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Question 1

Oxidation-Reduction Reactions Characteristics

Answer 1

• Occurs simultaneously when electrons are transferred from one species to another

The reactant species that gains electrons is said to be reduced.
* That which loses electrons is oxidized.
* The species from which electrons are taken is oxidized (and is the reducing agent).
* The species that takes the electrons is reduced
(and is the oxidizing agent).

Question 2

Oxidation States

Answer 2

a measure of the degree of oxidation of an element in a compound compared with when it is uncombined

Question 3

Oxidation

Answer 3

accompanied by an increase in the oxidation state of an element.

ANODE

Question 4

Reduction

Answer 4

accompanied by a decrease in the oxidation state of an element.

CATHODE

Question 5

GUIDELINES to Redox reactions

Answer 5

• The oxidation state of each atom in a pure element is zero.
• For a simple, monatomic ion, the oxidation state is equal to the charge on the ion.
• The algebraic sum of oxidation states in an ion is equal to the ionic charge. In a molecule, the sum is zero.
• In all of its compounds and ions, fluorine is in oxidation state -1.
• In most compounds and ions, hydrogen is in oxidation state +1.
• In most compounds, O is in oxidation state -2.

Like oxygen, the halogens (F2, Cl2, Br2, I2) are always oxidizing agents in their reactions with metals and non-metals.

Question 6

Electrochemistry

Answer 6

a field of study about the interaction between electricity and chemistry.

Question 7

Voltaic cells (or galvanic cells)

Answer 7

direct the transfer of electrons in a spontaneous redox reaction from one compartment of the cell, through a conductor, to another compartment.

Question 8

Electrolysis cells (or electrolytic cells)

Answer 8

in which the application of an electrical potential forces a non-spontaneous redox reaction to occur. This forced non-spontaneous process is called electrolysis

Question 9

Voltaic Cells

Answer 9

Contains anode, cathode, salt bridge, voltmeter and stuff idk

• The electrons transferred during a spontaneous reaction are harnessed as an electrical current—electrons move from the anode, the site of oxidation, through the external circuit to the cathode, the site of reduction.
• Charge balance in each half-cell is achieved by migration of ions through the salt bridge (internal circuit). Anions move toward the anode, and cations move toward the cathode.

Question 10

Inert electrodes

Answer 10

made of conducting materials that are too unreactive to be oxidized or reduced.

Question 11

Cell EMF

Answer 11

a measure of the difference between the abilities of species in the half-cells to compete for electrons.

Standard Conditions
– Solutes in aqueous solution have a conc of 1.0 mol L-1
– Gaseous substances have a pressure of 1.0 bar.
– Solids and liquids (not solutions) are pure.

Question 12

Standard Reduction Potentials

Answer 12

Different combinations of half cells will give rise to a different cell EMF.

The EMF is the difference between the two electrode potentials.
E(cell) = E(cathode) – E(anode)

E(cathode) and E(anode) are half-cell reduction potentials orelectrode potentials (Ehalf-cell)

Question 13

Potential ladder for reduction half-reactions

Answer 13

The higher ranked the half equation is, the more powerful an oxidizing agent the species is.

At bottom: stronger reducing agents

The tendency for the rxn to occur as a reduction (as written) DECREASES down the table.

For any cell:
* The ½ reaction that occurs highest up the table is written in the FORWARD direction (as is).
* The one that is lower goes in the REVERSE direction. It is the oxidation reaction.

Question 14

Primary batteries: Alkaline cell

Answer 14

Cannot be recharged

Anode (-):
Zn(s) + 2OH- (aq) → ZnO(s) + H2O + 2e

Cathode (+): 2MnO2(s) + H2O + 2e- → Mn2O3(s) + 2OH- (aq)

Advantages:
No gas buildup, more suitable for high drain applications. last ~50% longer than older lead acid batteries

Question 15

secondary batteries: Lead Storage Battery

Answer 15

Reactions can be reversed (rechargable)

PbO2(s) + Pb(s) + 2H2SO4(aq) -> 2PbSO4(s) + 2H2O(l)

6 cells each generating 2.04 V = 12 V output

The reactants Pb and PbO2 are electrodes:

• as both are solids no need to separate into
  different half cells
• because both are solids have no effect on rxn quotient (Q)
• Thus the EMF is constant during discharge.

Question 16

Voltaic Cells under Nonstandard Conditions

Answer 16

Dependence of Cell emf on Concentrations
* When the species in a voltaic cell are not in their standard states, cell emf can be calculated using the Nernst equation:

E (cell) = E (nought cell) - (RT/nF) lnQ
= E (nought cell) - (0.0257V/n) lnQ at 25 degrees

• R is the gas constant (8.3144 J K mol-1);
• T is the temperature (K);
• n is the moles of electrons transferred
• F is the Faraday constant (9.6485338 x 10¹ C mol-1)
• Q is the reaction quotient

Question 17

pH Meters and Ion-Selective Electrodes: Nerst equation

Answer 17

used to determine an unknown concentration by using a measured cell potential, using a pH meter.

