Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 91

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$ \mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q) $$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+} .\) )

Short Answer

Expert verified
The standard electrode potential for the given half-reaction can be calculated using the Nernst Equation and the given values for \(E^{\circ}_{\text{Ag}^+}\) and \(K_{\text{sp}}\). First, calculate the equilibrium constant for the half-reaction using \(K = \frac{1}{K_{\mathrm{Ag}^{+}}} \times K_{sp}\). Then, use the Nernst Equation with n = 1: \(\mathscr{E}^{\circ} = -0.0592 \log K\). Plug in the calculated K value to find the standard electrode potential \(\mathscr{E}^{\circ}\).

Step by step solution

01

Write the related half-reactions for Ag+

We are given the standard reduction potential for \(\mathrm{Ag}^{+}\). Here are the standard half-reaction and its reverse reaction: $$ \mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) \quad \quad \text{Reduction: E}^{\circ} = E_{\mathrm{Ag}^{+}} $$ $$ \mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} \quad \quad \text{Oxidation} $$
02

Derive the given half-reaction from Ag-related reactions and calculate K

To get to our target half-reaction, take the reverse of the oxidation reaction and add it to the target half-reaction: $$ \mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} $$ $$ \mathrm{Ag}^{+}(aq) + \mathrm{I}^{-}(aq) \longleftrightarrow \mathrm{AgI}(s) \quad \quad K = K_{sp} $$ Adding them, we get: $$ \mathrm{AgI}(s) + \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s) + \mathrm{I}^{-}(aq) $$ Now we have the given half-reaction, and we can calculate the equilibrium constant (K) for it using the equilibrium constants of the reactions we used: $$ K = \frac{1}{K_{\mathrm{Ag}^{+}}} \times K_{sp} $$
03

Apply Nernst Equation to calculate the standard electrode potential

Now we can use the Nernst Equation to find the standard electrode potential (E°) for the given half-reaction: $$ \mathscr{E}^{\circ} = \frac{-0.0592}{n} \log K $$ Here n = 1, as there is only one electron involved in the half-reaction. So, we can plug in the values we calculated in step 2: $$ \mathscr{E}^{\circ} = -0.0592 \log K $$ Finally, you can use the given values for \(E^{\circ}_{\text{Ag}^+}\) and \(K_{\text{sp}}\) to find the standard electrode potential for the given half-reaction.

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

Under standard conditions, what reaction occurs, if any, when each of the following operations is performed? a. Crystals of \(\mathrm{I}_{2}\) are added to a solution of NaCl. b. \(\mathrm{Cl}_{2}\) gas is bubbled into a solution of NaI. c. A silver wire is placed in a solution of \(\mathrm{CuCl}_{2}\) d. An acidic solution of \(\mathrm{FeSO}_{4}\) is exposed to air. For the reactions that occur, write a balanced equation and calculate \(\mathscr{E}^{\circ}, \Delta G^{\circ},\) and \(K\) at \(25^{\circ} \mathrm{C}\)

Consider the following galvanic cell: a. Label the reducing agent and the oxidizing agent, and describe the direction of the electron flow. b. Determine the standard cell potential. c. Which electrode increases in mass as the reaction proceeds, and which electrode decreases in mass?

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{ll} & \mathscr{E}^{\circ}(\mathrm{V}) \\ \mathrm{M}^{4+}+4 \mathrm{e}^{-} \longrightarrow \mathrm{M} & 0.66 \\ \mathrm{N}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{N} & 0.39 \\ \hline \end{array}$$ What is the value of \(\Delta G^{\circ}\) and \(K\) for this cell?

A zinc-copper battery is constructed as follows at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.10 M)\right|\left|\mathrm{Cu}^{2+}(2.50 M)\right| \mathrm{Cu} $$ The mass of each electrode is \(200 .\) g. a. Calculate the cell potential when this battery is first connected. b. Calculate the cell potential after 10.0 A of current has flowed for \(10.0 \mathrm{h}\). (Assume each half-cell contains \(1.00 \mathrm{L}\) of solution.) c. Calculate the mass of each electrode after \(10.0 \mathrm{h}\). d. How long can this battery deliver a current of 10.0 A before it goes dead?

Consider the following galvanic cell at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Pt}\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co} $$ The overall reaction and equilibrium constant value are $$\begin{aligned} 2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) & \longrightarrow \\ 2 \mathrm{Cr}^{3+}(a q)+\mathrm{Co}(s) & K=2.79 \times 10^{7} \end{aligned}$$ Calculate the cell potential, \(\mathscr{E}\), for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

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