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Galvanic Cells. Electrochemistry = the interchange of chemical and electrical energy = used constantly in batteries, chemical instruments, etc… Galvanic Cells Definitions

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galvanic cells
Galvanic Cells

Electrochemistry = the interchange of chemical and electrical energy

= used constantly in batteries, chemical instruments, etc…

  • Galvanic Cells
    • Definitions
      • Redox Reaction = oxidation/reduction reaction = chemical reaction in which electrons are transferred from a reducing agent (which gets oxidized) to an oxidizing agent (which gets reduced)
      • Oxidation = loss of electron(s) to become more positively charged
      • Reduction = gain of electron(s) to become more negatively charged
    • Using Redox Reactions to generate electric current (moving electrons)
      • 8H+(aq) + MnO4-(aq) + 5Fe2+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
        • Fe2+ is oxidized and MnO4- is reduced
        • Half Reaction = oxidation or reduction process only

Reduction: 8H+ + MnO4- + 5e- Mn2+ + 4H2O

Oxidation: 5(Fe2+ Fe3+ + e-)

Sum = Redox Rxn

slide2
C. Balancing Redox Equations: Half-Reaction Method in Acidic Solution
  • MnO4-(aq) + Fe2+(aq) -----> Fe3+(aq) + Mn2+(aq) (acidic solution)
  • Identify and write equations for the two half-reactions
    • MnO4- -----> Mn2+ (this is the reduction half-reaction)

(+7)(-2) (+2)

    • Fe2+ -----> Fe3+ (this is the oxidation half-reaction)

(+2) (+3)

  • Balance each half-reaction
    • Add water if you need oxygen
    • Add H+ if you need hydrogen (since we are in acidic solution)
    • Balance the charge by adding electrons
    • MnO4- -----> Mn2+ + 4H2O
    • 8H+ + MnO4- -----> Mn2+ + 4H2O

(+7) (+2)

    • 5e- + 8H+ + MnO4- -----> Mn2+ + 4H2O Balanced!
    • Fe2+ -----> Fe3+ + 1e- Balanced!
slide3
3. Equalize the number of electrons in each half-reaction and add reactions
    • 5(Fe2+ -----> Fe3+ + 1e-) = 5Fe2+ -----> 5Fe3+ + 5e-
    • 5e- + 8H+ + MnO4- -----> Mn2+ + 4H2O
    • 5Fe2+ + 8H+ + MnO4- ------> 5Fe3+ + Mn2+ + 4H2O
    • Species (including e-) on each side cancel out (algebra)
    • Check that the charges and elements all balance: DONE!
  • Example: H+ + Cr2O72- + C2H5OH -----> Cr3+ + CO2 + H2O

D. Balancing Redox Equations: Half-Reaction Method in Basic Solution

  • Follow the Acidic Solution Method until you have the final balance eqn
  • H+ can’t exist in basic solution, so add enough OH- to both sides to turn all of the H+ in H2O
  • Example: Ag + CN- + O2 -----> Ag(CN)2- (basic solution)
    • Ag + CN- -----> Ag(CN)2- (oxidation half-reaction)
    • Becomes: Ag + 2CN- -----> Ag(CN)2- + 1e- Balanced
    • O2 -----> (reduction half-reaction)
    • Becomes: 4e- + 4H+ + O2 -----> 2H2O Balanced
slide4
Equalize the electrons in each half-reaction and add reactions
  • 4(Ag + 2CN- -----> Ag(CN)2- + 1e-)
  • Becomes: 4Ag + 8CN- -----> 4Ag(CN)2- + 4e-
  • Add to: 4e- + 4H+ + O2 -----> 2H2O
  • Gives: 4Ag + 8CN- + 4H+ + O2 -----> 4Ag(CN)2- + 2H2O DONE!

j. Add OH- ions to both sides to remove H+ ions

k. 4Ag + 8CN- + 4H+ + O2 + 4OH- -----> 4Ag(CN)2- + 2H2O + 4OH-

l. 4Ag + 8CN- + 4H2O + O2 -----> 4Ag(CN)2- + 2H2O + 4OH-

m. Cancel water molecules appearing on both sides of the equation

4Ag + 8CN- + 2H2O + O2 -----> 4Ag(CN)2- + 4OH-

n. Check that everything balances REALLY DONE!

slide5
4) In solution:
  • Fe and MnO4- collide and electrons are transferred
  • No work can be obtained; only heat is generated

5) In separate compartments, electrons must go through a wire = Galvanic Cell

  • Generates a current = moving electrons from Fe2+ side to MnO4- side
  • Current can produce work in a motor
  • Salt Bridge = allows ion flow without mixing solutions (Jello-like matrix)
slide6
Chemical reactions occur at the Electrodes = conducting solid dipped into the solution
      • Anode = electrode where oxidation occurs (production of e-)
      • Cathode = electrode where reduction occurs (using up e-)

