electrochemistry

Chemistry 30

• What is

Science

? Why do we care and

why do you take it?

• The course is really quite

simple

. The

Electrochemistry

Oxidation and Reduction

• define oxidation and reduction operationally and theoretically

• define oxidizing agent, reducing agent, and half reactions

• Reduction

– When metals are refined (purified) a large mass of ore is reduced to a small mass of pure metal

• Operational definition of reduction:

– producing pure metals from ores

• This is an operational definition because it doesn’t explain why or how

Oxidation and reduction

• Oxidation

• Operational definition

– the reaction of a substance with oxygen (ie burning,rusting)

• Example

• 2 Mg(s) + O₂(g) ---> 2 MgO(s)

• in time scientists realized that other non metals could undergo similar reactions

• Example

Oxidation and reduction theoretical

definitions

• Reduction

– The gain of electrons GER

• Example

• Cu

(aq) + 2 e

--> Cu(s)

• Oxidation

– The loss of electrons LEO

• Example

-½ Reactions

• A reaction that shows either oxidation or

reduction

• Shows the electrons involved (in the

equation)

• The net charge must be equal on each

side of a half reaction

Half Reactions

• Oxidation 1/2 reaction

– Electrons are a product (on the right side) – 2 Cl-(aq) ---> Cl

2(g) + 2

e-– Metals and anions tend to be oxidized

• Reduction ½ reactions

– Reduction half reaction

– Electrons are a reactant (on the left side) – Fe3+ + 3e- --> Fe(s)

Agents

• Reducing agent

– Promotes reduction by donating electrons and is oxidized

• Oxidizing agent

Redox reactions

• Oxidation and reduction must always take

place together

• these reactions are called

redox

reactions

• In redox reaction the electrons lost by the

Redox equations

•

There are three ways of writing a

reaction

1. Chemical equation

– Cu(NO₃)_2(aq) + Zn_(s)  Zn(NO₃)_2(aq) + Cu_(s)

2. Total ionic equation

– Cu2+

(aq) + 2NO3-(aq) + Zn(s)  Zn2+(aq) + 2NO3-(aq) + Cu(s)

3. Net ionic

– Cu2+

Oxidation vs reduction

• Then we can decide what has been

oxidized (loses electrons) and what has

been reduced gains electrons)

• which is the reducing agent (undergoes

oxidation) and which is oxidizing agent

(undergoes reduction) and we can come

up with the two ½ reactions:

• LEO:

Zn



Zn

(aq)

+2e- (RA)

• GER:

Cu

More equations

•

Reaction 2:

• Cu

+ Zn(NO

)



no reaction

• Cu

+ Zn

(aq)



no reaction

• Why?

•

Reaction 3

Still more equations

•

_{Reaction 4:}

• Ag

+ Cu(NO

)



no reaction

•

Reaction 5:

• Zn

+ 2AgNO



Zn(NO

)

+ 2 Ag

•

Reaction 6:

Using these equations to rank electron attractiveness

compare the relative strengths of oxidizing and reducing agents, using empirical data

• Ag+

(aq) + e-  Ag(s)

• _Cu2+

(aq) + 2e-  Cu(s)

• _Zn2+

(aq) + 2e-  Zn(s)

• What good is this table?

• Read from top to bottom and left to right, it tells us the strongest oxidizing agents.

• Read from bottom to top right to left, it tells us the strongest reducing agents

Practice problems

Compare the relative strengths of oxidizing and reducing

Practice problems

Predicting Redox Reactions

• Text pg. 360 – 362

• Now we know how to construct a redox ½

reaction table, we are going to use the

published table to predict reactants, products and whether a reaction proceeds

• _{Rules for predicting Redox reactions}

1. Create a species list (write down everything,

splitting up soluble ionic and strong acid and bases) 2. Using pg 7 label everything as an OA (left side of

Rules for predicting net redox

reactions

3. Find your strongest OA (top left) and strongest RA (bottom right)

4. Write down the OA’s reduction ½ reaction and the RA’s oxidation (flip ½ reaction

5. Balance moles of electrons lost and gained 6. Add the two equations together

7. Calculate the net potential (E0

net / voltage) E0net

= E0

SOA – E0SRA

8. The equation is spontaneous if the SOA is above the SRA (if the potential is +ve)

• _Example:

