Chapter 19, Problem 81GQ

Interpretation Introduction

Interpretation:

The celll potential for the given cell has to be determined in which ${\text{[H}}^{\text{+}}\text{(aq)] is1}{\text{.0×10}}^{\text{-7}}\text{ M}$ and has to be identify whether the reaction is to be more or less favourable at the lower pH.

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.

$\begin{array}{c}\text{ΔU = Q - W}\\ \text{ΔU = Change in internal energy}\\ \text{Q = Heat added to the system}\\ \text{W=Work done by the system}\end{array}$

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

$\begin{array}{c}{\text{ΔG}}^{\text{0}}{\text{= -nFE}}^{\text{0}}\\ \text{n = Number of moles transferred per mole of reactant and products}\\ \text{F = Faradayconstant=96485C/mol}\\ {\text{ E}}^{\text{0}}\text{= Volts = Work(J)/Charge(C)}\end{array}$

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

$\text{lnK = }\frac{{\text{nE}}^{\text{0}}}{\text{0}\text{.0257}}\text{ at 298K}$

The relation between solubility product ${\text{K}}_{\text{sp}}$ and equilibrium constant is as follows.

${\text{K}}_{\text{sp}}{\text{= e}}^{\text{+lnK}}$