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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

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Galvanic Cells

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  1. 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

  2. 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!

  3. 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

  4. 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!

  5. 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)

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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)

  11. 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)

  12. 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)

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