• The electrodes used to measure ion conc are known as ion-selective electrodes.

Question 18

pH-Dependence of Oxidizing
Power of Oxoanions

Answer 18

ions with the generic formula AxOy ^z−

The more acidic the solution is, the more powerful oxoanions like nitrate and permanganate etc are as oxidixing agents.

If the half-equation for the reduction process involves H+ (aq) ions, then the half-cell reduction potential depends on the pH.

When pH is low, the concentration of H+ is high, which pushes the equilibrium to the right. This makes nitrate a more powerful oxidizing agent.

In general:
The lower the pH of solution, the more powerful oxoanions are as oxidizing agents.

Question 19

Electrolysis: Chemical Change Using Electrical Energy

Answer 19

reactions are forced the non-spontaneous direction by electrical energy.

In the schematic, electrons flow in the spontaneous direction in Galvanic cells (left).
A potential drives electrons in the non-spontaneous direction in the electrolytic cell (right).

Widely used in the refining of metals such as
aluminum and in the production chlorine gas.

Electrolysis of water makes hydrogen and oxygen gases.

Question 20

Electrolysis of Molten Salts

Answer 20

The melting point of table salt, NaCl, is 803 C.
Above 803 C sodium ions NaCl melts, and forms a molten salt of Na+ and Cl ions.
Electrodes are places in the molten NaCl and a
(sufficiently high) external voltage applied.

Sodium ions accept electrodes and form sodium metal at the cathode.
At the anode, electrons are taken from chloride ions, and chlorine gas is formed.

The production of sodium metal and chlorine gas is not spontaneous. It is driven by the electric current.

Question 21

Electrolysis of Aqueous Solutions

Answer 21

Electrolysis of aqueous solutions is more complicated than in molten salts, due to the presence of water.

The water molecules is dissolved ions could gain electrons at the cathode, or lose electrons at the anode. The water and ions compete.

Reduction decreases the H3O+ concentration,
so the concentration of OH- ions increases.

A drop of phenolphthalein has been added to
the solution so that the formation of OH-(aq) can be detected (by the red colour).

Question 23

Overvoltage

Answer 23

In industrial situations, large currents are used to produce large quantities of products quickly.
Gaseous products can form electrically insulating layer on the electrode.

A potential above that predicted by standard
reduction potentials must be applied to overcome the insultaing effect of the gas, called an overvoltage.

The overvoltage cannot be theoretically calculated.

Question 24

Spontaneous Direction of Change and Equilibrium

Answer 24

the direction of reaction that would take a reaction mixture closer to a state of chemical equilibrium.

• may be extremely fast or so slow that change is undetectable.
• A decrease of enthalpy of the reaction mixture is not a sufficient criterion to determine whether a process is spontaneous.

Question 25

Entropy: Dispersal of Energy and
Matter

Answer 25

Entropy: using a thermodynamic property to predict whether a process is spontaneous
- built on the idea that spontaneous change results in dispersal of energy
- The Second Law of Thermodynamics: Any
spontaneous process is accompanied by an
increase in the entropy of the universe.

• also observed in solitions.
• formation of a mixture does not always lead to greater disorder

Question 26

The Boltzmann Equation for Entropy

Answer 26

S = k log W

where k is the Boltzmann constant, and W represents the number of different ways that the energy can be distributed over the available energy levels.

Question 27

Entropy and the Laws of Thermodynamics

Answer 27

The Third Law of Thermodynamics: There is no disorder in a perfect crystal at 0 K, or S = 0.

All substances have positive entropy values at temperatures above 0 K.

Question 28

Standard molar entropy (S°)

Answer 28

the entropy of 1 mol of a substance in its
standard state at the specific temperature

Large molecules&raquo_space; smaller molecules.

Substances whose molecules have complex structures have more entropy than those with simpler molecules.

increases as thetemperature is raised. There are large increases in entropy when the state changes.

At a temperature T above the boiling point, the standard molar entropy includes contributions from:
1.The heat capacity of the solid between 0 K and M.P.
2.The heat of fusion of the solid
3.The heat capacity of the liquid from M.P. to B.P.
4.The heat of vaporization of the liquid
5.The heat capacity of the gas from the boiling point to T K.

