Electrochemistry
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Electrochemistry. Oxidation-Reduction Reactions. Oxidation and Reduction Metal undergoes corrosion, it loses electrons to form cations:Ca( s ) +2H + ( aq )  Ca 2+ ( aq ) + H 2 ( g ) Voltaic Cell:Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

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Electrochemistry

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Electrochemistry

Electrochemistry


Electrochemistry

Oxidation-Reduction Reactions

Oxidation and Reduction

  • Metal undergoes corrosion, it loses electrons to form cations:Ca(s) +2H+(aq)  Ca2+(aq) + H2(g)

  • Voltaic Cell:Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

  • Oxidized: atom, molecule, or ion becomes more positively charged.

    • Oxidation is the loss of electrons.

  • Reduced: atom, molecule, or ion becomes less positively charged.

    • Reduction is the gain of electrons.


Electrochemistry

  • Metal undergoes corrosion, it loses electrons to form cations:Ca(s) +2H+(aq)  Ca2+(aq) + H2(g)

  • Voltaic Cell:Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)


Electrochemistry

Oxidation-Reduction Reactions

Oxidation and Reduction


Electrochemistry

Oxidation-Reduction Reactions

Oxidation Numbers

  • Oxidation numbers are assigned by a series of rules:

    • If the atom is in its elemental form, the oxidation number is zero. e.g., Cl2, H2, P4.

    • For a monatomic ion, the charge on the ion is the oxidation state.

    • Nonmetal usually have negative oxidation numbers:

      (a)Oxidation number of O is usually –2. The peroxide ion, O22-, has oxygen with an oxidation number of –1.

      (b)Oxidation number of H is +1 when bonded to nonmetals and –1 when bonded to metals.

      (c)Halogens generally -1, oxid. # of F is always –1.

      4.The sum of the oxidation numbers for the atom is the charge on the molecule (zero for a neutral molecule).


Electrochemistry

Alkali metals

The Periodic Table

Noble Gases

Alkaline Earth metals

Halogens

Chalcogens

Transition Metals

Inner Transition Metals


Electrochemistry

Ions and Ionic Compounds

Predicting Ionic Charge


Electrochemistry

Common (Type II) Cations


Electrochemistry

Oxidation-Reduction Reactions

Oxidation Numbers

1. Determine the oxidation numbers (oxidation states) on the Sulfur atom for:(a) H2S(b) S8(c) SCl2

(d) Na2SO3(e) SO42-

  • Determine o.s. for the Redox reaction in Ni-Cd Batteries:

    Cd(s) + NiO2(s) + 2H2O(l)  Cd(OH)2(s) + Ni(OH)2(s)

  • Determine o.s. for the following Redox reaction:

    Al(s) + MnO4-(aq) + 2H2O(l)  Al(OH)4-(aq) + MnO2(s)

HW-Answers


Electrochemistry

Balancing Redox Reactions

  • Law of conservation of mass: the amount of each element present at the beginning of the reaction must be present at the end.

  • Conservation of charge: electrons are not lost in a chemical reaction.

    Half Reactions

  • Half-reactions are a convenient way of separating oxidation and reduction reactions.


Electrochemistry

Balancing Redox Reactions

Half Reactions

  • The half-reactions for

    Sn2+(aq) + 2Fe3+(aq)  Sn4+(aq) + 2Fe2+(aq)

    are ?

  • Oxidation: electrons are products.

  • Reduction: electrons are reagents.


Electrochemistry

Balancing Redox Reactions

Balancing Equations by the Method of Half Reactions

1. Write down the two half reactions.

2. Balance each half reaction:

a. First with elements other than H and O.

b. Then balance O by adding water.

c. Then balance H by adding H+.

d. Finish by balancing charge by adding electrons.

3. Multiply each half reaction to make the number of electrons equal.

4. Add the reactions and simplify.

5. Check for balanced atoms and charges!

Balance: MnO4-(aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (Acidic)


Electrochemistry

Balance: MnO4- (aq) + C2O42-(aq)  Mn2+(aq) + CO2(g) (Acidic)


Electrochemistry

Balancing Redox Reactions

Balancing Equations for Reactions Occurring in Basic Solution

  • Use OH- and H2O rather than H+ and H2O.

