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A voltaic cell is set up utilizing the reaction Cu(s) + 2 Ag + (aq) → Cu 2+ (aq) + 2 Ag(s) Cu(s) | Cu 2+ (aq, 1.0 M) || Ag + (aq, 0.001 M)|Ag(s) Under standard conditions, the expected potential is 0 45 V predict whether the potential for the voltaic cell will be higher, lower, or the same as the standard potential. Verify your prediction by calculating the new cell potential.

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Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

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Chapter
Section
BuyFindarrow_forward

Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 19, Problem 80GQ
Textbook Problem
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A voltaic cell is set up utilizing the reaction

Cu(s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag(s)

Cu(s) | Cu2+(aq, 1.0 M) || Ag+(aq, 0.001 M)|Ag(s)

Under standard conditions, the expected potential is 0 45 V predict whether the potential for the voltaic cell will be higher, lower, or the same as the standard potential. Verify your prediction by calculating the new cell potential.

Interpretation Introduction

Interpretation:

It has to be predicted whether the potential for the voltaic cell will be higher, lower or same as the standard potential and the verification of the prediction also has to be done.

Concept introduction:

According to the first law of thermodynamics, the change in internal energy of a system is equal ti the heat added to the sysytem minus the work done by the system.

The equation is as follows.

ΔU = Q - WΔU = Change in internal energyQ = Heat added to the systemW=Work done by the system

In voltaic cell, the maximum cell potential is directly related to the free energy difference between the reactants and products in the cell.

ΔG0= -nFE0n = Number of moles transferred per mole of reactant and productsF = Faradayconstant=96485C/mol  E0= Volts = Work(J)/Charge(C)

The relation between standard cell potential and equilibrium constant is as follows.

lnK = nE00.0257 at 298K

The relation between solubility product Ksp and equilibrium constant is as follows.

Ksp= e+lnK

Explanation of Solution

The given over all reaction is as follows.

Cu(s) + 2Ag+(aq) Cu2+(aq) + 2Ag(s)

At standard conditions, the ions in solution are at concentrations of 1.0 M. In the above reaction, Ag+ is at a concentration of 0.001 M. Ag+ is a reactant in the above reaction.

Because there is less of a reactant that is needed to drive the reaction., the measure potential will be lower than the standard potential.

Let’s calculate the reduction potential of voltaic cell:

lnK = nE00.0257 at 298K

Rearrange the above formula as follows.

E = E0 - 0

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Chapter 19 Solutions

Chemistry & Chemical Reactivity
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Ch. 19.3 - Which of the following statements is correct for...Ch. 19.3 - 2. You can determine if an automobile lead-based...Ch. 19.4 - (a) Rank the following metals in their ability to...Ch. 19.4 - The net reaction that occurs in a voltaic cell is...Ch. 19.4 - 2. Which metal in the following list is easiest to...Ch. 19.4 - Determine which of the following redox equations...Ch. 19.4 - A lithium ion battery produces a voltage of 3.6 V....Ch. 19.4 - Use the energy produced by the combustion of...Ch. 19.4 - What mass of lithium ion batteries would produce...Ch. 19.5 - A voltaic cell is set up with an aluminum...Ch. 19.5 - Check Your Understanding The half-cells Fe2+(aq,...Ch. 19.5 - 1. Calculate Ecell at 298 K for a cell involving...Ch. 19.6 - The following reaction has an E value of 0.76 V:...Ch. 19.6 - Calculate the equilibrium constant at 25 C for the...Ch. 19.6 - 1. 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