Chapter 14, Problem 14.5P

a)

Interpretation Introduction

Interpretation:

The oxidizing agent on the left side of the reaction and balanced half reaction has to be written.

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.

Half Reactions:

A half reaction can be defined as component of redox reaction that is either Oxidation or reduction reaction. The change in oxidation states of individual substances present in redox reaction gives the half reaction. Half reactions are common methods of balancing redox reactions.

b)

Interpretation Introduction

Interpretation:

The reducing agent of the left side of the reaction and balanced half reaction has to be written.

Concept Introduction:

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.

Half Reactions:

A half reaction can be defined as component of redox reaction that is either Oxidation or reduction reaction. The change in oxidation states of individual substances present in redox reaction gives the half reaction. Half reactions are common methods of balancing redox reactions.

c)

Interpretation Introduction

Interpretation:

The number of Coulombs of charge passed from reductant to oxidant when $\text{1}\text{.00}\text{g}$ of Thiosulphate reacts has to be calculated.

Concept Introduction:

Electric charge:

Electric charge (q) is given in Coulombs (C). The magnitude of charge of a single electron or proton is $\text{1}{\text{.602×10}}^{\text{-19}}\text{C}$, hence a mole of electrons or protons will have charge of $\left(\text{1}{\text{.602×10}}^{\text{-19}}\text{C}\right)\left(\text{6}{\text{.022×10}}^{\text{23}}{\text{mol}}^{\text{-1}}\right)\text{=9}{\text{.649×10}}^{\text{14}}\text{C}$, which is known as Faraday constant F.

The electric charge in coulombs can be expressed as,

$\text{q=n}\text{.N}\text{.F}$

Where,

q=coulombs

n=unit charges per molecule

N=moles

F= $\frac{\text{Coulombs}}{\text{mole}{\text{e}}^{\text{-}}}$

The units work since the number of unit charges per molecules, n is dimensionless.

d)

Interpretation Introduction

Interpretation:

The flow of current (in amperes) from reductant to oxidant has to be calculated.

Concept Introduction:

Electric current:

The quantity of charge that flows in each second through a circuit is called as electric current. Ampere is the unit of electric current and is expressed as A.

A current of one Ampere tells about a charge of one coulomb per second flowing across a point in a circuit.