   Chapter 19, Problem 70GQ

Chapter
Section
Textbook Problem

What is the value of E° for the following half-reaction?Ag2CrO4(s) + 2 e− → 2 Ag(s) + CrO42−(aq)

Interpretation Introduction

Interpretation:

The E0 for the following reaction has to be determined.

Ag2CrO4(s) + 2e- 2Ag(s) + CrO42-(aq)

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

The given reaction is as follows.

Ag2CrO4(s) + 2e- 2Ag(s) + CrO42-(aq)

E0 for the reaction is as follows.

lnK = nE00.0257 at 298K

Rearrange the is as follows.

E00

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