Net Redox Reaction examples

• A redox reaction occurs when an iron nail is placed in a solution of copper (II) sulphate. Elemental copper begins to form, and the colour of the solution changes. In this reaction, the reducing agent is

– Fe(s) – Cu(s) – Fe2+(aq)

Practice problems

Oxidation numbers

define oxidation number

• It is easy to see if e- are lost or gained in ions

• Example 2FeO(s)  2Fe(s) + O₂(g) (4 moles of e- moved)

• It is not so easy to see if e- were lost/gained in molecular compounds

• Example CH₄ +O₂ CO₂ + H₂O

• Oxidation numbers are “pretend” charges we give based on electro negativity

• Even though e- are shared they’re shared unevenly, we simply pretend the e- are

Oxidation numbers

• ON is a positive or negative number corresponding to the oxidation state assigned to an atom. In a covalently

bonded molecule or polyatom,

• the more electronegative atoms are considered negative and the less electronegative atoms are positive. In other words, its as if the covalent bond was an ionic bond –

remember it is only imaginary.

• Eg. In a water molecule oxygen has electronegativity of 3.5, hydrogen 2.1. ON of oxygen is –2 and that for

hydrogen is +1. To differentiate between ON’s and

Oxidation number rules

1. All atoms in elements have an ON of zero. EG in O₂, oxygen has ON = 0, in S₈ sulfur has ON of 0

2. Hydrogen is +1 when connected to a more

electronegative species (all non metals) and -1 when connected to a less electronegative

species (most metals)

3. Oxygen is -2 unless connected to F (more electronegative) or in H₂O₂

Rules for assigning oxidation

numbers

• NOTE: in some rare circumstances, ON may be a fraction

5. If the molecule has 2 or more non metals that are not O or H give the more electronegative its charge where possible

6. Assign all the fixed ON’s ex O is –2, His +1 etc. 7. Total of ON’s of atoms in a molecule is zero

8. The sum of the ON’s in a polyatomic ion is the polyatomic ion’s charge

9. Atoms of the same element in the same

ON examples

• CH

,CO

Oxidation numbers and redox reactions

• differentiate between redox reactions and other reactions, using half-reactions and/or oxidation numbers

• Oxidation numbers are used to identify

1. the OA and RA

2. the number of electrons lost and gained

per atom

3. The number of electrons lost and gained

per molecule

Oxidation numbers in Redox

reactions

• 2H

O

+ 2I

-(aq)



I

+ H

+ 2OH

• A good rule of thumb in this unit is

whenever they give you an equation

give everything an ON

Identify electron transfer, oxidizing agents and reducing agents in redox

Practice problems

Balancing Redox reactions in acidic or

neutral solutions

• Write and balance equations for redox reactions in acidic and

neutral solutions by using half-reaction equations obtained from a standard reduction potential table developing simple half-reaction equations from information provided about redox changes assigning oxidation numbers, where appropriate, to the species undergoing chemical change

• Sometimes equations redox ½ reactions

or equations are given unbalanced and

you are asked to balance them

• You can not use inspection here to

balance you MUST insure electrons lost =

electrons gained

Balancing Redox reactions in acidic or

neutral solutions

1. Balance redox reactions using

oxidation numbers (ON)

• Example C₂H₅OH + MnO₄-  CH

3CO2- + MnO4

2-1. Ignore H

O and H

2. assign ON’s to all species in the reaction

mixture.

Balancing Redox reactions in acidic or

neutral solutions

4. Balance the number of electrons lost/gained per molecule

5. Balance the number of electrons lost with the number gained

6. Balance number of O’s in the compound by adding H₂O’s.

7. If acidic balance number of H’s by adding H+

• Answer

• H₂O + C₂H₅OH + 4MnO₄-  CH

example

• They may include the H

O and H

(just

ignore the H+ and the H

O)

Practice problem

Balancing Redox equations

using ½ Reactions

2. Balance the following reaction in an acid solution • _{Ni(s) + NO3}-(aq)  NH

4+(aq) + Ni+2(aq)

1. Split the unbalanced equation into two ½ reactions. 2. Ni(s)  Ni2+(aq)

3. NO₃-(aq)  NH

4+(aq)

4. Balance all atoms except H and O • Ni(s)  Ni2+(aq)

• _NO3-(aq)  NH 4+

Balancing Redox reactions

• Ni(s)  Ni2+(aq)