Question 29

Entropy Change of Reaction (∆(r)S°)

Answer 29

∆S° (univ) = ∆S°(sys) + ∆S° (surr)

(∆(r)S°) = (sum of the entropies of the products) - (sum of the entropies of the reactants)

= ∑n(i) S°(products) - ∑n (i) S°(reactants)

S is the entropy per mole of each product and reactant
N is the number of moles of each reactant or product in the equation

Question 30

Contributions of ∆(r)S° and ∆rH° to
Spontaneity of Reaction

Answer 30

The entropy change of the surroundings is
∆S° (surr) = q (surroundings) / T = -∆H (sys) / T

There are 4 categories of reaction, depending on the signs of ∆S° (sys) and ∆S° (surr)

Question 31

Gibbs Free Energy

Answer 31

e used determine whether reaction is
spontaneous from just the system.
At a specified temperature T, the free energy is defined mathematically as:
G = H – TS
The free energy of a substance is usually not important.

∆rG = ∆rH ̶T∆(r) S

Question 32

∆(r)G and Spontaneity of Reaction

Answer 32

The second law states that a reaction is spontaneous if ΔSuniv > 0.
Reaction spontaneity can be worked out more conveniently from:

∆S (univ) = -∆(r) G / T

If ∆(r)G < 0, the reaction is spontaneous as written.
If ∆(r)G = 0, the reaction is at equilibrium.
If ∆(r)G > 0, the reaction is not spontaneous (but the reverse reaction is).

Question 33

Standard Free Energy Change of
Reaction (∆rG°)

Answer 33

∆rG° <0, the spontaneous direction of reaction is the direction of the chemical equation.

∆rG° is directly related to the value of the equilibrium constant, thus whether reactions are reactant-favoured or product-favoured.

The free energy change is the maximum energy available for work:
∆rG = w(max)

Question 34

Calculating ∆G0 (rxn)

Answer 34

1. Determine ∆Ho (rxn) and ∆So (rxn) and use Gibbs equation:
   ∆G = ∆H - T∆S
   ΔHrxn = ΔHprod – ΔHreact
   ΔSrxn = ΔSprod – ΔSreact

Values for H and S can be found in SI data book.

2. Use tabulated values of free energies of formation, ∆Gfo
   ∆Gfo is the standard free energy per mole for the formation of a compound from its
   elements in their most stable form under standard conditions (298 K).

∆G0 (rxn) = Σ∆G0 f(products) - Σ∆G0 f(reactants)

Question 35

Dependence of ∆rG° and Spontaneity
on Temperature: if ∆H is negative, ∆S is positive

Answer 35

reaction is always spontaneous, reverse is
always non-spontaneous

∆S positive favours reaction
∆H negative favours reaction

Question 36

Dependence of ∆rG° and Spontaneity
on Temperature: if ∆H is negative, ∆S is negative

Answer 36

reaction is spontaneous at low T, becomes
non-spontaneous as T increases

∆H negative favours reaction

Have to do the calculation to determine spontaneity.

Question 37

Dependence of ∆rG° and Spontaneity
on Temperature: if ∆H is positive, ∆S is positive

Answer 37

non-spontaneous at low T, spontaneous at
high T

∆S positive favours reaction

Have to do the calculation to determine spontaneity.

Question 38

Dependence of ∆rG° and Spontaneity
on Temperature: if ∆H is positive, ∆S is negative

Answer 38

reaction is non-spontaneous at all
temperatures, reverse is spontaneous

Question 39

∆rG for Non-standard Reaction Mixtures

Answer 39

∆rG = ∆rG° + RT ln Q

where R is the gas constant, 8.314 J K-1 mol-1, and T is the temperature (in K).

Question 40

The Relationship Between ∆rG° and K

Answer 40

When equilibrium is reached Q = K and ∆rG = 0. If we substitute into:
∆rG = ∆rG° + RT lnQ
0 = ∆rG° + RT ln K
∆rG° = ̶RT ln K

This equation allows calculation of:
* an equilibrium constant from thermochemical data in tables.
* the standard free energy change of a reaction an experimentally equilibrium constants.

Question 41

∆rG° and E°cell for Voltaic Cell Reactions

Answer 41

Recall the cell emf (E° cell) is related to K by:

ln K = + (nFE°cell) / RT

We can rearrange ∆rG° = ̶RT ln K to get:

ln K = -∆rG° / RT

Equate these two requations equations (at constant T):

∆rG° = ̶nFE° cell

Which relates the standard free energy of reaction to emf.

Question 42

For a spontaneous reaction in a standard mixture

Answer 42

(Q = 1): K > 1,
E°cell > 0, and ∆rG° < 0

Question 43

Dependence of Equilibrium Constants on Temperature

Answer 43

The dependence of the equilibrium constant on temperature is given by the following relationship:

ln K = (-∆rH° / RT) + (∆rS°/R)

R is independent of temperature changes and enthalpy and entropy are approximately constant over small temperature ranges.

The equation is linear

Question 44

The van’t Hoff equation again i think

Answer 44

used to estimate K at one temperature from an experimentally measured value at another temperature:

ln K {1} - lnK {2} = -∆rH°/R (1/T{1} - 1/T {2})