19. Redox HL

Describe the standard hydrogen electrode

Standard Hydrogen Electrode- A platinum black (unreactive) electrode pumped with H₂ and placed 1moldm^-3 H⁺ ions (usually from HCl) that catalyzes the reaction H­₂ ßà 2H⁺ + 2e^-

Define the term “standard hydrogen potential” (E^o)

Standard Electrode Potential- The potential difference (voltage) of an electrode (a.k.a. a half-cell) when it is put in a voltaic cell with the Standard Hydrogen Electrode (the SHE is used as a relative comparison)

E°- Symbol for standard electrode potential

Standard Electrodes are measured in volts at 298K, 101.3 kPa, and using 1moldm^-3_­ sol’ns

Calculate cell potentials using standard electrode potentials

Cell Potential (E­_cell­)- is the cell potential found by combining the Eº of  the two half equations

E_cell = E_reduction + (- E_oxidation)

It’s the OPPOSITE of E­_oxidation because you’re using the reverse reaction (from left à right of the series), so the sign has to flip

EXAMPLE: Find the E_cell for the reaction Sn²⁺ + 2Fe³⁺ à 2Fe²⁺ + Sn⁴⁺

Sn⁴⁺ à Sn²⁺ = + 0.154

Fe³⁺ à Fe²⁺ = + 0.771

(Sn is more negative, so that is E_oxidation)

E_cell = 0.771 + (- 0.154)

E_cell = + 0.617

Predict whether a reaction will be spontaneous using standard electrode potential values

As E_cell increases in the positive direction, the reaction increases in spontaneity

As E_cell increases in the negative direction, the reaction decreases in spontaneity.

            (E_cell < 0 = not spontaneous!)

Take the E_cell value from the previous example. Since it is + 0.617, it will be spontaneous

Predict and explain the products of electrolysis of aqueous solutions

Electrolysis of water…

At the cathode, water and H⁺ ions are reduced to form hydrogen gas.

An equation for the reduction of water is 2H₂O + 4e^- à 2 H­₂ + O₂

At the anode, water and OH^- ions are oxidized to form oxygen gas.

An equation for the oxidation of water is 2 H­₂O à O₂ + 4H⁺ + 4e^-

Electrolysis of aqueous solution…

Reactions at the cathode- Metal or H₂ will form                                

            Hydrogen gas will form unless- Cu²⁺ or Ag⁺ is present

Reactions at the anode- Non-metal or oxygen will form. Reactive conductors will degrade

            Oxygen gas will form unless- Br^- or I^- is present

Reactive metals can be used as conductors if you want to purify a metal ore into the elemental form. Metal ions will be ionized (oxidized) and reduced while impurities are removed into solution

Describe the use of electrolysis in electroplating

Electroplating- is the process by which a think layer of metal is deposited on the other electrode (used for decorating + corrosion production).

            Usually, metal ions in solution are used to electroplate the cathode (reduction).

Practice Problems:

http://alpha.chem.umb.edu/chemistry/ch115/Mridula/CHEM%20116/documents/Chapter20PracticeQuestions.pdf

9. Redox SL

Define oxidation and reduction in terms of electron loss and gain

Reduction- the process of gaining electrons

Oxidation- the process of losing electrons

Deduce the oxidation number of an element in a compound

The oxidation number of…

Flourine      -1

Hydrogen   +1       (except when bonded with a metal à NaH, CaH₂, etc…

Oxygen      -2         (except when bonded in H₂O₂ OR F₂O)

Halogens     -1        (except when bonded to O OR a halogen higher up in the group)

When you’re working with ionic compounds- It’s the charge of the ion

When you’re working with covalent compounds- Use the tricks above to help you figure it out. Keep in mind, POLYATOMICS are covalent compounds (Their total charge just doesn’t equal zero)

State the names of compounds using oxidation numbers

Oxidation numbers are used in compounds involving transition metal ions. They are added as roman numerals in parenthesis

            E.g. FeO is Iron(II) Oxide
            E.g. V­₂O₅ is Vanadium (V) Oxide

Deduce whether an element undergoes oxidation or reduction in reactions using oxidation numbers

