18.2 Voltaic Cell. Generating Voltage (Potential) Dr. Fred Omega Garces. Chemistry 201. Miramar College. 1 Voltaic Cell.

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1 18.2 Voltaic Cell Generating Voltage (Potential) Dr. Fred Omega Garces Chemistry 201 Miramar College 1 Voltaic Cell

2 Redox Between If Zn (s) and Cu 2+ (aq) is in the same solution, then the electrons transfer directly between the Zn and Cu. No useful work is obtained. However if the reactants are separated and the electrons shuttle through an external path... 2 Voltaic Cell

3 Electrochemical Cells Voltaic / Galvanic Cell Apparatus which produce electricity Electrolytic Cell Apparatus which consumes electricity Consider: Zn Cu Initially there is a flow of e- after some time, the process stops. Electron transport stops because of charge build up. Build up of positive charge Build up of negative charge The charge separation leads to a process that requires too much energy for transfer electron to continue. 3 Voltaic Cell

Voltaic Cell (Movie) Electron transfer can occur if the circuit is closed Parts: Two conductors Electrolyte solution Salt Bridge / Porous membrane 3 process must happen if e - is to flow. A.

4 Voltaic Cell (Movie) Electron transfer can occur if the circuit is closed Parts: Two conductors Electrolyte solution Salt Bridge / Porous membrane 3 process must happen if e - is to flow. A. e- transport through external circuit B. In the cell, ions a must migrate C. Circuit must be closed (no charge build up) Anode (-) Black Negative electrode generates electron Oxidation Occur Cathode (+) Red Positive electrode accepts electron Reduction Occur Anode/Anion (-) Cathode/Cation(+) 4 Voltaic Cell

5 Salt Bridge: Function In order for electrons to move through an external wire, charge must not build up at any cell. This is done by the salt bridge in which ions migrate to different compartments thereby neutralizing any charge build up. Oxidation Reaction Zn g Zn e - As the oxidation reaction produces Zn +2 ions, negatively charged SO 4 2- ions flow in from the salt bridge, preventing a buildup of positive charge. Reduction Reaction Cu e - g Cu As the reduction reaction removes Cu+2 ions from the solution, Na+ ions flow in from the salt bridge, preventing a buildup of negative charge. 5 Voltaic Cell

6 Sign Convention of Voltaic Anode: Negative Terminal (anions). Source of electron then repels electrons. Oxidation occurs. Zn (s) g Zn +2 (aq) + 2e- : Electron Cathode: Positive Terminal (cation) Attracts electron and then consumes electron. Reduction occurs. Electron target: 2e - + Cu +2 (aq) g Cu (s) Overall: Zn (s) + Cu +2 (aq) g Zn+2 (aq) + Cu (s) E = 1.10 V Note when the reaction is reverse: Cu (s) + Zn+2 (aq) g Cu+2 (aq) + Zn (s) Sign of E is also reversed E = V Oxidation: Zn (s) g Zn +2 (aq) E = 0.76 V Reduction: Cu +2 (aq) g Cu (s) E = 0.34 V 1.10 V = E CELL or E CELL = E red (Red-cathode) - E red (Oxid-anode) (-.76) V 6 Voltaic Cell

E Cell Evaluation E Cell Function of the reaction g Oxidation Process (Anode reaction) g Reduction Process (Cathode reaction) or E Cell = E Cathode & E Anode Cathode (+) Anode (-)

7 E Cell Evaluation E Cell Function of the reaction g Oxidation Process (Anode reaction) g Reduction Process (Cathode reaction) or E Cell = E Cathode & E Anode Cathode (+) Anode (-) Most Negative Reduction reaction Therefore, E Cell = E red (Cathode) - E red (anode) Very Positive Very Spontaneous Neg Minus (Large negative) (Very Positive Value) 7 Voltaic Cell

Other Voltaic Cell (Movie) Zn (s) + 2H + (aq) g Zn+2 (aq) + H 2 (g) E = 0.76 V @ Anode: Negative Terminal (anions): Zn (s) g Zn +2 (aq) + 2e- : Source of electron then repels electrons.

8 Other Voltaic Cell (Movie) Zn (s) + 2H + (aq) g Zn+2 (aq) + H 2 (g) E = 0.76 Anode: Negative Terminal (anions): Zn (s) g Zn +2 (aq) + 2e- : Source of electron then repels electrons. Oxidation Cathode: Positive Terminal (cation): 2e - + 2H + (aq) g H 2 (g) Attracts electron and then consumes electron. Reduction occurs. Net: Zn (s) + 2H + (aq) g Zn 2+ (aq) + H 2 (g) 8 Voltaic Cell

Line Notation Convention Line notation: Convenient convention for electrochemical cell Schematic Representation 1. Anode g Cathode [oxidation (-) ] [reduction (+)] 2.

9 Line Notation Convention Line notation: Convenient convention for electrochemical cell Schematic Representation 1. Anode g Cathode [oxidation (-) ] [reduction (+)] 2. phase boundary (where potential may develop) 3. Liquid junction 4. Concentration of component 1 4 Zn (s) ZnSO 4 (aq,1.0m) CuSO 4 (aq,1.0m) Cu (s) Voltaic Cell

10 Line Notation Examples Consider : Zn (s) + Cu +2 (aq) g Zn+2 (aq) + Cu (s Anode: Zn g Zn e - Cathode: Cu e - g Cu Shorthand Line notation 2nd Example : Zn (s) + 2H + (aq) g Zn+2 (aq) + H 2(g) Anode: Zn g Zn e - Cathode: 2H + + 2e - g H 2 (g) Shorthand Line notation 10 Voltaic Cell

