Chapter 14.4, Problem 6TY

Interpretation Introduction

Interpretation:

The Nernst equation for the given reaction has to be written.

Concept Introduction:

Nernst Equation:

For Half-reaction,

${\text{aA + ne}}^{\text{-}}\stackrel{}{\rightleftharpoons }\text{ bB}$

The Nernst equation for the half-cell potential E is given as:

$\text{E = E° - }\frac{\text{RT}}{\text{nF}}\text{ ln }\frac{{\text{A}}_{\text{B}}{}^{\text{b}}}{{\text{A}}_{\text{A}}{}^{\text{a}}}$

Where,

$\text{E°}$= standard reduction potential ( ${\text{A}}_{\text{A}}{\text{=A}}_{\text{B}}\text{=1}$ )

$\text{R}$= gas constant $\left(\text{8}\text{.314}\text{J/}\left(\text{K}\text{.mol}\right)\right)\text{=8}\text{.314}\left(\left(\text{V}\text{.C}\right)\text{/}\left(\text{K}\text{.mol}\right)\right)$

T = Temperature (in K)

n = number of electrons in half-reaction

F = Faraday constant ( $\text{9}{\text{.649×10}}^{\text{4}}\text{C/mol}$)

A = Activity of species, i

At $25°\text{C}$, by converting the Nernst equation to natural logarithm and by substituting the gas constant and temperature, we get the Nernst equation as:

$\text{E = E°}-\frac{0.05916\text{ V}}{n}\log \frac{{\text{A}}_{\text{B}}{}^{\text{b}}}{{\text{A}}_{\text{A}}{}^{\text{a}}}$

To Write: The Nernst equation for the given reaction