Ex. FIND the oxidation number of H₂ à H­₂O

In H₂, there is no net charge so H has no oxidation number (0)

In H₂O, oxygen has a -2 charge, which means that the 2 H molecules each have a +1 charge in order to balance it out

Therefore, H is REDUCED

A reaction is not redox if the oxidation number does not change for any of the atoms

Disproportionation- a reaction in which the same element is both reduced and oxidized (ex. 2Cu⁺ à Cu­²⁺ + Cu)

            +1      +2        0

Deduce simple oxidation and reduction half-equations given the species involved in a redox reaction

In the equation Zn(s) + I₂(s) à Zn²⁺(aq) + 2I^- (aq) which is reduced?

The easiest way to do this would be to create half equations (deal with one element)

Zn à Zn²⁺                        Add electrons to one side or the other to create an equal charge

Zn à Zn²⁺ + 2e^-               Since the Zinc ion LOST electrons, it was oxidized

I₂ à 2I^-                              In this case, the products have two more electrons, soooo

I₂ + 2e^- à 2I^-                    Since the Iodide ions GAINED electrons, they were reduced

Define the terms oxidizing agent and reducing agent

Oxidizing agent- a substance that GAINS electrons in a redox reaction

            (i.e. it takes the electrons from the other substance, OXIDIZING it)

Reducing agent- a substance that LOSES electrons in a redox reaction

            (i.e. it gives the electrons to the other substance, REDUCING it)

Deduce redox equations using half-equations AND

Identify the oxidizing and reducing agents in redox equations

How does one balance a reaction between Zinc and acidified dichromate ions to produce chromium(III) ions?

Use Half Equations!

Chromium

            The oxidation number of chromium in dichromate is +6

            The oxidation number of chromium in Cr³⁺ is + 3

            THEREFORE, Chromium is being reduced (It is the oxidizing agent)

           

            CrO₇^2- + 3e^- à Cr³⁺

            BUT, realize there are now 7 Oxygen atoms on the left side of the e                                    equation. SO we have to balance it using water molecules (we assume the

            reaction (we assume the reaction is aqueous)

            CrO₇^2- + 3e^- à Cr³⁺ + 7H₂O

            A good start, but we now have 14 hydrogens on the right side of the eq.

            Luckily, the dichromate solution is ACIDIFIED meaning there is plenty

            Of our buddy H⁺ to save us from this disaster

            CrO₇^2- + 14H⁺ à Cr³⁺ + 3e^- + 7H₂O

Zinc

            Since Chromium is being reduced (as we said above), we can take a

            Guess that Zinc must be oxidized (It is the reducing agent)

            Zn à Zn²⁺ + 2e^-

           

            And that’s all you need! (SIMPLE)

Uh oh, we’re not done yet!

            We’re gonna need to balance the number of electrons on each side

            Lets go back to the simple half equations

            The easiest way is to make sure there are 6 electrons on each side

                        2[CrO₇^2- + 3e^- à Cr³⁺] = 2CrO₇^2- + 6e^- à 2Cr³⁺

                        3[Zn à Zn²⁺ + 2e^-] = 3Zn à 3Zn²⁺ + 6e^-

So re-add those coefficients to the original whole equation

            2CrO₇^2- + 14H⁺ + 3Zn + 6e^-à 2Cr³⁺ +6e^-  + 3Zn²⁺ + 7H₂O

Get rid of the electrons (they cancel each other out) and viola!!!

            2CrO₇^2- + 14H⁺ + 3Zn à 2Cr³⁺  + 3Zn²⁺ + 7H₂O

Deduce a reactivity series based on the chemical behavior of a group of oxidizing and reducing agents

Deduce the feasibility of a redox reaction from a given reactivity series

Explain how a redox reaction is used to produce electricity in a voltaic cell. AND

Function of Voltaic cell- convert chemical energy à electrical energy

Draw a voltaic cell-

How does a voltaic cell work? Each electrode establishes equilibrium with the solution, creating a concentration of ions in the solution and excess electrons. The more reactive metal (anode) will be more oxidized than the positive, creating a negative charge. These electrons from the anode travel through the wires to the cathode, where they reduce the ions in solution. This movement of electrons is the electric energy. In order to complete the circuit, the salt bridge is added which will allow negative ions to move in accordance with the increasing/decreasing numbers of positive ions

State that oxidation occurs at the negative electrode (anode) and reduction occurs at the positive electrode (cathode).

