Electrochemistry

1
Electrochemistry

• (Applications of Redox)

2
Unit Essential Questions

• What does electrochemistry study?
• How are cell potentials calculated?

3
Review- Redox

• Redox oxidation/reduction reaction.
• What occurs during a redox reaction?
• Changes in oxidation states oxidation losing
  e-, reduction gaining e-.
• OIL- RIG
• Oxidation Involves Loss
• Reduction Involves Gain
• LEO-GER
• Lose Electrons Oxidation
• Gain Electrons Reduction

4
Practice

• Which of the reactions below exhibits oxidation
  and which exhibits reduction?
• Cu2(aq) 2e- ? Cu(s)
• Zn(s) ? Zn2(aq) 2e-
• What is the reducing agent?
• What is the oxidizing agent?

reduction
oxidation
Zn(s)
Cu2
5
Electrochemical Cells

• The two half reactions can be combined from the
  previous example
• Cu2(aq) Zn(s) ? Cu(s) Zn2(aq)
• Notice the e- are not shown because they
  cancelled out on both sides.
• The e- are directly transferred from the copper
  to the zinc.
• These e- can be used to do work if they are
  indirectly transferred between substances!

6
Electrochemical Cells

• Two half reactions are separated in two different
  beakers with a wire connecting them.
• Wire allows the current (e-) to travel between
  beakers.
• Flow of e- through the wire can be used to do
  work.
• Electrochemical cells can be used to produce
  electricity from a redox reaction or can use
  electricity to produce a redox reaction.

7
Section 17.1

• Galvanic (Voltaic) Cells

8
What is a Galvanic Cell?

• A type of electrochemical cell that allows
  chemical energy to be changed into electrical
  energy.
• Chemical energy comes from change in oxidation
  states (redox reaction).
• Wire used in the galvanic cell allows the
  electrical energy to be used for work.

9
Making a Galvanic Cell

• Continue with previous example
• Cu2(aq) Zn(s) ? Cu(s) Zn2(aq)
• Zinc metal is placed in one beaker and copper
  metal is placed in the other.
• These are the electrodes- the part that conducts
  e- in the redox reaction.
• Zinc sulfate solution is added to the zinc metal
  and copper (II) sulfate is added to the copper
  metal.
• Called electrode compartments.

10

• Electrodes are connected with a wire so the
  reaction can start. But nothing happens! Why?
• Charge would build up, and solutions (electrode
  compartments) must remain neutral!

Negativeb/c lose Cu2 ions
Positive b/c form Zn2 ions
Cu
Zn
Cu2 SO4-2
Zn2 SO4-2
11
Galvanic Cell
Salt bridge is added to maintain neutrality.
e-
e-
e-
NO3-
K
Cu
Zn
SO4-2
SO4-2
Cu2
Zn2
12
Salt Bridge

• Any electrolyte can be used to keep charges
  neutral, as long as the ions dont interfere with
  the redox reaction.
• KNO3 used in previous example.
• K ions flow into the copper beaker (to make up
  for the negative charge).
• NO3- ions flow into the zinc beaker (to make up
  for the positive charge).
• Salt bridge allows the circuit to be complete and
  the redox reaction to occur.

13
Galvanic Cell Components

• A voltmeter can be attached to the wire between
  the electrodes to measure the current that can do
  work.
• Oxidation half reaction is always shown in the
  left beaker and the reduction half reaction is
  always shown in the right beaker.
• Anode oxidation half reaction,
• Cathode reduction half reaction
• Also have anode and cathode compartments.

14
Galvanic Cell
e-
Notice the e- travel from the anode to the
cathode!
e-
e-
NO3-
K
Cu
Zn
SO4-2
SO4-2
Cu2
Zn2
anode
cathode
15
Inert Electrodes

• Solid electrodes that do not participate in the
  redox reaction can be used.
• These electrodes are only there to complete the
  circuit/allow the electricity to flow.
• They supply/accept e- as needed.
• Graphite and platinum are often used.

