1. Electrochemistry
Experiment 12

2. Oxidation – Reduction Reactions
Consider the reaction of Copper wire and AgNO3(aq)
AgNO3(aq)
Cu(s)
Ag(s)

3. Oxidation – Reduction Reactions
If you leave the reaction a long time the solution goes blue!
The blue is due to Cu2+(aq)

4. Oxidation-Reduction Reactions
So when we mix Ag+(aq) with Cu(s) we get Ag(s) and Cu2+(aq)
Ag+(aq) + 1e-  Ag(s)
Cu(s)  Cu2+(aq) + 2e-
The electrons gained by Ag+ must come from the Cu2+
Can’t have reduction without oxidation (redox)
Each Cu can reduce 2 Ag+
2Ag+(aq) + 2e-  2Ag(s)
2Ag+(aq) + 2e- + Cu(s) 2Ag(s)+ Cu2+(aq) + 2e-
gain electrons = reduction
lose electrons = oxidation

5. Redox Cu/Ag E Cu
electron flow
Ag+
Ag+

6. Redox Cu/Ag E Cu2+ ΔE = e.V Ag Ag e = charge on an electron
V = Voltage in a electrochemical cell
If we could separate the two reactions we could use the energy gained by the e to do work E
Cu2+
ΔE = e.V Ag Ag

7. Electrochemical/Voltaic Cell Cu/Ag
Where reduction
happens
Where oxidation happens
Cu(s)  Cu2+(aq) + 2e-
2Ag+(aq) + 2e-  2Ag(s)

8. Tro, Chemistry: A Molecular Approach
Cell Potential Eocell
the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential = Ecell
the cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode
the cell potential under standard conditions is called the standard emf, E°cell = Eoox + Eored
25°C, 1 atm for gases, 1 M concentration of solution
sum of the cell potentials for the half-reactions
If we could independently measure Eoox and Eored (the half cell potentials) separately we would have a way to predict what redox reactions would happen and how much work the electrons could do
Tro, Chemistry: A Molecular Approach

9. Measuring the Tendency to Reduce: Standard Reduction Potential
when two half-cells are connected, one side will oxidize and one side will reduce, but which will do what?
clearly the electrons will flow so that the half-reaction with the stronger tendency to reduce will reduce, making the other half-cell reaction oxidize
we cannot measure the absolute tendency of a half-reaction to reduce, we can only measure it relative to another half-reaction
To measure the tendency to reduce we measure the E (the voltage) in a voltaic cell where one of the electrodes is a standard hydrogen electrode (see right)
We assign a potential difference = 0.0 V to the following reaction
2H+(aq) +2e-  H2(g) Eored = 0.0V
E°cell = Eoox + Eored = Eoox V = Eoox
In so doing scientists have come up with the following standard reduction potential tables.

10. Tro, Chemistry: A Molecular Approach

11. The measured voltage can then be placed in a table like this one, where we right the voltage for the reduction reaction

12. Tro, Chemistry: A Molecular Approach
Half-Cell Potentials
SHE reduction potential is defined to be exactly 0 V
half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red
half-reactions with a stronger tendency toward oxidation than the SHE have a - value for E°red
ΔE°cell = E°oxidation + E°reduction
E°oxidation = -E°reduction
when adding E° values for the half-cells, do not multiply the half-cell E° values (voltage is the energy per unit charge), even if you need to multiply the half-reactions to balance the equation
ΔGocell=-nFΔE°cell
Since ΔGocell < 0 to be spontaneous ΔE°cell > 0 for a redox reaction to be spontaneous
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13. red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
Calculate E°cell for the reaction at 25°C Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l)
Separate the reaction into the oxidation and reduction half-reactions
ox: Al(s)  Al3+(aq) + 3 e−
red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l)
find the Eo for each half-reaction from the standard reduction table. If you need the oxidation potential of one Eoox=-Eored and sum to get Eocell
Eoox = −Eored = V
Eored = V
Eocell = (+1.66 V) + (+0.96 V) = V

14. Redox Reactions & Current
redox reactions involve the transfer of electrons from one substance to another
therefore, redox reactions have the potential to generate an electric current
in order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring

15. Electric Current Flowing Directly Between Atoms

16. Voltaic (Electrochemical) Cell
Tro, Chemistry: A Molecular Approach

17. Electrochemical Cells
electrochemistry is the study of redox reactions that produce or require an electric current
the conversion between chemical energy and electrical energy is carried out in an electrochemical cell
spontaneous redox reactions take place in a voltaic cell
aka galvanic cells
nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy
Tro, Chemistry: A Molecular Approach