OXidation occurs at the ANode (AN OX)

            The anode is NEGATIVE in a voltaic cell

REDuction occurs at the CATthode (RED CAT)

            The cathode is POSITIVE in a voltaic cell

Describe using a diagram, the essential components of an electrolytic cell

Electrolysis- Non-spontaneous reaction driven by application of electrical energy

Electrolyte- Solution containing ions to transport electric current

Draw an Electrolytic Cell

State that oxidation occurs at the positive electrode (anode) and reduction occurs at the negative electrode (cathode)

OXidation occurs at the ANode (AN OX)

            The anode is POSITIVE in an electrolytic cell

REDuction occurs at the CATthode (RED CAT)

            The cathode is NEGATIVE in an electrolytic cell

Explain how current is conducted in an electrolytic cell

How does an electrolytic cell work?

            Two inert conductors are placed in an ionic solution and a current is applied (usually through a battery.) Electrons leave the battery and move to the negative electrode, where they are used to reduce cations in solution to anions. This negative conductor becomes the Cathode. These negative ions then travel through the solution to the positive ion, where the anions are oxidized back into cations. These electrons move through the wires back to the battery and the cycle is repeated.

Inert conductors are used so that they do not become oxidized/reduced by the reaction.

Examples of inert conductors include- platinum

Deduce the products of the electrolysis of a molten salt

EXAMPLE: CuBr₂

Solid ions form an ionic lattice

Liquid ions are free to move within a solution

At the cathode, the positive ions will be reduced

            Cu²⁺ + 2e^- à Cu

            Cu (s) will bind to the electrode

At the anode, the negative ions will be oxidized

            2Br^- à Br₂

Practice Questions:

http://alpha.chem.umb.edu/chemistry/ch115/Mridula/CHEM%20116/documents/Chapter20PracticeQuestions.pdf

15. Energetics HL

Define and apply the terms standard state, standard enthalpy change of formation, and standard enthalpy change of combustion AND

Determine the enthalpy change of a reaction using standard enthalpy changes of combustion and formation

Standard State- state in which matter has an enthalpy of zero and is under standard thermochemical conditions

(i.e. is it a solid, liquid, or gas at room temperature?)

Standard enthalpy of formation (ΔH_f^°)- the amount of energy required to form one mole of a compound in its standard state from its elements in their standard states.

(i.e. Like using Hess’ Law, except working with individual compounds and not the intermediate reactions)

            ΔH_f^°= ΣΔH_products - ΣΔH_reactants

EXAMPLE: Find the ΔH_f^° for the reaction 2NaHCO₃ à NaCO₃ + CO₂ + H­₂O using the following data: NaHCO₃= -948, NaCO₃ = -1131, CO₂ = -395, H₂O= -286

ΔH_f^°= ΣΔH_products - ΣΔH_reactants

ΔH_f^°= (-1131 + -395 + -286) – (2 x -948)

ΔH_f^°= -1815 + 1896

ΔH_f^°= +81 kJ mol^-1

Standard enthalpy of combustion (ΔH_comb^°)- the amount of energy required for one mole of a substance to undergo complete combustion in excess oxygen under standard conditions

            (i.e. Like finding ΔH_f^°, except something gets burned)

            ΔH_c^°= ΣΔH_products - ΣΔH_reactants

Define and apply the terms lattice enthalpy and electron affinity

Lattice Enthalpy (ΔH_lat^°)- The enthalpy change when one mole of a solid ionic compound is separated into gaseous ions

            (i.e. The energy gained/lost when a lattice breaks down into component ions)

            Because it is being broken down, it will always be ENDOTHERMIC

Electron Affinity- The amount of energy needed or released when a mole of a substance to gain an additional electron (each)

Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides and use it to calculate an enthalpy change