  • Follow same method as in Acidic Solution, but OH- is added to “neutralize” the H+ used.

  • Consider:

    CN-(aq) + MnO4-(aq)  CNO-(aq) + MnO2(s) [ Basic Solution ]


Electrochemistry

CN-(aq) + MnO4-(aq)  CNO-(aq) + MnO2(s) [ Basic Solution ]

Solution Key


Electrochemistry

Voltaic Cells

  • Energy released in a spontaneous redox reaction is used to perform electrical work.

  • Voltaic or galvanic cells are devices in which electron transfer occurs via an external circuit.

  • Voltaic cells are spontaneous.

  • If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves by forming Zn2+.

Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)


Electrochemistry

Cell EMF

  • Flow of electrons from anode to cathode is spontaneous.

  • Electrons flow from anode to cathode because the cathode has a lower electrical potential energy than the anode.

  • Potential difference: difference in electrical potential. Measured in volts.

  • One volt is the potential difference required to impart one joule of energy to a charge of one coulomb:


Electrochemistry

Cell EMF

  • Electromotive force (emf) is the force required to push electrons through the external circuit.

  • Cell potential: Ecell is the emf of a cell.

  • For 1M solutions at 25 C (standard conditions), the standard emf (standard cell potential) is called Ecell.


Electrochemistry

Cell EMF

Standard Reduction (Half-Cell) Potentials

  • Convenient tabulation of electrochemical data.

  • Standard reduction potentials, Ered are measured relative to the standard hydrogen electrode (SHE).


Electrochemistry

Cell EMF

Standard Reduction (Half-Cell) Potentials

  • Reactions with Ered < 0 are spontaneous oxidations relative to the SHE.

  • The larger the difference between Ered values, the larger Ecell.

  • In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode).

Calculate Eocell for: 2 Al(s) + 3 I2(s)  2 Al3+(aq) + 6 I-(aq)


Electrochemistry

Calculate Eocell for: 2 Al(s) + 3 I2(s)  2 Al3+(aq) + 6 I-(aq)


Electrochemistry

Calculate Eocell for: 2 Al(s) + 3 I2(s)  2 Al3+(aq) + 6 I-(aq)


Electrochemistry

Spontaneity of Redox Reactions

  • In a voltaic (galvanic) cell (spontaneous) Ered(cathode) is more positive than Ered(anode) since

  • More generally, for any electrochemical process

  • A positive E indicates a spontaneous process (galvanic cell).

  • A negative E indicates a nonspontaneous process.


Electrochemistry

Spontaneity of Redox Reactions

EMF and Free-Energy Change

  • Can show that under Standard Conditions:

  • G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and Eo is the emf of the cell.

  • Since n and F are positive, if G > 0 then E < 0.

Calculate Eo and ΔGo for:

  • 4Ag(s) + O2(g) + 4H+(aq)  4Ag+(aq) + 2H2O(l)

  • 2Ag(s) + ½O2(g) + 2H+(aq)  2Ag+(aq) + H2O(l)


Electrochemistry

Calculate Eo and ΔGo for:

  • 4Ag(s) + O2(g) + 4H+(aq)  4Ag+(aq) + 2H2O(l)

  • 2Ag(s) + ½O2(g) + 2H+(aq)  2Ag+(aq) + H2O(l)


Electrochemistry

Calculate Eo and ΔGo for:

  • 4Ag(s) + O2(g) + 4H+(aq)  4Ag+(aq) + 2H2O(l)

  • 2Ag(s) + ½O2(g) + 2H+(aq)  2Ag+(aq) + H2O(l)


Electrochemistry

Calculate Eo and ΔGo for:

  • 4Ag(s) + O2(g) + 4H+(aq)  4Ag+(aq) + 2H2O(l)

  • 2Ag(s) + ½O2(g) + 2H+(aq)  2Ag+(aq) + H2O(l)


Electrochemistry

Effect of Concentration on Cell EMF

The Nernst Equation

  • The Nernst equation can be simplified by collecting all constants together using a temperature of 298 K:

  • n is number of moles of electrons.

