/*! 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 150 A galvanic cell has a silver ele... [FREE SOLUTION] | 91影视

91影视

A galvanic cell has a silver electrode in contact with \(0.050 \mathrm{M}\) \(\mathrm{AgNO}_{3}\) and a copper electrode in contact with \(1.0 \mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) (a) Write a balanced equation for the cell reaction, and calculate the cell potential at \(25^{\circ} \mathrm{C}\). (b) Excess \(\mathrm{NaBr}(a q)\) is added to the \(\mathrm{AgNO}_{3}\) solution to precipitate AgBr. What is the cell potential at \(25^{\circ} \mathrm{C}\) after the precipitation of \(\mathrm{AgBr}\) if the concentration of excess \(\mathrm{Br}^{-}\) is \(1.0 \mathrm{M}\) ? Write a balanced equation for the cell reaction under these conditions. \(\left(K_{\mathrm{sp}}\right.\) for \(\mathrm{AgBr}\) at \(25^{\circ} \mathrm{C}\) is \(5.4 \times 10^{-13}\).) (c) Use the result in part (b) to calculate the standard reduction potential \(E^{\circ}\) for the half-reaction $$ \mathrm{AgBr}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Br}^{-}(a q) $$

Short Answer

Expert verified
(a) Cell reaction: 2Ag鈦 + Cu 鈫 2Ag + Cu虏鈦, E掳 = 0.46 V; (b) E = 0.288 V after AgBr precipitation; (c) E掳 for AgBr = 0.071 V.

Step by step solution

01

Understanding the Galvanic Cell Setup

The given galvanic cell consists of a silver electrode with AgNO3 and a copper electrode with Cu(NO3)2. We need to write a balanced equation for the cell reaction based on the standard reduction potentials. The standard reduction potential for Ag鈦/Ag is +0.80 V, and for Cu虏鈦/Cu is +0.34 V.
02

Writing the Cell Reaction

Silver ion (Ag鈦) will be reduced to silver (Ag), and copper will be oxidized. Therefore, the half-reactions are: 1. Ag鈦 + e鈦 鈫 Ag (reduction) 2. Cu 鈫 Cu虏鈦 + 2e鈦 (oxidation) The balanced cell reaction is: 2Ag鈦 + Cu 鈫 2Ag + Cu虏鈦.
03

Calculating Cell Potential using Nernst Equation

The standard cell potential E掳cell is calculated as: E掳cell = E掳(reduction) - E掳(oxidation) = 0.80 V - 0.34 V = 0.46 V. The Nernst equation at 25掳C for the cell potential is: E = E掳 - (RT/nF)ln(Q), where Q is the reaction quotient. Q = [Cu虏鈦篯/[Ag鈦篯虏. Substituting, we have: E = 0.46 V - (0.0257/n)ln(1.0/0.050虏). Calculating gives the cell potential.
04

Cell Conditions after Precipitation of AgBr

After adding NaBr, Ag鈦 precipitates as AgBr, shifting the reaction. The Ksp for AgBr is given, and the reaction is Ag鈦 + Br鈦 鈫 AgBr(s). The new concentration of Ag鈦 is Ksp/Br鈦 = 5.4 脳 10鈦宦孤/1.0 M = 5.4 脳 10鈦宦孤 M.
05

Calculating New Cell Potential after Precipitation

Using the adjusted concentration of Ag鈦, calculate the new cell potential using the Nernst equation: E = 0.46 V - (0.0257/2) ln(1.0/(5.4 脳 10鈦宦孤)虏). Calculate the value for the new E.
06

Balanced Equation for Precipitation-Situation Cell Reaction

The balanced equation with precipitated AgBr and a high Br鈦 concentration is: AgBr(s) + Cu 鈫 Cu虏鈦 + Ag + Br鈦.
07