• _NO3-_{(aq)  NH}

4+(aq) + 3H2O(l)

6. Balance H with H+ (in an acidic solution) • Ni(s)  Ni2+(aq)

• _10H+_{(aq) + NO}

3-(aq)  NH4+(aq) + 3H2O(l)

7. Balance the charge with electrons: • Ni(s)  Ni2+(aq) +

2e-• _{8e- + 10 H}+ _{+ NO}

3-  NH4+ + 3H2O(l)

8. Balance electrons in each ½ by multiplying through by a coefficient

• 4 X LEO : 4 Ni(s)  4Ni2+_{(aq) +}

8e-• _{8e- + 10 H}+ + NO

3-  NH4+ + 3H2O(l)

9. Add the two ½ reactions • _{4Ni(s) + 10H}+(aq) +NO

3-(aq)  NH4+ + 3H2O + 4Ni2+

Example

• Balance the following equation anyway

possible

Disproportion/autoxidation

Reactions

• define disproportionation

• In a disproportion reaction some atoms from an element are oxidized while some others are

reduced.

• In other words, a single chemical species oxidizes itself.

• use ON’s to spot and ½ reactions to balance • Eg. 3Br_2(l) + 3H₂O  HBrO_3(aq) + 5HBr_(g)

ONs

Disproportionation Reactions

• 3Br2 + 3H2OHBrO3 + 5HBr

• 0 +5 -1

• MnO₄2- MnO

2 + MnO4- (acidic solution)

• NO₂  NO₃- + NO (acidic solution) • HNO₂ NO + NO₃- (basic solution) • Cl₂  Cl- + ClO- (acidic solution) • S  S2- + SO

32- (basic solution) • NO + NO₃-  N

Practice problems

Electrochemistry Stoichiometry

• perform calculations to determine quantities of substances involved in redox titrations.

• Stoichiometry

• Once you have a balanced equation, eitehr from a net redox reaction or the balancing techniques we just learnt) you can calculate the moles of

something (given m=nM or n=cv) and move it to a different compound in the equation by using the molar ratio

• 2Na + Cl₂  2NaCl 1mol of Na is ½ mol of cl₂ • Usually involves titration, which is the

Elecrochemistry stoich

• The endpoint of the titration is usually a

colour change.

• Two OAs commonly used are

permanganate ions in aq. solution and

dichromate ions in aq. solution.

• the ½ reactions involved are:

• MnO

(aq) + 8H

+ 5e-



Mn

(aq)+ 4H

2

O

Electrochemistry stoich

• And

• Cr

O

+ 14H

= + 6e-  2Cr

+ 7H

2

O

• orange

green

• In a titration, the concentration of the

solution being added from the buret is

usually known and

• the concentration of the sample (solution in

the conical flask) is usually to be

Electrochemistry stoich example

•

10.0 mL samples of acidic 0.0500 M tin

(II) chloride solution is used to titrate

Electrochem stoich example

• Step 1

• Equation list the species present, • _{Step 2}

• state the RA/OA and find SOA/SRA • Sn2+ Cl- H

2O K+ MnO4- H+

• MnO₄ with H+ turns out to be the SOA

• Sn2+ turns out to be the SRA

• Step 3

• ½ rxns to net redox

• 2(MnO₄-(aq) +8H+ + 5e-  Mn2+ + 4 H

2O)

Electrochem stoich example

•

_{Step 4}

• do stoich

• balance the overall equation and do the

stoich the answer is 119 mmol/L

• 12.4 mL of 0.0832 M permanganate

solution is used to titrate 10.00 mL of tin (II)

solution. What is concentration of tin (II)

Stoich examples

• In an experiment, a student used 11.33 mL of H₂O₂(aq)

to titrate a 17.00 mL sample of acidified 8.0 X 10–3 mol/L

KMnO₄(aq). If Mn2+(aq) is one of the products, then the

concentration of the H₂O₂(aq) is – 1.2 X 10–2 mol/L

– 1.5 X 10–2 mol/L

– 3.0 X 10–2 mol/L

Practice problems

Electrochemical Cells

• define anode, cathode, anion, cation, salt bridge/porous cup, electrolyte, external circuit, voltaic cell

