   Chapter 19.6, Problem 19.9CYU

Chapter
Section
Textbook Problem

Calculate the equilibrium constant at 25 °C for the reaction2 Ag+(aq) + Hg(ℓ) ⇄ 2 Ag(s) + Hg2+(aq)

Interpretation Introduction

Interpretation:

The equilibrium constant for the given reaction has to be calculated.2Ag+(aq) + Hg(l) 2Ag(s) + Hg2+(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.

2Ag+(aq) + Hg(l) 2Ag(s) + Hg2+(aq)

Let’s write the half reactions:

At anode:Oxidaion : Hg(l) Hg2+(aq) + 2e-At cathodeReduction: 2Ag+(aq) + 2e- 2Ag(s)___________________________________Net : 2Ag+(aq) + Hg(l) 2Ag(s) + Hg2+(aq)

Let’s calculate the Ecello of the reaction.

Ecello= ECathode0- EAnode0= 0

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