Chapter 17, Problem 17.135MP

(a)

Interpretation Introduction

Interpretation:

The values of $\text{ΔG°}\text{and}\text{E°}$ at $\text{95°C}$ has to be calculated.

Concept Introduction:

Change in free energy:

The standard cell potential related to the free-energy is given by the equation,

$\text{ΔG°}\text{=}\text{-nFE°}$

Where,

$\begin{array}{l}\text{n}\text{=}\text{number}\text{of}\text{moles}\text{of}\text{electrons}\text{transferred}\\ \text{F}\text{=}\text{Faraday}\text{constant}\text{(96,500}\text{C/mol}{\text{e}}^{\text{-}}\text{)}\\ \text{E°}\text{=}\text{Standard cell potential}\\ \text{ΔG°}\text{=}\text{Change in free-energy}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The cell voltage at $\text{95°C}$ when the partial pressures of Hydrogen and Oxygen are $\text{25}\text{atm}$ has to be calculated.

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,

$\begin{array}{l}{\text{E=E}}^{\text{°}}\text{-}\frac{\text{RT}}{\text{nF}}\text{ln}\frac{{\text{A}}_{\text{B}}{}^{\text{b}}}{{\text{A}}_{\text{A}}{}^{\text{a}}}\\ \\ {\text{E=E}}^{\text{°}}\text{-}\frac{\text{RT}}{\text{nF}}\text{lnQ}\\ \\ \text{E=E°-}\frac{\text{0}\text{.0592V}}{\text{n}}\text{logQ}\end{array}$

Here,

$\begin{array}{l}\text{E°}\text{=}\text{standard}\text{reduction}\text{potential}\\ \text{R}\text{=}\text{Gas}\text{constant}\text{(8}\text{.314J/}\left(\text{K}\text{.mol}\right)\text{)=8}\text{.314}\left(\left(\text{V}\text{.C}\right)\text{/}\left(\text{K}\text{.mol}\right)\right)\\ \text{T}\text{=}\text{Temperature(in}\text{K)}\\ \text{n}\text{=}\text{number}\text{of}\text{electrons}\text{in}\text{half-reaction}\\ \text{F}\text{=}\text{Faraday}\text{constant(9}{\text{.649×10}}^{\text{4}}\text{C/mol)}\\ \text{Q}\text{=}\text{Reaction}\text{quotient}\end{array}$