16
Alternative Galvanic Cell
A porous disk is used instead of a salt bridge
to maintain the circuit/neutrality.
Porous disk allows ions to flow between
solutions.
17
Cell Potential

• Oxidizing agent pulls the e- from reducing
  agent.
• The pull/driving force cell potential Ecell, or
  electromotive force (emf).
• Unit volts (V)
• 1 volt 1 joule of work/coulomb of charge
• Measured with a voltmeter

18
Section 1 Homework
Pg. 830 15, 25
19
Section 17.2

• Standard Reduction Potential

20
Standard Reduction Potentials

• Show the number of volts produced from a half
  reaction.
• All values are based on the reduction half
  reaction (thus the name standard reduction
  potential).
• This value is for substances in their standard
  states 25C(298K), 1M, 1atm, and pure solid
  electrodes.
• Standard hydrogen electrode- all other values
  based off of H!

21
AP Practice Question

• 2BrO3-(aq) 12H(aq) 10e- ? Br2(aq) H2O (l)
• Which of the following statements is correct for
  the above reaction?
• The BrO3- is oxidized at the anode.
• Br goes from a -1 to a 0 oxidation state.
• Br2 is oxidized at the anode.
• The BrO3- is reduced at the cathode.

22
Standard Hydrogen Electrode

• This is the reference all other
  oxidations/reductions are compared to.
• Eº 0
• º indicates standard states of 25ºC, 1 atm, 1
  M solutions.

H2 in
H Cl-
1 M HCl
23
Since H 0, this value is assigned to the
oxidation of Zn.
0.76V
H2 in
Anode
Cathode
H Cl-
Zn2 SO4-2
1 M HCl
1 M ZnSO4
24
Cell Potential

• Zn(s) Cu2 (aq) Zn2(aq) Cu(s)
• The total cell potential is the sum of the
  potential at each electrode.
• E ºcell E ºZn Zn2 E ºCu2 Cu
• Look up reduction potentials in a table.
• Since potentials are given in terms of reduction,
  the sign is switched for the oxidation reaction
  since the reaction is the reverse of reduction.

25
Cell Potential

• Zn(s) Cu2 (aq) Zn2(aq) Cu(s)
• Reduction potentials are as follows
• Zn2 2e- ? Zn E -0.76V
• Cu2 2e- ? Cu E 0.34V
• But Zn is not being reduced- its being oxidized!
  So the sign must be changed
• E ºcell 0.76V 0.34V 1.10V
• Alternate formula E ºcell E ºcathode - E
  ºanode
• Cell potentials are always gt 0 because they run
  spontaneously in the direction that produces a
  positive potential.

26
Cell Potential

• Note that even if a half reaction must be
  multiplied by an integer to keep e- transfer
  equal, the standard reduction potential
  associated with that half reaction IS NOT
  multiplied by that same integer!
• This is because the standard reduction potential
  depends upon the reduction that is occurring, not
  how many times it occurs.
• Like how the density of a sample of a substance
  is always the same, regardless of the size. It
  only depends on identity!

27
Practice Cell Potential

• Determine the cell potential for a galvanic cell
  based on the redox reaction.
• Cu(s) Fe3(aq) Cu2(aq) Fe2(aq)
• Fe3(aq) e- Fe2(aq) Eº 0.77 V
• Cu2(aq)2e- Cu(s) Eº 0.34 V
• Eºcell 0.43V

28
Cell/Line Notation

• Shorthand way of representing galvanic cells.
  Format
• solid½Aqueous½½Aqueous½solid
• Anode on the left, cathode on the right.
• Single line separates different phases at each
  electrode.
• Double line indicates porous disk or salt bridge.
• Concentrations in ( ) may be added after the
  aqueous species, if known.

