Chapter 14, Problem 14.25P

Interpretation Introduction

Interpretation:

The cell voltage and direction flow of electrons has to be determined.

Concept introduction:

Nernst Equation:

For Half-reaction,

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

The Nernst equation results in the half-cell potential E as,

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

Here,

$\text{E°}$ = 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

The voltage of a battery is calculated as

Cell voltage = potential of right hand electrode( ${\text{E}}_{\text{+}}$) – potential of left hand electrode( ${\text{E}}_{-}$).