Electrochemistry Experiment 12

Oxidation – Reduction Reactions Consider the reaction of Copper wire and AgNO 3 (aq) AgNO 3 (aq) Ag(s) Cu(s)

Oxidation – Reduction Reactions If you leave the reaction a long time the solution goes blue! The blue is due to Cu 2+ (aq)

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

Redox Cu/Ag Cu Ag + E electron flow

Redox Cu/Ag Cu 2 + Ag E ΔE = e.V 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

Electrochemical/Voltaic Cell Cu/Ag 2Ag + (aq) + 2e -  2Ag (s) Cu (s)  Cu 2+ (aq) + 2e -

8 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

9 Electric Current Flowing Directly Between Atoms

Tro, Chemistry: A Molecular Approach10 Electric Current Flowing Indirectly Between Atoms

Tro, Chemistry: A Molecular Approach11 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 Approach12 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 Approach13 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 Approach14 Voltaic Cell the salt bridge is required to complete the circuit and maintain charge balance

Tro, Chemistry: A Molecular Approach15 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 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 Approach16 Cell Potential the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential 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 – 25°C, 1 atm for gases, 1 M concentration of solution – sum of the cell potentials for the half-reactions

17 Standard Reduction Potential when two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency to reduce will reduce we cannot measure the absolute tendency of a half- reaction to reduce, we can only measure it relative to another half-reaction we select as a standard half-reaction the reduction of H + to H 2 under standard conditions, which we assign a potential difference = 0 V 2H + (aq) +2e -  H 2 (g) – standard hydrogen electrode, SHE a half-reaction with a strong tendency to occur has a large +ve half-cell potential

Tro, Chemistry: A Molecular Approach18

Tro, Chemistry: A Molecular Approach19 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, even if you need to multiply the half-reactions to balance the equation ΔG o cell =-nFΔE° cell

Tro, Chemistry: A Molecular Approach21 Electrochemical Cell Summary salt bridge e-e- anode Zn (s)--> Zn 2+ (aq)+ 2e - cathode Cu 2+ (aq)+ 2e - --> Cu(s ) The differing stability of reactants, (Zn(s), Cu 2+ (aq)), and products (Zn 2+, and Cu(s)), creates a potential energy gradient through which the charges migrate (from high energy to low). This manifests as a potential difference E cell, across the electrodes. Where -qE cell is the change in potential energy when an amount of negative charge (-q) passes from the anode to the cathode The cell potential is related to the free energy of the reaction according to the relation  G cell = -nFE cell The cell potential can be calculated knowing the standard reduction potentials. These can be used to find E o red for the reaction at the cathode, and E o ox (= - E o red ). Then E o cell = E o ox + E o red Zn 2+ (aq) + 2e - --> Zn(s)-0.76V Cu 2+ (aq) + 2e - --> Cu(s) E red =0.34V Zn(s) --> Zn 2+ (aq) + 2e - E ox = 0.76V E cell =1.1 V E cell = 0.76V+0.34V = 1.1V

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

Tonight complete Parts B, C and D

Part B: Effect of Concentration on Cell Potential Δ G = Δ G° + RT ln Q Δ E = Δ E° - (0.0592/n) log Q at 25°C (since Δ G=-nF  Δ E) use to calculate E when concentrations not 1 M n = number of electrons in balanced redox reaction

Part B M Copper (II) sulfate0.1 M Zinc (II) sulfate

26 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 E tot, called the overvoltage

27 Battery Voltage > 1.1 V to make this spontaneous

28 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 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)  Cu 2+ (aq) + 2e = Cathode: Cu 2+ (aq) + 2e =  Cu(s)

Part C To Power Amplifier Nickel Plate 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 D M cathode 1M Copper Sulfate Calculate the mass of Cu, M cu, 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 M Cu = g/mol * Q / (2F)

30 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 H 2 2 H 2 O + 2 e -1  H OH -1 E° = stand. cond. E° = pH 7 possible anode reactions – oxidation of anion to element – oxidation of H 2 O to O 2 2 H 2 O  O 2 + 4e H +1 E° = stand. cond. E° = pH 7 – oxidation of electrode particularly Cu graphite doesn’t oxidize half-reactions that lead to least negative E tot will occur – unless overvoltage changes the conditions

31 Part D: Electrolysis of NaI (aq) with Inert Electrodes possible oxidations 2 I -1  I e -1 E° = −0.54 v 2 H 2 O  O 2 + 4e H +1 E° = −0.82 v possible reductions Na e -1  Na 0 E° = −2.71 v 2 H 2 O + 2 e -1  H OH -1 E° = −0.41 v possible oxidations (anode –ve terminal) 2 I -1  I e -1 E° = −0.54 v 2 H 2 O  O 2 + 4e H +1 E° = −0.82 v possible oxidations (anode –ve terminal) 2 I -1  I e -1 E° = −0.54 v 2 H 2 O  O 2 + 4e H +1 E° = −0.82 v possible reductions (cathode +ve terminal) Na e -1  Na 0 E° = −2.71 v 2 H 2 O + 2 e -1  H 2(g) + 2 OH -1 E° = −0.41 v possible reductions (cathode +ve terminal) Na e -1  Na 0 E° = −2.71 v 2 H 2 O + 2 e -1  H 2(g) + 2 OH -1 E° = −0.41 v overall reaction 2 I − (aq) + 2 H 2 O (l)  I 2(aq) + H 2(g) + 2 OH -1 (aq) overall reaction 2 I − (aq) + 2 H 2 O (l)  I 2(aq) + H 2(g) + 2 OH -1 (aq) Write down ions/molecules/metals present and don’t forget water Here Na +, I -, H 2 O are what is present (the electrodes here are inert) Write down the reduction reactions involving these species I 2 (aq)+2e-  2I - (aq)E red = 0.54V 2 H 2 O(l)  O 2(g) + 4e H +1 (aq)E red = -0.82V H 2 (g)+OH-(aq)  2H 2 O(l) + 2e - E red = 0.41V Na + (aq) + e-  Na(s)E red = -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 E cell = -0.54V – 0.41V = -0.95V The battery needs to at least be 1V to make this happen Brown liquid bubbles

Part D copper(II) bromide (aq) + 3 drops phenolphthalein 8 mL of 0.5 M KI + 3 drops phenolphthalein NaCl(aq) + 3 drops phenolphthalein 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 =