• predict and write the half-reaction equation that occurs at each electrode in an electrochemical cell

• Species list species list species list

• Let’s look at one of the simpler redox reactions • Cu2+ + Zn  Zn2+ + Cu

• E- are moving from Zn(s) (SRA) to Cu2+ (SOA) • This generates a potential (voltage)

• Society wanted to capture these e- to use as batteries

Voltaic cells

• Cells can be divided into 2 ½ cells

• The cathode ½ cell is the ½ cell where reduction takes place. This is the side of the SOA

• Red cat

• The anode ½ cell is the ½ cell where oxidation takes place, it contains the SRA

• An Ox

• In our example the beaker containing the Zn(s)/Zn2+_{(aq) is the}

anode ½ cell

• Each half consists of electrolyte (solution) and electrode (solid)

• The electrical circuit is completed with a wire (a voltmeter is shown here and shown here is a slat bridge.

Voltaic cells

• A short hand way of representing this cell

is:

• Cu

│Cu(NO

)

║ Zn(NO

)

│Zn

• A single vertical line indicates the interface

between electrode and electrolyte, the

Voltaic cell description

• In this zinc half-cell, the Zn of the electrode loses electrons and the Zn2+ goes into solution. Where do the electrons go?

• _{Into the wire and over into the copper half cell.} The electrons travel to the copper electrode and electrons are taken by copper ions in solution and Cu(s) is produced. Solid copper plates on the

copper electrode

• What else happens in the zinc half-cell? Consider the electrolyte. Charges are balanced for every

Zn2+(aq) we have 2 nitrates. But when the reaction

starts extra cations (Zn2+) go into solution. We

Voltaic cells

• Consider the copper half. Before the reaction begins, the electrolyte is electrically balanced. But Cu2+(aq) are taken out of solution. Because positives are taken away we are going to have an excess of negatives

• The imbalances – build up of negatives in the copper half cell and build up of positives in the zinc half – cell can be tolerated for a time. However, if allowed to build up unchecked, the imbalances would eventually cause current (electrons) to stop flowing.

Voltaic cell example

• Describe an electrochemical cell with one half comprising a silver electrode in silver nitrate and the other half comprising a

copper electrode in copper nitrate.

• Identify SOA and SRA: • Present

• Ag(s) Ag+(aq) Cu(s) Cu2+(aq) H 2O

• RA OA RA OA OA/RA

Workbook question

Cell potentials

• calculate the standard cell potential for electrochemical cells • predict the spontaneity of a redox reaction, based on standard

reduction potentials, and compare their predictions to experimental results

• _{Identify ½ reactions (add electrical potential):}

• Red cat Ag+ + e-  Ag(s) 0.80V

• An oxCu(s)  Cu2+ + 2e- -0.34V

• Balance and add ½ reactions:

• 2Ag+ + Cu(s)  2Ag(s) + Cu2+ 0.46V

Cells

• Draw Diagram

• Evidence table

•

Evidence: Interpretation:

• Copper electrode decreases in mass

– Oxidation: Cu(s)  Cu2+ + 2e

-• Colour of electrolyte in anode ½ cell

– _{Deepening (bluer)}

• Silver electrode (cathode) increase in size

Cells

• Blue colour moves up salt bridge

– Cu2+ migrate to cathode. Nitrate moves to

anode

• Voltmeter shows Ag to be cathode (+ve)

– electrons move from Cu to Ag

• And Cu to be anode (-ve)

– _{electrons move from Cu to Ag}

Cells

• Things we should test or look for as evidence • Electro conductivity

– Examples include molecular substances or non metals becoming ions or vice versa

– PbSO₄(s) + 2e- Pb(s) + SO

42-(aq)

• pH test

– Examples include the production or removal of H+(aq)

or OH-(aq)

– H₂(g)  2H+(aq) + 2e

-• gas formation (bubbles)

– Examples include the production of a gas – 2H+(aq) + 2e-  H

2(g)

Cells

• Precipitate

– Examples include the production of a solid – Fe2+(aq) + 2e- Fe(s)

• corrosion

– Examples include the removal of a solid – Fe(s)  Fe2+(aq) + 2e

-• ion colour

– Examples include ions produced found on page 11

-Cell potentials

• Using net potentials to identify unknowns

• If you are given the net potential of a cell

and either the cathode or anode you can

use the equation

• Net potential = CAT

–AN

• to solve for the other potential and look it

up on page 7

– Draw a voltaic cell including copper/copper solution X/X solution show an arrow pointing from X to Cu and a net potential of 0.60V.

