Chapter 17, Problem 17.114CHP

(a)

Interpretation Introduction

Interpretation:

The materials used for electrodes, cathode and anode, and the direction of electron and ion flow has to be given.

(b)

Interpretation Introduction

Interpretation:

The overall balanced cell reaction and the standard cell potential has to be given.

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,

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

(c)

Interpretation Introduction

Interpretation:

The cell potential at $\text{25°C}$ 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,

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

(d)

Interpretation Introduction

Interpretation:

The change in free energy and the equilibrium constant has to be calculated.

Concept Introduction:

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,

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

(e)

Interpretation Introduction

Interpretation:

The change in mass of Aluminium electrode has to be calculated.

Concept Introduction:

The mass of a substance is calculated from the constant current and the time duration of the reaction.  Conversion factors are used along with the atomic weight of the substance.  The general formula for obtaining the mass using the conversion factor is given as,

$\text{Mass}\text{=}\text{Current×time(in min)×}\frac{\text{60s}}{\text{1min}}\text{×}\frac{\text{1}\text{mole}{\text{e}}^{\text{-}}}{\text{96,500C}}\text{×}\frac{\text{1}\text{mole}\text{of}\text{substance}}{\text{n}\text{mole}{\text{e}}^{\text{-}}}\text{×}\frac{\text{Atomic}\text{weight}}{\text{1}\text{mole}\text{of}\text{substance}}$

The flow chart below is useful in the calculation of grams or litres of the product.

Figure 2