Chapter 17, Problem 17.112CHP

(a)

Interpretation Introduction

Interpretation:

The strongest oxidizing agent and the strongest reducing agent has to be given.

Concept Introduction:

Oxidizing agent:

An oxidizing agent or oxidant is a substance that gains electrons and gets reduced in a chemical reaction.  An oxidizing agent is also called as electron acceptor. 

Strongest oxidizing agent carries most positive standard reduction potential.

Reducing agent:

A reducing agent or reductant is a substance that loses electrons and gets oxidized in a chemical reaction.  A reducing agent is also called as electron donor.

Strongest reducing agent carries most negative standard reduction potential.

(b)

Interpretation Introduction

Interpretation:

The cell that delivers the highest voltage has to be given along with identification of anode, cathode and direction of electron and ion has to be given.

(c)

Interpretation Introduction

Interpretation:

The cell reaction of part (b) has to be written and the values of $\text{E°,ΔG°}\text{and}\text{K}$ for the reaction at $\text{25°C}$ has to be calculated.

Concept Introduction:

Cell potential (EMF):

The maximum potential difference between two electrodes of voltaic cell is known as cell potential.

If standard reduction potentials of electrodes are known, the cell potential (EMF) is given by,

${\text{E}}_{\text{cell}}\text{=}{\text{E}}_{\text{cathode}}{\text{-E}}_{\text{anode}}$

Where,

$\begin{array}{l}{\text{E}}_{\text{cathode}}\text{is}\text{the}\text{reduction}\text{half}\text{cell potential}\\ {\text{E}}_{\text{anode}}\text{is}\text{the}\text{oxidation}\text{half}\text{cell potential}\end{array}$

Standard cell potential and equilibrium constant:

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

$\text{ΔG°}\text{=}\text{-nFE°}$

The free-energy change is also related to equilibrium constant by the equation,

$\text{ΔG°}\text{=}\text{-RTlnK}$

Combine both the equations,

$\begin{array}{l}\text{-nFE°}\text{=}\text{-RTlnK}\\ \\ \text{E°}\text{=}\frac{\text{RT}}{\text{nFlnK}}\text{=}\frac{\text{2}\text{.303RT}}{\text{nF}}\text{logK}\\ \\ \text{(or)}\\ \\ \text{E°}\text{=}\frac{\text{0}\text{.0592V}}{\text{n}}\text{logK}\end{array}$

Where,

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

(d)

Interpretation Introduction

Interpretation:

The cell voltage for part (b) if both ion concentration are $\text{0}\text{.010}\text{M}$ 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,

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