cells

• In this reaction • Identify our

• Cathode (red)

Inert electrodes

• If one of the ½ reactions used in the cell

does not contain an electrode then one

must be added in

cells

• Describe the following cell

• K

Cr

O

(aq),H

(aq)/C(s)//FeSO

4

(aq)/Fe(s)

• Your description should include

1.a labeled drawing

1. Including arrows for e- movement and ion movement

2.½ reactions

3.Cell potential

Workbook questions

Reference cell

• explain that the values of standard reduction potential are all relative to 0 volts, as set for the hydrogen electrode at standard conditions

• It is impossible for something to gain/lose e- on its own

• A ½ reaction cant “really” have its own potential • All potentials were calculated base on their

relationship to hydrogen

• On your exams, you will be asked to change the reference cell from hydrogen to something else • Simply make that new ½ reaction have a

Reference cell

• Whatever you do to make the new

reference zero you will do to all the other

½ reactions

• This will have no effect on the potential of

a cell

• If fluorine is the new reference ½ reaction

what’s the new potential for zinc?

• F

goes from 2.87 – 0 (-2.87) so subtract

2.87 from everything

Using voltages to predict

reactions

• Often ½ reactions not found on page 7 are

used

• Cell potentials are given and these ½

reactions can be placed into page 7

example

• Which of the oxidizing agents would be able to react

Workbook question

Cell terms

• Standard voltaic cell; operate at 25 degrees Celsius and electrolytes are 1.0 M and all gases at standard pressure. These are designated

with the Eo symbol. They are comprised of two

½ cells, and converts chemical energy to electrical

• Standard ½ cell: contains a redox couple –

redox couple has reduced and oxidized forms of a chemical entity, Eg. Ag+ is oxidized form of

Ag; Ag is reduced from of Ag+. Also contains a

Cell terms

•

_{Inert electrode:}

_{If a ½ cell redox couple}

does not contain a solid

ex. H

(g)



2H

(aq) + 2e

then a spectator solid is added to the ½

cell, usually Pt(s) or C(s)

•

_{Gas electrodes}

• Gas is supplied as one of the members of

the species list, typically these use an inert

electrode

Electrolytic cells

• identify the similarities and differences between the operation of a voltaic cell and that of an electrolytic cell

• define power supply, and electrolytic cell

• There are two types of electrochemical cells 1. Voltaic cells (chemical to electrical energy)

2. Electrolytic cells (electrical to chemical energy)

• _{When the SOA is above SRA in the ½ reaction}

table, then there is a spontaneous reaction. This process when connected by a wire

Electrolytic cells

• When the SOA is below the SRA electrical

energy has to be supplied to the cell in order for a reaction to occur.

• The process of supplying electrical energy to force a non-spontaneous reaction to occur is electrolysis.

• This process converts electricity into usable chemicals or chemical energy

• An electrolytic cell is the opposite of a voltaic cell.

• In fact when a voltaic cell is recharged it is

Electrochemical cells

• Voltaic cell electrolytic cell

• Spontaneity spontaneous reaction non-spontaneous reaction

• Standard cell positive negative Potential Δ E0

• Cathode positive electrode negative electrode

Strongest oxidizing agent strongest oxidizing agent Undergoes reduction undergoes reduction

• Anode negative electrode positive electrode Strongest reducing agent strongest reducing agent Undergoes an oxidation undergoes oxidation

• Electron movement anode  cathode anode  cathode

Electrolysis

• Electrolysis typically occurs in only one beaker or container.

