) & E^{\circ}=0.92 \mathrm{~V} \\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=0.85 \mathrm{~V} \end{array} $$ calculate \(\Delta G^{\circ}\) and \(K\) for the following process at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Hg}_{2}^{2+}(a q) \longrightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l) $$ (The preceding reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)

Short Answer

Expert verified
\(\Delta G^{\circ} = -13.508 kJ/mol\), \(K = 408.411\)

Step by step solution

01

Write the given reactions in the form required

First, the reaction for which we are finding \(\Delta G^{\circ}\) and \(K\) is a combination of the two given reactions. Therefore, write the combination of the two half-reactions: \(\mathrm{Hg}_{2}^{2+}(a q) \rightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l)\). However, this reaction is the reverse of the sum of the two given half-reactions, so note the need to reverse the sign of the E° when we add them together.
02

Calculate the standard potential for the overall reaction

The standard cell potential \(E^{\circ}\) for an overall reaction is the sum of the standard potentials for the half-reactions. Mathematically, \(E_{cell}^\circ = E_{cathode}^\circ - E_{anode}^\circ\). In this case, the standard cell potential for the overall reaction \(E_{cell}^{\circ}\) is equal to \(E^{\circ}_{Hg^{2+}|Hg} - E^{\circ}_{Hg_{2}^{2+}|Hg^{2+}} = 0.85V - 0.92V = -0.07V\). However, you must remember to change the sign because the overall reaction is the reverse of the sum of these two half-reactions. Therefore, we will use \(E_{cell} = +0.07V\).
03

Calculate the standard Gibbs free energy change

Next, calculate the standard Gibbs free energy change \(\Delta G^{\circ}\) using the following equation: \(\Delta G^{\circ} = -nFE_{cell}^\circ\). Where \(n\) is the number of moles of electrons transferred (which is 2 in this case), \(F\) is the Faraday constant (96485 C/mol), and \(E_{cell}^\circ\) is the standard cell potential. We plug these values into the equation to get: \(\Delta G^{\circ} = -2 \times 96485 \times 0.07 = -13508.3 J/mol = -13.508 kJ/mol\) (after converting from J to kJ).
04

Calculate the equilibrium constant

Finally, using the relationship between the standard Gibbs free energy change and the equilibrium constant, \( \Delta G^{\circ} = -RT \ln K\), solve for \(K\) to find the equilibrium constant of the reaction. This will require rearranging the formula to \( K = e^{-\Delta G^{\circ}/RT} \) where \(R\) is the universal gas constant (8.314 J/mol·K), \(T\) is the temperature in Kelvin (298K), and \(\Delta G^{\circ}\) is the standard Gibbs free energy change. Substitute the values into the equation to get \( K = e^{-(-13508.3)/(8.314\times298)} \), which simplifies to \( K = 408.411 \).
05

Final Answer

The value of \(\Delta G^{\circ}\) and \(K\) for the reaction \(\mathrm{Hg}_{2}^{2+}(a q) \rightarrow \mathrm{Hg}^{2+}(a q) + \mathrm{Hg}(l)\) at \(25^{\circ}C = 298K\) are \(-13.508 kJ/mol\) and 408.411, respectively. These values indicate that the reaction is spontaneous and shifts towards the reaction products.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A sample of iron ore weighing \(0.2792 \mathrm{~g}\) was dissolved in an excess of a dilute acid solution. All the iron was first converted to \(\mathrm{Fe}(\mathrm{II})\) ions. The solution then required \(23.30 \mathrm{~mL}\) of \(0.0194 \mathrm{M} \mathrm{KMnO}_{4}\) for oxidation to Fe(III) ions. Calculate the percent by mass of iron in the ore.

The hydrogen-oxygen fuel cell is described in Section 19.6 . (a) What volume of \(\mathrm{H}_{2}(g)\), stored at \(25^{\circ} \mathrm{C}\) at a pressure of 155 atm, would be needed to run an electric motor drawing a current of \(8.5 \mathrm{~A}\) for \(3.0 \mathrm{~h} ?\) (b) What volume (liters) of air at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) will have to pass into the cell per minute to run the motor? Assume that air is 20 percent \(\mathrm{O}_{2}\) by volume and that all the \(\mathrm{O}_{2}\) is consumed in the cell. The other components of air do not affect the fuel-cell reactions. Assume ideal gas behavior.

What is Faraday's contribution to quantitative electrolysis?

For a number of years it was not clear whether mercury(I) ions existed in solution as \(\mathrm{Hg}^{+}\) or as \(\mathrm{Hg}_{2}^{2+}\). To distinguish between these two possibilities, we could set up the following system: $$ \mathrm{Hg}(l) \mid \text { soln } \mathrm{A} \| \text { soln } \mathrm{B} \mid \mathrm{Hg}(l) $$ where soln A contained 0.263 g mercury(I) nitrate per liter and soln B contained 2.63 g mercury(I) nitrate per liter. If the measured emf of such a cell is \(0.0289 \mathrm{~V}\) at \(18^{\circ} \mathrm{C},\) what can you deduce about the nature of the mercury(I) ions?

What is the potential of a cell made up of \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cells at \(25^{\circ} \mathrm{C}\) if \(\left[\mathrm{Zn}^{2+}\right]=0.25 \mathrm{M}\) and \(\left[\mathrm{Cu}^{2+}\right]=0.15 \mathrm{M} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free