Calculate the emf for the Zn-Cu Voltaic cell with [Cu2+] = 1.50 M and [Zn2+] = 0.050 M.


Electrochemistry

Calculate the emf for the Zn-Cu Voltaic cell with [Cu2+] = 1.50 M and [Zn2+] = 0.050 M.

Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)


Electrochemistry

Effect of Concentration on Cell EMF

Concentration Cells


Electrochemistry

Effect of Concentration on Cell EMF

Cell EMF and Chemical Equilibrium

  • A system is at equilibrium when G = 0.

  • From the Nernst equation, at equilibrium and 298 K (E = 0 V and Q = Keq):


Electrochemistry

Batteries

  • A battery is a self-contained electrochemical power source with one or more voltaic cell.

  • When the cells are connected in series, greater emfs can be achieved.


Electrochemistry

Batteries

Lead-Acid Battery

  • A 12 V car battery consists of 6 cathode/anode pairs each producing 2 V.

  • Cathode: PbO2 on a metal grid in sulfuric acid:

    PbO2(s) + SO42-(aq) + 4H+(aq) + 2e- PbSO4(s) + 2H2O(l)

  • Anode: Pb:

    Pb(s) + SO42-(aq)  PbSO4(s) + 2e-


Electrochemistry

Batteries

Lead-Acid Battery

  • The overall electrochemical reaction is

    PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l)

    for which

    Ecell = Ered(cathode) - Ered(anode)

    = (+1.685 V) - (-0.356 V)

    = +2.041 V.

  • Wood or glass-fiber spacers are used to prevent the electrodes from touching.


Electrochemistry

Batteries

Alkaline Battery

  • Anode: Zn cap:

    Zn(s)  Zn2+(aq) + 2e-

  • Cathode: MnO2, NH4Cl and C paste:

    2NH4+(aq) + 2MnO2(s) + 2e- Mn2O3(s) + 2NH3(aq) + 2H2O(l)

  • The graphite rod in the center is an inert cathode.

  • For an alkaline battery, NH4Cl is replaced with KOH.


Electrochemistry

Batteries

Fuel Cells

  • Direct production of electricity from fuels occurs in a fuel cell.

  • On Apollo moon flights, the H2-O2 fuel cell was the primary source of electricity.

  • Cathode: reduction of oxygen:

    2H2O(l) + O2(g) + 4e- 4OH-(aq)

  • Anode:

    2H2(g) + 4OH-(aq)  4H2O(l) + 4e-


Electrochemistry

Corrosion

Corrosion of Iron

  • Since Ered(Fe2+) < Ered(O2) iron can be oxidized by oxygen.

  • Cathode: O2(g) + 4H+(aq) + 4e- 2H2O(l).

  • Anode: Fe(s)  Fe2+(aq) + 2e-.

  • Dissolved oxygen in water usually causes the oxidation of iron.

  • Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3.xH2O(s).


Electrochemistry

Corrosion

Corrosion of Iron

  • Oxidation occurs at the site with the greatest concentration of O2.

    Preventing Corrosion of Iron

  • Corrosion can be prevented by coating the iron with paint or another metal.

  • Galvanized iron is coated with a thin layer of zinc.


Electrochemistry

Corrosion

Preventing Corrosion of Iron

  • To protect underground pipelines, a sacrificial anode is added.

  • The water pipe is turned into the cathode and an active metal is used as the anode.

  • Often, Mg is used as the sacrificial anode:

    Mg2+(aq) +2e- Mg(s), Ered = -2.37 V

    Fe2+(aq) + 2e- Fe(s), Ered = -0.44 V


Electrochemistry

Electrolysis

Electrolysis of Aqueous Solutions

  • In electrolytic cells the anode is positive and the cathode is negative. (In galvanic cells the anode is negative and the cathode is positive.)


Electrochemistry

Electrolysis

Electroplating

  • Active electrodes: electrodes that take part in electrolysis.

  • Example: electrolytic plating.


Electrochemistry

Electrochemistry

Oxidation is the loss of electrons


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