Electrochemistry

Experiment 12

Oxidation – Reduction Reactions

• Consider the reaction of Copper wire and AgNO3(aq)

AgNO3(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 Cu2+(aq)

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)

Cu(s) Cu2+(aq) + 2e-

2Ag+(aq) + 2e- + Cu(s) 2Ag(s)+ Cu2+

(aq) + 2e-

lose electrons = oxidation

gain electrons = reduction

Redox Cu/Ag

Cu

Ag+ Ag+

E

electron flow

Redox Cu/Ag

Cu2+

Ag Ag

E

ΔE = e.V

e = charge on an electronV = Voltage in a electrochemical cell

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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

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Electric Current Flowing Directly Between Atoms

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Electric Current Flowing Indirectly Between Atoms

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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

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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

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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

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Voltaic Cell

the salt bridge is required to complete the circuit and maintain charge balance

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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 = 6.242 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

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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

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Standard Reduction Potential

• a half-reaction with a strong tendency to occur has a large + half-cell potential

• when two half-cells are connected, the electrons will flow so that the half-reaction with the stronger tendency will occur

• we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction

• we select as a standard half-reaction the reduction of H+ to H2 under standard conditions, which we assign a potential difference = 0 V

– standard hydrogen electrode, SHE

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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

• ΔGocell=-nFΔE°cell

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Electrochemical Cell Summary

salt bridgee-

anode

Zn (s)--> Zn2+ (aq)+ 2e-

cathode

Cu2+(aq)+ 2e- --> Cu(s)

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).

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 is related to the free energy of the reaction according to the relation Gcell = -nFEcell

The cell potential can be calculated knowing the standard reduction potentials. These can be used to find Eo

red for the reaction at the cathode, and Eo

ox (= - Eored). Then Eo

cell = Eoox+ Eo

red Zn2+(aq) + 2e- --> Zn(s) -0.76V

Cu2+(aq) + 2e- --> Cu(s) Ered=0.34VZn(s) --> Zn2+(aq) + 2e- Eox= 0.76V

Ecell=1.1 VEcell = 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

Trial electrodes Trial electrodes

1 Cu/Zn 4 Zn/Pb

2 Cu/Pb 5 Zn/Ni

3 Cu/Ni 6 Pb/Ni