E. Cell Potential

  • Think of the Galvanic Cell as an oxidizing agent “pulling” electrons off of the reducing agent. The “pull” = Cell Potential
    • ecell = Cell Potential = Electromotive Force = emf
    • Units for ecell = Volt = V 1 V = 1 Joule/1 Coulomb
  • Voltmeter = instrument drawing current through a known resistance to find V

Potentiometer = voltmeter that doesn’t effect V by measuring it

slide7
Standard Reduction Potentials
    • Standard Hydrogen Electrode
      • When measuring a value, you must have a standard to compare it to
      • Cathode = Pt electrode in 1 M H+ and 1 atm of H2(g)

Half Reaction: 2H+ + 2e- H2(g) e1/2 = 0

3) We will use this cathode to find ecell of other Half Reactions

slide8
What is the e1/2 of Zno Zn2+ + 2e-
    • Cathode = SHE Anode = Zn(s) in 1 M Zn2+(aq) (standard states)
    • Voltmeter reads ecell = 0.76 V
    • Define eocell = ecell at standard states so everyone can compare data
    • eocell = eo(H+ H2) + eo(Zno Zn2+)

+0.76 V = 0.00 V + x

    • x = eo(Zno Zn2+) = +0.76 V
    • Combine known values in other Galvanic Cells to determine other eo1/2 values

a) Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s)

b) Anode: Zn Zn2+ + 2e-

c) Cathode: Cu2+ + 2e- Cu

d) eocell = 1.10 V = eo(Zn) + eo(Cu2+)

e) 1.10 V = +0.76 V + eo(Cu2+)

f) eo(Cu2+ + 2e- Cu) = +0.34 V

slide9
Standard Reduction Potentials can be found in your text appendices
    • Always given as a reduction process
    • All solutes are 1M, gases = 1 atm
  • Combining Half Reactions to find Cell Potentials
    • Reverse one of the half reactions to an oxidation; this reverses the sign of e1/2
    • Don’t need to multiply for coefficients = Intensive Property (color, flavor)
    • Example: 2Fe3+(aq) + Cuo 2Fe2+(aq) + Cu2+(aq)
      • Fe3+ + e- Fe2+ e1/2 = +0.77 V
      • Cu2+ + 2e- Cuoe1/2 = +0.34 V
      • Reverse of (ii) added to (i) = -0.34 V + +0.77 V = +0.43 V = e1/2
slide10

Pt(s)

Pt(s)

Mn2+,

H+, MnO4-

ClO3-, ClO4-,H+

  • Line Notation = shorthand way to draw Galvanic Cells
    • 2Al3+(aq) + 3Mg(s) 2Al(s) + 3Mg2+(aq)
    • Line Notation: Mg(s) | Mg2+(aq) || Al3+(aq) | Al(s)
      • Left side = Anode, Right side = Cathode
      • | = phase change, || = salt bridge
      • Far left = anodic electrode material, Far right = cathodic electrode material
    • 2MnO4-(aq) + 6H+(aq) + 5ClO3-(aq) 2Mn2+(aq) + 3H2O(l) + 5ClO4-(aq)
      • Anode: ClO3- + H2O ClO4- + 2H+ + 2e-
      • Cathode: MnO4- +5e- + 8H+ Mn2+ + 4H2O
      • Pt(s) | ClO3-(aq), ClO4-(aq), H+(aq) || H+(aq), MnO4-(aq), Mn2+(aq) | Pt(s)
slide11
Direction of electron flow in a cell
    • Cell always runs in a direction to produce a positive ecell
    • Fe2+ + 2e- Feoe1/2 = -0.44 V

MnO4- + 5e- + 8H+ Mn2+ + 4H2O e1/2 = +1.51 V

    • We put the cell together to get a positive potential
      • 5(Feo Fe2+ + 2e-)e1/2 = +0.44 V
      • 2(MnO4- + 5e- + 8H+ Mn2+ + 4H2O) e1/2 = +1.51 V

16H+(aq) + 2MnO4-(aq) + 5Feo(s) 2Mn2+(aq) + 5Fe2+(aq) + 8H2O(l) ecell = 1.95V

4) Line Notation: Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s)

slide12
The complete description of a Galvanic Cell
    • Items to include in the description
      • Cell potential (always +) and the balanced overall reaction
      • Direction of electron flow
      • Designate the Anode and the Cathode
      • Identity of the electrode materials and the ions present with concentration
    • Example: Completely describe the Galvanic Cell based on these reactions

Ag+ + e- Ag eo = +0.80 V

Fe3+ + e- Fe2+eo = +0.77 V

Ag+(aq) + Fe2+(aq) Ago(s) + Fe3+(aq) eocell = +0.03 V

Pt(s) | Fe2+(aq), Fe3+(aq) || Ag+(aq) | Ag(s)