Chapter 17, Problem 17.131MP

(a)

Interpretation Introduction

Interpretation:

The balanced net ionic equation for the reaction has to be written.

Concept Introduction:

The soluble ionic compounds are shown as separate ions and the precipitate is shown together, such an equation is called as complete ionic equation.

An ionic equation in which the spectator ions cancels each other is called as net ionic equation.

(b)

Interpretation Introduction

Interpretation:

$\text{E°}$ for the 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,

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

(c)

Interpretation Introduction

Interpretation:

The completion of the reaction has to be proved by calculating the $\text{ΔG°}\text{and}\text{K}\text{at}\text{25°C}\text{.}$

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,

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

(d)

Interpretation Introduction

Interpretation:

The molarity of ${\text{KMnO}}_{\text{4}}$ solution has to be calculated.

Concept Introduction:

Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,

$\text{Molarity(M)=}\frac{\text{moles}\text{of}\text{solute(g)}}{\text{volume}\text{of}\text{solution(L)}}$