Calculating Standard Reduction Potential for AgBr Reaction

For the reaction AgBr(s) + e鈦 鈫 Ag(s) + Br鈦, based on part (b) and the Nernst equation to relate standard potentials. The condition is at infinite dilution (standard state), so use the data from part b to find E掳. Use: E = E掳(AgBr) - 0.0592/1 log(Ksp/(Br鈦)). Calculate E掳(AgBr).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cell Potential
In a galvanic cell, the cell potential, also known as electromotive force (EMF), is the measure of the potential difference between two electrodes. This potential difference drives the electrons from the anode to the cathode. In our exercise, silver and copper electrodes are used, and the cell potential is determined by their standard reduction potentials.To find the cell potential (\(E_{cell}\)), we use the formula:
  • \(E_{cell} = E_{red} - E_{ox}\)
where \(E_{red}\) is the reduction potential of the cathode (silver in this case), and \(E_{ox}\) is the oxidation potential of the anode (copper here). Based on the given standard reduction potentials for Ag and Cu 鈥 0.80 V and 0.34 V, respectively 鈥 we can calculate the standard cell potential: \(E_{cell}^掳 = 0.80 \, \text{V} - 0.34 \, \text{V} = 0.46 \, \text{V}\).This value indicates how forcefully the electrons are driven from one electrode to the other under standard conditions (1 M concentration, 25掳C). Changes in concentration, like those resulting from chemical reactions, can also affect this potential.
Nernst Equation
The Nernst Equation allows us to calculate the cell potential under non-standard conditions. When the reaction conditions deviate from the standard state, such as when concentrations change or precipitates form, the Nernst equation provides a way to adjust the standard cell potential to account for these changes.The equation is given by:
  • \(E = E掳_{cell} - \left(\frac{RT}{nF}\right)\ln Q\)
where:
  • \(E\) is the cell potential under non-standard conditions,
  • \(E掳_{cell}\) is the standard cell potential,
  • \(R\) is the ideal gas constant (8.314 J/mol K),
  • \(T\) is the temperature in Kelvin,
  • \(n\) is the number of moles of electrons transferred in the reaction,
  • \(F\) is Faraday's constant (\(96485 \, C/mol\)),
  • \(Q\) is the reaction quotient.
For instance, in the exercise when the concentration of \(Ag^+\) changes after AgBr precipitates, the concentration term in \(Q\) is significantly altered, thus changing the potential calculated by the Nernst equation. This demonstrates how cell conditions, like the formation of a precipitate, impact the overall cell potential through shifts in concentration affecting the Nernst equation.
Standard Reduction Potential
Standard reduction potential (\(E^掳\)) is the measure of the tendency of a chemical species to gain electrons under standard conditions. It's essential in calculating the cell potential of galvanic cells. In the standard state, conditions involve 1 M concentration for each ion at 25掳C with a standard hydrogen electrode, which is arbitrarily assigned a potential of 0 V, used as the reference.For metallic elements, such as in our problem with silver (\(Ag^+/Ag\)) and copper (\(Cu^{2+}/Cu\)), we use standard reduction potentials to predict which metals are more likely to oxidize or reduce:
  • Silver: \(E^掳_{red}= +0.80 \, \text{V}\)
  • Copper: \(E^掳_{red}= +0.34 \, \text{V}\)
The higher the \(E^掳\) value, the greater the species' tendency to be reduced. For the complete cell reaction, we subtract the lower from the higher standard reduction potential to determine the standard cell potential. Moreover, calculating the standard reduction potential for reactions involving precipitates, such as \(AgBr\) being reduced to \(Ag\) and \(Br^-\), uses similar principles derived from measured concentrations and equilibrium constants like the solubility product (\(K_{sp}\)). Determining these potentials helps us understand and design better electrochemical cells.

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

Write balanced net ionic equations for the following reactions in acidic solution: (a) \(\mathrm{Zn}(s)+\mathrm{VO}^{2+}(a q) \rightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{V}^{3+}(a q)\) (b) \(\mathrm{Ag}(s)+\mathrm{NO}_{3}^{-}(a q) \rightarrow \mathrm{Ag}^{+}(a q)+\mathrm{NO}_{2}(g)\) (c) \(\mathrm{Mg}(s)+\mathrm{VO}_{4}{ }^{3-}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{V}^{2+}(a q)\) (d) \(\Gamma(a q)+\mathrm{IO}_{3}^{-}(a q) \rightarrow \mathrm{I}_{3}^{-}(a q)\)

Given the following half-reactions, combine the two that give the cell reaction with the most positive \(E^{\circ} .\) Write a balanced equation for the cell reaction, and calculate \(E^{\circ}\) and \(\Delta G^{\circ}\). $$ \begin{array}{ll} \mathrm{Co}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}(s) & E^{\circ}=-0.28 \mathrm{~V} \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}(a q) & E^{\circ}=0.54 \mathrm{~V} \\ \mathrm{Cu}^{2+}(a q)+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) & E^{\circ}=0.34 \mathrm{~V} \end{array} $$

What is rust? What causes it to form? What can be done to prevent its formation?

What is meant by cathodic protection? (a) Steel is coated with a layer of paint. (b) Iron in steel is oxidized to form a protective oxide coating. (c) Steel is coated with zinc because zinc is more easily oxidized than iron. (d) A strip of magnesium is attached to steel because the magnesium is more easily oxidized than iron.

Which of the following metals can offer cathodic protection to iron? Select all the correct choices. \(\mathrm{Mn} \mathrm{Ni} \mathrm{Ph}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.