Chapter 17, Problem 17.116CHP

(a)

Interpretation Introduction

Interpretation:

The overall balanced cell reaction has to be given.

Concept Introduction:

In a balanced equation the number of atoms of each element as a reactant is equal to the number of atoms of that element as a product.

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,

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

(b)

Interpretation Introduction

Interpretation:

The $E^0$ and $\Delta G^0$ of the given reaction 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,

$\text{ΔG°}\text{=}-\text{nFE°}$

Where,

$\begin{array}{c}\text{n}\text{=}\text{number}\text{of}\text{moles}\text{of}\text{electrons}\text{transferred}\\ \text{F}\text{=}\text{Faraday}\text{constant}\text{(96,500}\text{C/mole}{\text{e}}^{\text{-}}\text{)}\\ \text{E°}\text{=}\text{Standard cell potential}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The quantity (in litres) of chlorine at $\text{740}\text{mm}\text{Hg}$ and $\text{25°C}$ resulting from the reaction of $179gKMn{O}_4$ has to be given.

Concept Introduction:

The number of moles can be calculated by,

$Numberofmoles=\frac{Massingram}{Molarmass}$

Ideal gas law is,

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{where,}\\ \text{P}\text{=}\text{Pressure}\\ \text{n}\text{=}\text{number of moles}\\ \text{T}\text{=}\text{Temperature}\\ \text{R}\text{=}\text{Gas}\text{constant}\\ V=Volume\end{array}$