Born-Haber Cycle- Hess cycle used to measure lattice enthalpies

Steps to a Born Haber Cycle- Atomization, Ionization, Determining bonds (AID)

1. Determine the enthalpy of atomization for both substances.

Enthalpy of Atomization (ΔH_lat^°)- Energy change when one mole of gaseous atoms are formed from the element in its standard states

* This can be calculated using info from your data booklet

2. Determine the ionization/electron affinity enthalpy change

Ionization enthalpy change (ΔH_i^°)- Enthalpy change for one mole of gaseous atoms or cations to LOSE an elec. to form a mole of gaseous positive ions.

Electron affinity enthalpy change (ΔH_e^°)- Enthalpy change for one mole of gaseous atoms or anions to GAIN an elec to form a mole of gaseous negative ions

* This can be calculated using info from the periodic table

**If you need an ion with a multiple charge (ex. Mg²⁺), you need to use the sum of two values (ΔH_e^° when Mg à Mg⁺ AND ΔH_e^° when Mg⁺ à Mg²⁺)

3. Determine the lattice enthalpy of the bond

ΔH_lat^° =  (ΔH_atom^°[ A] + ΔH_atom^°[B] + ΔH_i^°[A] + ΔH_e^°[B] ) - ΔH_f^° [Compound AB]

ΔH_lat^°= (Everything) – (Standard Enthalpy of Formation)

EXAMPLE!!! Construct a Born-Haber Cycle for the reaction: Na (s) + ½Cl­₂ (g) à NaCl

Given that ΔH_f[NaCl] = -411kJ and ΔH_atom[Na] = +109 kJ

1. Atomization

ΔH_a [Na] = +109        *This was stated above

ΔH_a [Cl] = + 121         *This was found by taking the Cl-Cl bond enthalpy and dividing it by two (1/2 mole of Cl₂)

2. Ionization

ΔH_i [Na] = + 494        *These were found using the data booklet

ΔH_e [Cl] = -364

3. Determining Lattice

ΔH_lat^° =  (ΔH_atom^°[ Na] + ΔH_atom^°[Cl] + ΔH_i^°[Na] + ΔH_e^°[Cl] ) - ΔH_f^° [NaCl]

ΔH_lat^° = (109 + 121 + 494 -364) – 411

ΔH_lat^° = 360 - (- 411)

ΔH_lat^° = + 711 kJ mol^-1

***Sometimes, you may be asked to Draw or Construct a Born Haber Cycle. You will still need to do the mathematics behind it, but you will need to draw one as well. A sketch is included on the following page. The 3-steps listed above are written in red.

Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds AND

Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character

A theoretical method for predicting lattice enthalpy- the Ionic Model

Ionic model- assumes that ions are perfect spheres and the only interaction is due to attraction between ions

A decrease in ionic radius means an increase in the force of attraction between ions

An increase in ionic radius means a decrease in the force of attraction between ions.

Ionic model is more accurate for smaller ions than larger ions. Larger ions have a larger percent difference as compared to experimental (Born-Haber) models

***Works best for small, highly charged, atoms

State and explain the factors that increase the entropy of a system.

Entropy- A measure of the degree of disorder or randomness in a system. A.k.a.

ENTROPY IS DISORDER!!!

If ΔS^° is positive- entropy increases

If ΔS^° is negative- entropy decreases

Predict whether the entropy change (ΔS) for a given process is positive or negative

Less moles à More moles= increase in entropy

More moles à Less moles= decrease in entropy

If those are the same, use the following tricks:

• From solid à liquid à gas, there is more entropy
• Production of gas = increase in entropy
• Using up gas = decrease in entropy

Calculate the standard entropy change for a reaction (ΔS) using standard entropy values.