• It does not have a salt bridge, porous cup or two ½ cells. It typically has a very small species list and often water is either the SOA or SRA

• _{How to describe the electrolysis of}

potassium iodide:

• List: K+(aq) I-(aq) H

2O(l)

OA RA OA/RA

Electrolysis

• ½ reactions: • 2I-  I

2(s) +2e- SRA anode 0.54V

• 2H₂O + 2e-H₂(g) + 2OH-(aq)

SOA-GER-cathode -0.83V

• 2H₂O + 2I- I

2(s) + H2(g) + 2OH

-• Net E0 = E0

cat –E0an

-.83 - .54 = -1.37

• negative sign indicates a non-spontaneous reaction – electricity must be supplied – a

Electrolysis

•

Evidence/ interpretation:

• Gas bubbles from cathode,

– Hydrogen gas is produced

• pH around cathode increases

– OH-(aq) is being produced

• Yellow-brown substance produced at

anode

– _{iodine produced in oxidation ½ rxn}

Workbook questions

Electrolysis suprise

• recognize that predicted reactions do not always occur; e.g., the production of chlorine gas from the electrolysis of brine

• Sometimes the electrolysis of solutions produce unusual results, this is due to an over potential necessary to actually carry out the results

• In the chlor alkali process, the electrolysis of sodium and chloride (actually any metal and

chloride), we find that despite pg 7’s predictions chlorine actually behaves as our SRA. The

Uses for electrolysis

• Electrolysis is often used to plate a metal

on another substance (electroplating).

• Since electrons are being sent to the

cathode that is where the plating takes

place

• By dipping a substance such as a vase or

jewelry in a solution of gold ions and

Uses for electrolysis

• A good example of this is the electrolysis

of impure copper into pure copper

• At the anode we take an impure copper

electrode and remove its electrons

• Cu(blister)



Cu

+2e-

• At the cathode we take pure copper and

attract the cation ions

Types of cells

•

_{Electrochemical cells can be divided into}

three types of cells

1. Primary cells, which is a typical

electrochemical cell and is not

rechargeable

2. Secondary cells, which be recharged by

using electricity to reverse the redox

Types of cells

3. Fuel cells; a fuel cell produces electricity

by the redox reaction of a fuel that is

continually supplied to maintain cell

operation.

•

_{Example; describe how a Hydrogen fuel}

cell works in the presence of NaOH

– H₂, O₂, and OH

-– Anode 2H₂(g) + 4OH-(aq)  4H

2O(l)

Workbook

Faradays equations and ½ cell

stoich

• calculate mass, amounts, current and time in single voltaic and electrolytic cells by applying Faraday’s law and stoichiometry

• Stoichiometry allows us to compare moles

of one substance in a reaction to those of

another substance

• If done on a ½ cell it can help us figure out

the number of moles of electrons

• Ex

• Ag

(aq) + e



Ag(s)

Faradays equations and ½ cell

stoich

• Cu2+(aq) + 2e-  Cu(s)

• 2 mole of e- for 1 mol of Cu

• The amount of charge passing through a circuit is measured in coulombs

• q-It q is coulombs I is amps and T is seconds

• Faraday discovered that 1 mole of e- is 9.65 x 104 C

• Faraday’s constant found on pg 3 of your data book states

e-Faradays equations and ½ cell

stoich

• This allows us to convert between current,

time and moles of

e-• n

=It / F

• Combine this with the ½ reaction of the

metal plating or donating electrons and we

can predict the mass required at the

Faradays equations and ½ cell

stoich

•

What mass of aluminum is produced at

the cathode during the electrolysis of

molten (liquid) AlCl

if 10.0 amps of

current is supplied for 1.0 hours?

1. ½ reactions and n

=It / F

• n_e = (10.0A)(3600s) / 9.65 X104 c/mol

e-• n_e =0.373 mol

Faradays equations and ½ cell

stoich

•

Ex. 8.1 g of copper is produced using

12.7 amps of electricity during the

electrolysis of copper 2

ions. How long

did the electrolysis take?

1. ½ reactions and n

=It / F

examples

• Four electrolytic cells contain CsCl(aq),

AgCl(aq), PbCl

(aq), and AuCl

(aq). If a

current of 5.0 A flows through each of

workbook

Corrosion protection

• We use metals outside all the time, the

problem is the corrode/rust/oxidize Why?

• Metals are R.A. and most react

spontaneously with O

and H

O (oxidizing

agents)

• Gold and silver are okat but most are not

• When a metal rusts, it oxidizes

2e-Corrosion protection

• There are 5 ways to stop corrosion

1. Paint/barriers – don’t let the RA touch the OA

2. Oil/grease – also a barrier used on moving parts

3. Coat the metal with a SRA – this technique is often called chrome or galvanize

4. Sacrificial anode – on large objects you simply attach a SRA periodically around the object, do not use group 1 metals as they explode with water

workbook