18. Electrochemical Cells
oxidation and reduction reactions kept separate
half-cells
electron flow through a wire along with ion flow through a solution constitutes an electric circuit
requires a conductive solid (metal or graphite) electrode to allow the transfer of electrons
through external circuit
ion exchange between the two halves of the system
electrolyte
Tro, Chemistry: A Molecular Approach

19. Tro, Chemistry: A Molecular Approach
Electrodes
Anode (donates electrons to the cathode)
electrode where oxidation occurs
anions attracted to it
connected to positive end of battery in electrolytic cell
loses weight in electrolytic cell
Cathode (attracts electrons from the anode)
electrode where reduction occurs
cations attracted to it
connected to negative end of battery in electrolytic cell
gains weight in electrolytic cell
electrode where plating takes place in electroplating
Tro, Chemistry: A Molecular Approach

20. Tro, Chemistry: A Molecular Approach
Voltaic Cell
the salt bridge is required to complete the circuit and maintain charge balance
Tro, Chemistry: A Molecular Approach

21. Tro, Chemistry: A Molecular Approach
Current and Voltage
the number of electrons that flow through the system per second is the current
unit = Ampere
1 A of current = 1 Coulomb of charge flowing by each second
1 A = x 1018 electrons/second
Electrode surface area dictates the number of electrons that can flow
the difference in potential energy between the reactants and products is the potential difference (the potential for an electric field to cause an electrical current)
unit = Volt
1 V of force = 1 J of energy/Coulomb of charge
the voltage needed to drive electrons through the external circuit
amount of force pushing the electrons through the wire is called the electromotive force, emf
Tro, Chemistry: A Molecular Approach

22. Electrochemical Cell Summary
This manifests as a potential difference Ecell, across the electrodes. Where -qEcell is the change in potential energy when an amount of negative charge (-q) passes from the anode to the cathode
The cell potential can be calculated knowing the standard reduction potentials. These can be used to find Eored for the reaction at the cathode, and Eoox (= - Eored). Then Eocell = Eoox+ Eored
The differing stability of reactants, (Zn(s), Cu2+(aq)), and products (Zn2+, and Cu(s)), creates a potential energy gradient through which the charges migrate (from high energy to low).
The cell potential is related to the free energy of the reaction according to the relation DGcell = -nFEcell
Zn(s) --> Zn2+(aq) + 2e- Eox= 0.76V
Zn2+(aq) + 2e- --> Zn(s) V
Cu2+(aq) + 2e- --> Cu(s) Ered=0.34V
Ecell = 0.76V+0.34V = 1.1V
Ecell=1.1 V e-
anode
Zn (s)--> Zn2+ (aq)+ 2e-
salt bridge
cathode
Cu2+(aq)+ 2e- --> Cu(s)
Tro, Chemistry: A Molecular Approach

23. Tonight
Construction of Voltaic Cells and Measurement of Cell Potentials
Use the corresponding 0.1 M metal sulfate of the same metal as the electrode in the half cell
Construct a salt bridge
Measure the voltage
Trial
electrodes 1
Cu/Zn 4
Zn/Pb 2
Cu/Pb 5
Zn/Ni 3
Cu/Ni 6
Pb/Ni

24. Tonight complete Parts B, C and D

25. Part B: Effect of Concentration on Cell Potential
As long as the concentrations of the ionic solutions in both cells is the same and it is measured at room temperature the measured 𝐸 𝑐𝑒𝑙𝑙 = 𝐸 𝑐𝑒𝑙𝑙 𝑜
When non-standard conditions are present then we must use the Nernst Equation
ΔG = ΔG° + RT ln Q
ΔE = ΔE° - (0.0592/n) log Q at 25°C (since ΔG=-nF ΔE)
use to calculate E when concentrations not 1 M
n = number of electrons in balanced redox reaction

26. Part B 0.005 M Copper (II) sulfate 0.1 M Zinc (II) sulfate
𝐶𝑢 (𝑎𝑞) 2+ + 𝑍𝑛 (𝑠) +2 𝑒 − → 𝐶𝑢 (𝑠) + 𝑍𝑛 (𝑎𝑞) 𝑒 −
Δ 𝐸 𝑐𝑒𝑙𝑙 =Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 𝑙𝑜𝑔 𝑍𝑛 𝑎𝑞 𝐶𝑢 𝑎𝑞 =Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 𝑙𝑜𝑔 0.1𝑀 𝑀