Change in entropy (ΔS^°)= ΔS^°_products - ΔS^°_reactants

I won’t go through an example. It’s the same general idea as ΔH

Units of entropy- J kg^-1 mol^-1

***WATCH THE UNITS!!!

Predict whether a reaction or a process will be spontaneous using the sign for ΔG

Spontaneous- A change that tends to happen

            Ex. Salt dissolves in water, gas expands to fill its container

Gibbs Free Energy (ΔG^°)- The change in the amount of energy available to do usable work

If ΔG^° is positive- the process IS NOT spontaneous

If ΔG^° is negative- the process IS spontaneous

Calculate the Δhmm.

if it was great sex, excusable. if he was otherwise normal average regular all around

nope

he's getting kicked out of bed.

why, is the guy you've been screwing a smoker? :P

G for a reaction using the equation ΔG^° = ΔH^° - ΔS^°T

EXAMPLE!!!

For the reaction of PCl₃ and Chlorine gas to form PCl₅ at 25^oC, the entropy change is -85J mol^-1 K^-1 and the enthalpy change is -124 kJ mol^-1. What is the approximate value of the Gibbs free energy? Is this reaction spontaneous?

 ΔG^° = ΔH^° - ΔS^°T

ΔG^° = -124 – (85/1000) x (25 + 273)

REMEMBER to change J mol^-1 K^-1 à kJ mol^-1 K^-1 by dividing the value by 1000

REMEMBER ALSO to change Celcius à Kelvin by adding 273

REMEMBER REMEMBER the 5^th of November the gun…nevermind. Back to chem.

ΔG^° = -124 – (298 x 0.085)

ΔG^° = -124 – 25.33

ΔG^° = -98.67

It’s negative, then it’s spontaneous!!!!

Predict the effect of a change in temperature on the spontinety of a reaction using standard entropy and enthalpy changes

Favored reactions are exothermic ones that increase entropy, though neither is entirely sufficient

Exothermic reactions with a negative entropy change will always be spontaneous

Exothermic reactions with a positive entropy change will be spontaneous at high temperatures because only there will ΔH > ΔST

Endothermic reactions with a negative entropy change will be spontaneous at low temperatures because only there will ΔH <ΔST

Endothermic reactions with a positive entropy change will never be spontaneous

ΔG^° = ΔH^° - ΔS^°T

            If you’ve seen the HP series, remember this

Gryffindor = Harry – Slytherin Tendencies

EXAMPLE!!!

For the reaction of PCl₃ and Chlorine gas to form PCl₅ at 25^oC, the entropy change is -85J mol^-1 K^-1 and the enthalpy change is -124 kJ mol^-1. What is the approximate value of the Gibbs free energy? Is this reaction spontaneous?

 ΔG^° = ΔH^° - ΔS^°T

ΔG^° = -124 – (85/1000) x (25 + 273)

REMEMBER to change J mol^-1 K^-1 à kJ mol^-1 K^-1 by dividing the value by 1000

REMEMBER ALSO to change Celcius à Kelvin by adding 273

REMEMBER REMEMBER the 5^th of November the gun…nevermind. Back to chem.

ΔG^° = -124 – (298 x 0.085)

ΔG^° = -124 – 25.33

ΔG^° = -98.67

It’s negative, then it’s spontanteous!!!!

Follow this link for some good practice problems (E-H)
http://coffman.dublin.k12.oh.us/teachers/teacherpages/brown/mr._browns_chemistry_pages/Topic_5_files/practiceenergetics.pdf

5. Energetics SL

STANDARD TEMPERATURE AND PRESSURE- 298 K (25º C) and 101.3 kPA

Define the terms exothermic reaction, endothermic reaction, and standard enthalpy change of a reaction

Standard enthalpy change of a reaction (ΔH)- The change in the amount of energy that exists within a system

Endothermic- Reaction in which energy goes INTO a system from the surroundings (ΔH+)

Exothermic- Reaction in which energy EXITS a system into the surroundings (ΔH-)

State that combustion and neutralization are exothermic processes.

Combustion Reaction: Reaction between a hydrocarbon and oxygen to produce water vapor and carbon dioxide (Exothermic Process)

Neuralization Reaction between an acid and a base that produces water and a salt (Exothermic Process)

Apply the relationship between temperature change, enthalpy change, and the classification of a reaction as either endo- or exothermic

	Temperature	ΔH	Product Stability	Product Energy	Bond Strength
Endothermic	Decrease	Increase	Less	More	Less
Exothermic	Increase	Decrease	More	Less	More

Deduce, from an enthalpy level diagram, the relative stabilities of reactants and products, and the sign of the enthalpy change for the reaction.