29
Cell/Line Notation

• Examples
• (1) Mg(s)½Mg2(aq)½½Al3(aq)½Al(s)
• (2) Zn(s)½Zn2(1M)½½Cu2(1M)½Cu(s)
• If an inert electrode is used
• Ag(aq) Fe2(aq) ? Fe3(aq) Ag(s)
• Pt electrode used for Fe2 or Fe3
• Pt(s)½Fe2(aq),Fe3(aq)½½ Ag(aq)½Ag(s)

30
Practice Cell/Line Notation

• Given the cell reaction below, write the
  cell/line notation.
• Ni(s) 2Ag(aq) ? Ni2(aq) 2Ag(s)
• Ni(s)½Ni2(aq)½½Ag (aq)½Ag

31
Summing up Galvanic Cells

• Reaction always runs spontaneously in the
  direction that produces a positive cell potential
  (E ºcell gt 0).
• Four things for a complete description
• Cell Potential
• Direction of flow
• Designation of anode and cathode
• Species present in all components- electrodes and
  ions.

32
Section 2 Homework
I will do 31, part 25, with you. Pg. 830 27,
31
33
Review Balancing Redox Rxns. Using Half Rxn.
Method

• Write separate half reactions.
• For each half reaction balance all species except
  H and O.
• Balance O by adding H2O to one side.
• Balance H by adding H to one side.
• Balance charge by adding e- to the more positive
  side.

34
Review Balancing Redox Rxns. Using Half Rxn.
Method

• Multiply equations by a number to make electrons
  equal.
• Add equations together and cancel identical
  species. Reduce coefficients to smallest whole
  numbers.
• Check that charges and elements are balanced.

35
In Basic Solution

• Add enough OH- to both sides to neutralize the
  H.
• Any H and OH- on the same side form water.
  Cancel out any H2Os on both sides.
• Simplify coefficients, if necessary.

36
AP Practice Question

• What is the coefficient of H when the following
  reaction is balanced?
• MnO4-(aq) H(aq) C2O4-2(aq) ?
  Mn2(aq) H2O(l) CO2(g)
• 16
• 2
• 8
• 5

37
Homework

• Balance the following equation in acidic and
  basic solution
• NO3- Mn ? NO Mn2
• Balanced equations
• 3Mn 8H 2NO3- ? 2NO 4H2O 3Mn2
• 3Mn 4H2O 2NO3- ? 2NO 8OH- 3Mn2

38
Section 17.3

• Cell Potential, Electrical Work, and Free Energy

39
Linking Cell Potential to ?G

• Cell potential is directly related to the
  difference in free energy between reactants and
  products.
• This is shown in the following equation
• ?G -nFE
• n moles of e- in redox/half reaction
• F Faradays constant 96,485C/mole-
• E cell potential (V J/C)
• Units of G J

Coulomb unit of electric charge
40
Linking Cell Potential to ?G

• ?G -nFE
• Why negative?
• Galvanic cells are spontaneous!
• If ?G was positive, the redox reaction would be
  nonspontaneous.
• ?G can be calculated for half reactions and for
  entire redox reactions.

41
Sample Calculation

• Calculate ?G for the reaction
• Cu2(aq) Fe(s) ? Cu(s) Fe2(aq)
• Look up half reactions and determine individual E
  values. Then calculate E for the cell.
• E 0.78V 0.78J/C
• Make sure e- are correct!
• ?G -(2mole-)(96,485C)(0.78J) -1.5x105J
• Process is spontaneous.

mole- C
42
Practice Problem

• Consider the following reaction
• Mn2(aq) IO4-(aq) ? IO3-(aq) MnO4-(aq)
• Calculate the value of E cell and ?G. Be sure to
  balance the e- (you can use the half reactions
  and balance them from there).
• E cell 0.09V
• ?G -90kJ

43
Section 3 Homework
Pg. 831 37
44
Section 17.4

• Dependence of Cell Potential on Concentration

45
Qualitative Understanding

• The following reaction is under standard
  conditions
• Cu(s) 2Ce4(aq) ? Cu2(aq) 2Ce3(aq)
• What if Ce4 was greater than 1.0M?
• Use LeChateliers principle!
• This shifts the rxn. forward, which means more
  products are formed and therefore more e- are
  flowing/transferring to allow them to form.
• Cell potential increases.

46
Another Way of Looking at It

• Cu(s) 2Ce4(aq) ? Cu2(aq) 2Ce3(aq)
• What if Ce4 was greater than 1.0M?
• Use LeChateliers principle!
• The rxn. shifting more towards the products means
  the rxn. is more spontaneous, so ?G increases.
• Recall that ?G -nFEcell, so if -?G is
  increasing, so must Ecell because n and F are
  constant for this rxn.

