Chapter 14, Problem 14.29P

(a)

Interpretation Introduction

Interpretation:

The standard reduction potential of given cell has to be determined.

Concept introduction:

When the potentiometer is replaced with a wire in a cell there would be more current flows and concentration varies until the cell attains equilibrium.  At this situation there is nothing to drive reaction.  The cell voltage becomes zero.

Relation between ${\text{E}}^{\text{0}}\text{and}\text{K}$ is given as

$\begin{array}{l}{\text{E}}^{\text{0}}\text{=}\frac{\text{0}\text{.05916}}{\text{n}}\text{log}\text{K}\\ \\ \text{K}\text{=}{\text{10}}^{{\text{nE}}^{\text{0}}\text{/0}\text{.05916}}\\ \\ {\text{ΔG}}^{\text{0}}\text{=}{\text{-nFE}}^{\text{0}}\\ \\ {\text{E}}^{\text{0}}\text{is}\text{standard}\text{reduction}\text{potential}\\ \\ \text{K}\text{is}\text{equilibrium}\text{constant}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The standard reduction potential of given cell has to be determined.

Concept introduction:

When the potentiometer is replaced with a wire in a cell there would be more current flows and concentration varies until the cell attains equilibrium.  At this situation there is nothing to drive reaction.  The cell voltage becomes zero.

Relation between ${\text{E}}^{\text{0}}\text{and}\text{K}$ is given as

$\begin{array}{l}{\text{E}}^{\text{0}}\text{=}\frac{\text{0}\text{.05916}}{\text{n}}\text{log}\text{K}\\ \\ \text{K}\text{=}{\text{10}}^{{\text{nE}}^{\text{0}}\text{/0}\text{.05916}}\\ \\ {\text{ΔG}}^{\text{0}}\text{=}{\text{-nFE}}^{\text{0}}\\ \\ {\text{E}}^{\text{0}}\text{is}\text{standard}\text{reduction}\text{potential}\\ \\ \text{K}\text{is}\text{equilibrium}\text{constant}\end{array}$