27. Making non-spontaneous Reactions happen:Electrolysis
uses electrical energy to overcome the energy barrier and cause a non-spontaneous redox reaction to happen
must be DC source
the + terminal of the battery: anode (oxidation happens here)
Anions attacted to the anode and release electrons to the anode and are oxidized
the - terminal of the battery: cathode (reduction happens here)
Cations attracted to the cathode and pick up electrons from the cathode and are reduced
some electrolysis reactions require more voltage than Etot, called the overvoltage

28. Battery Voltage > 1.1 V to make this spontaneous
Δ 𝐸 𝑐𝑒𝑙𝑙 𝑜 V V = -1.1V

29. Part C: Electroplating and Refining
In electroplating, the work piece is the cathode.
Cations are reduced at cathode and plate to the surface of the work piece.
The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution
This process can be used to purify (refine) metals
Anode: Cu(s)  Cu2+(aq) + 2e=
Cathode: Cu2+(aq) + 2e=  Cu(s)

30. Part C
Calculate the mass of Cu, Mcu , deposited using Faraday’s law
Charge flowing through amp / F in 600s = moles Cu/2
Charge flowing through amp Q = area under I vs t curve
F = charge of one mole of protons = C/mol
MCu=  g/mol * Q / (2F)
To Power Amplifier
Nickel Plate
1M Copper Sulfate
Weigh the dry electrodes attach the Nickel to which terminal?
Connect Power Amplifier to the Science Workshop interface Channel A
Open Electroplating
Run for 600s (10 mins)
Plots current (amperes) vs time (s)
Dry and reweigh the electrodes and calculate the mass change DMcathode

31. Part D: Electrolysis of Aqueous Solutions
Complicated by more than one possible oxidation and reduction
possible cathode reactions
reduction of cation to metal
reduction of water to H2
2 H2O + 2 e-1  H2 + 2 OH-1 E° = stand. cond.
E° = pH 7
possible anode reactions
oxidation of anion to element
oxidation of H2O to O2
2 H2O  O2 + 4e-1 + 4H+1 E° = stand. cond.
E° = pH 7
oxidation of electrode
particularly Cu (where Cu(s) is oxidized to Cu2+(aq))
graphite doesn’t oxidize (inert electrode)
half-reactions that lead to least negative Etot will occur
unless overvoltage changes the conditions

32. Part D: Electrolysis of NaI(aq) with Inert Electrodes
Write down ions/molecules/metals present and don’t forget water
Here Na+, I-, H2O are what is present (the electrodes here are inert)
Write down the reduction reactions involving these species
I2(aq)+2e-  2I-(aq) Ered = 0.54V
O2(g) + 4e-1 + 4H+1(aq) 2 H2O(l) Ered = V
2 H2O + 2 e-1  H2 + 2 OH-1 Ered = 0.41V
Na+(aq) + e-  Na(s) Ered = -2.71V
Rearrange the equations so the reactants are on the left (reverse the sign of E if you flip the equation – these will be the oxidation reactions
possible oxidations (anode +ve terminal)
2 I-1  I2 + 2 e-1 E° = −0.54 v
2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v
possible oxidations
2 I-1  I2 + 2 e-1 E° = −0.54 v
2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v
possible reductions (cathode -ve terminal)
Na+1 + 1e-1  Na0 E° = −2.71 v
2 H2O + 2 e-1  H2(g) + 2 OH-1 E° = −0.41 v
possible reductions
Na+1 + 1e-1  Na0 E° = −2.71 v
2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v
bubbles
Brown liquid
overall reaction
2 I−(aq) + 2 H2O(l)  I2(aq) + H2(g) + 2 OH-1(aq)
Ecell = -0.54V – 0.41V = -0.95V
The battery needs to at least be 1V to make this happen

33. Part D
Before you do this experiment first figure out what should happen at the anode and cathode
Do the experiment and write out what you see at each electrode
Phenolphthalein will detect the presence of OH=
8 mL of 0.5 M KI + 3 drops phenolphthalein
NaCl(aq) + 3 drops phenolphthalein
copper(II) bromide (aq)
+ 3 drops phenolphthalein