Energy Diagram for Endothermic Process looks like:

ΔH IS POSITIVE

Energy Diagram for Exothermic Process looks like

Calculate the heat energy change when the temperature of a pure substance is changed

Equation for Calculating Heat Enthalpy: q = mcΔt

q = Enthalpy change

m = mass

c = specific heat capacity

Δt = temperature change

Practice: The temperature of a 2.0 g sample of aluminum (Specific heat = 0.90 J g^-1K^-1) increases from 25 degrees C to 30 degrees C. How many joules of heat were added

q = mc Δt

q = 2.0 g * 0.90 J g^-1K^-1 * 5 K

q = 1.8 * 5

q = 9.0 J

Calculate the enthalpy change for a reaction using experimental data on temperature changes, quantities of reactants and mass of water AND

Evaluate the results of experiments to determine enthalpy changes

Practice: The heat released from the combustion of 0.0500 g of phosphorous increases the temperature of 15.00 g of water from 25.0 °C to 31.5 °C. Calculate the ΔH of phosphorus in kJ mol^-1

*TAKE NOTE: When doing experiments which involve temperature change of water, mass refers to the mass of the water which is changing temperature

q = mc Δt

q = (15.00 g) (.0418 kJ g^-1K^-1) (6.5 K)

q = 4.0755 kJ

(This gives is the ΔH for 0.0500 grams of P)

4.0755 = 81.5 kJ mol^-1

0.0500

Design suitable experimental procedures for measuring the heat energy changes of reactions

Include the following:

• Known mass of solution in an insulated container (coffee cup, bomb calorimeter, etc…)
• Temperature measured continuously with thermometer
• Substance burned beneath it, transferring heat energy to the solution
• This creates a change in temperature. Use mcΔT to find the energy

In experiments, the results are not always accurate due to error from heat loss.

Determine the enthalpy change of a reaction that is the sum of two or three reactions with known enthalpy changes

Hess’ Law: In a chemical reaction, it takes the same amount of energy to go from Compound A à C as it does to go through intermediate steps from A à B à C.

EXAMPLE!!! Find the amount of energy needed to undergo the following reaction using the equations and Enthalpy change values listed

Define the term average bond enthalpy

Average Bond Enthalpy- the AVERAGE amount of energy needed to break one mole of bonds in the gaseous state. Because this is an AVERAGE value, it is not always reliable

Explain in terms of average bond enthalpies, why some reactions are exothermic and others are endothermic

When bonds are broken: Energy is absorbed [into the system]

When bonds are made: Energy is released [out of the system]

Trick to remember this: BREAKING IN and MAKING OUT

Endothermic reactions result when Energy going in < Energy coming out. It takes more energy to break the bonds than make the bonds, so energy goes INTO the system (creating a positive value)

Exothermic reactions result when Energy going in > Energy coming out. It takes less energy to break the bonds than make the bonds, so energy LEAVES the system (creating a negative value)

Equation for figuring out average bond enthalpy:

Total bonds = Bonds broken – Bonds made

(This is the one exception to the Products – Reactants Rule. Remember Breaking in and making out!)

EXAMPLE!!! What is the ΔH reaction for the following equation in kJ?

CS₂(g) + 3O₂(g) à CO₂(g) + 2SO₂(g)

Energy to make bonds: CS₂ = 110kJ, CO₂ = 390 kJ, SO₂ = 580 kJ

CS₂(g) + 3O₂(g) à CO₂(g) + 2SO₂(g)

1 CS₂                         1 CO₂      2 SO₂

110 * 1                       390 * 1   290 * 2             

   110                           390          580

       Total bonds = Bonds broken – bonds made

            Total =     110 – (  970)

                 Total = -860 kJ

Follow this link for good practice questions (PART A-D)
http://coffman.dublin.k12.oh.us/teachers/teacherpages/brown/mr._browns_chemistry_pages/Topic_5_files/practiceenergetics.pdf