47
Practice

• For the cell reaction
• 2Al(s) 3Mn2(aq) ? 2Al3(aq) 3Mn(s)
• Ecell 0.48V
• Predict whether Ecell is larger or smaller than
  Ecell for the following cases.
• Al3 2.0M, Mn2 1.0M
• Al3 1.0M, Mn2 3.0M

smaller
larger
48
Quantitative Understanding

• We will not cover the quantitative side of how
  cell potential changes with concentration, but I
  want to mention the Nernst Equation
• Ecell Ecell (RT)lnQ Ecell (0.0592)logQ
• Lets you calculate cell potential under
  non-standard conditions.
• Not tested on the AP exam.

nF
n
49
Homework

• Pg. 832 51

50
Section 17.7

• Electrolysis

51
Electrolytic Cells

• Electrolytic cells use electrolysis and are the
  opposite of galvanic cells.
• Reactions are not spontaneous.
• Electrolysis a voltage bigger than the cell
  potential is applied to the cell to force the
  non spontaneous redox reaction to occur.
• Can be used to decompose compounds.
• Water can be broken into hydrogen oxygen.
• Can also be used for (electro)plating.
• Forces metal ions in solution to plate out on
  electrodes in their solid form.

52
Galvanic Cell
Cell potential 1.10V
1.10
e-
e-
Zn
Cu
1.0 M Cu2
1.0 M Zn2
Cathode
Anode
53
Voltage applied gt 1.10V
Electrolytic Cell
A battery gt1.10V
e-
e-
?
?
Zn
Cu
1.0 M Zn2
1.0 M Cu2
Cathode
Anode
54
Electrolytic Cells

• Relationship exists between current, charge, and
  time I q/t
• I current (A, amp)
• q charge (C)
• t time (s)
• This formula can be used in conjunction with
  other conversion factors to solve for various
  items with respect to electrolytic cells.

55
Electrolytic Cells Stoichiometry

• The previous formula and stoichiometry can be
  used to determine the amount of chemical change
  that occurs when current is applied for a certain
  amount of time.
• Answers the following questions
• How much of a substance will be produced?
• How long will it take?
• How much current is needed?

56
Electrolytic Cells Stoichiometry

• No set way to solve these problems each time! In
  addition to the formula, the following conversion
  factors may be used
• 96,485C/mol e- (this is 1 Faraday)
• Molar mass
• Moles of electrons to moles of other species
  involved in the redox rxn.
• Make sure moles in the half rxn. are correct!

57
Example

• If liquid titanium (IV) chloride (acidified with
  HCl) is electrolyzed by a current of 1.000amp for
  2.000h, how many grams of titanium will be
  produced?
• Have amps time ? can use both to find q from
  previously discussed formula
• I q/t ? 1.000A q/2.000h ? convert h to s
• 1.000A q/7,200.s
• q 7,200.A?s ? q 7,200.C ? s ? q 7,200.C
• Now use stoichiometry to solve for grams!

s
58
Example Cont.

• If liquid titanium (IV) chloride (acidified with
  HCl) is electrolyzed by a current of 1.000amp for
  2.000h, how many grams of titanium will be
  produced?
• Use Faradays constant to get moles of e-, write
  balanced half rxn. to get moles of e- and moles
  of Ti Ti4 ? Ti 4e-
• 7,200.C x 1mole- x 1molTi x 47.88gTi

96,485C
4mole-
1molTi
0.8932g
59
Practice Problem

• What mass of copper is plated out when a current
  of 10.0amps is passed for 30.0min through a
  solution containing Cu2?
• Solve for q 10.0A q/1,800s ? q 18,000C
• 18,000C (1mole-)(1molCu)(63.55gCu)
• 96,485C 2mole- 1molCu
• 5.93gCu

60
Homework

• In class- 77(a)
• Given mass and I from mass we can find charge,
  q. Then use formula to solve for time.
• Pg. 834 77(bc),79(ab)