11 Line Notation Examples Consider : Zn (s) + Cu +2 (aq) g Zn+2 (aq) + Cu (s Anode: Zn g Zn e - Cathode: Cu e - g Cu Shorthand Line notation Zn (s) Zn +2 (aq) (1.0M) Cu +2 (aq) (1.0M) Cu(s) 2nd Example : Zn (s) + 2H + (aq) g Zn+2 (aq) + H 2(g) Anode: Zn g Zn e - Cathode: 2H + + 2e - g H 2 (g) Shorthand Line notation Zn (s) Zn +2 (aq) (1.0M) H + (aq) (1.0M), H 2 (g, 1atm) Pt (s) 11 Voltaic Cell

Other Voltaic Cell & Their Line Notation Oxidation half-reaction Zn (s) g Zn

(s) + 2e - Cr (s) g Cr +3 (aq) + 3e- Reduction half-reaction MnO 4 - (aq) +

e- g Ag (s) Zn (s) Zn +2 (aq) H + (aq), H 2 (g,1atm) Pt Cr (s) Cr +3 (aq) Ag

12 Other Voltaic Cell & Their Line Notation Oxidation half-reaction Zn (s) g Zn +2 (aq) + 2e- Oxidation half-reaction Oxidation half-reaction 2I- (aq) g I 2 (s) + 2e - Cr (s) g Cr +3 (aq) + 3e- Reduction half-reaction MnO 4 - (aq) + 8H + (aq) + 5e- g Mn 2+ (aq) +4H 2 O (l) Reduction half-reaction Ag + (aq) + e- g Ag (s) Zn (s) Zn +2 (aq) H + (aq), H 2 (g,1atm) Pt Cr (s) Cr +3 (aq) Ag + (aq) Ag (s) C (s) I - (aq), I 2 (s,) MnO 4 - (aq), Mn+2 (aq) C (s) 12 Voltaic Cell

13 Line Notation Examples Example 1: B&L Zn (s) + Ni 2+ (aq) g Zn+2 (aq) + Ni (s) Example 2: B&L Tl +3 (aq) + 2Cr2+ (aq) g Tl+ (aq) + Cr+3 (aq) 13 Voltaic Cell

14 Line Notation Examples Example 1: B&L Zn (s) + Ni 2+ (aq) g Zn+2 (aq) + Ni (aq) Zn (s) Zn +2 (aq) Ni+2 (aq) Ni (s) Example 2: B&L Tl +3 (aq) + 2Cr2+ (aq) g Tl+ (aq) + Cr+3 (aq) C (s) Cr +2 (aq), Cr +3 (aq) Tl+3 (aq), Tl + (aq) C (s) 14 Voltaic Cell

Voltage of Galvanic / Voltaic Cell Transport of

Pressure (h) Flow (Fluid Transport) Pressure

Δh Similarly, electron are transported through

15 Voltage of Galvanic / Voltaic Cell Transport of any object requires a net force. Consider water flowing through pipes. This occurs because of pressure gradient. Pressure (h) Flow (Fluid Transport) Pressure (i) Or Object falling or transport down due to Δh Similarly, electron are transported through wires because of the electromotive force EMF or Ecell. (-) e - (+) 15 Voltaic Cell

16 EMF - ElectroMotive Force Potential energy of electron is higher at the anode. This is the driving force for the reaction (e- transfer) Δ P.E. = V = J e - C V 2.87 More on this scale later F 2(g) + 2e - D F - (aq) Anode (-) e e- flow toward cathode Cu 2+ (aq) + 2e- D Cu (s) 2H + + 2e- D H 2 Zn 2+ (aq) + 2e- D Zn (s) Larger the gap, the greater the potential (Voltage) (+) Cathode Li + (aq) + e- D Li (s) 16 Voltaic Cell

17 ElectroMotive Force EMF - Electro Motive Force Potential energy difference between the two electrodes The larger the ΔP.E. the larger EMF value. The magnitude of P.E. for the reaction (half reaction) is an intensive property) i.e., Size independent: ρ, T bpt, C s. Therefore EMF is also an intensive property. Analogy: Size of rock not important, only the height from ground. (Electron all have the same mass) Unit: EMF: V - Volts : 1V - 1 Joule / Coulomb 1 Joule of work per coulomb of charge transferred. 17 Voltaic Cell

18 Stoichiometry Relationship to E EMF - Intensive Property E cell Standard state conditions 25 C, 1atm, 1.0 M E cell Intensive property, Size Independent Consider: Li + + e - g Li (s) E Cell = V x 2 2 Li e - g 2 Li (s) E Cell = ( V) x 2 =?? But E = Voltage per electron E = E x 2 =? g V 2 = V 2 e - Stoichiometry does not change E, but reversing the reaction does change the sign of E. 18 Voltaic Cell

19 Standard Reduction Potential Cell Potential is written as a reduction equation. M + + e - g M E = std red. potential Written as reduction Most spontaneous <Reduction occurs> Oxidizing Agent Most nonspontaneous Spontaneous in the reverse direction. <Oxidation occurs> Reducing Agent 19 Voltaic Cell

20 Side-Bar: Relative Scale Consider a baby whose weight is to be determine but will not remain still on top of a scale. How can the parents determine the babies weight? Carry the child in arms and weight both child and parent then subtract the weight of the parent from the total to yield the baby weight. 20 Voltaic Cell

21 Zoom View of Std. Reduction Potential Cell Potential is written as a reduction equation. Written as reduction M + + e - g M E = F 2 (g) + 2e - g 2 F - (aq) 2.87 V Ce 4+ + e - g Ce 3+ (aq) 1.61 V 2H + + 2e - g H 2 (g) 0.00 V Li + (aq) + e- g Li (s) V Most spontaneous Reduction Oxidizing Agent Most nonspontaneous. Spontaneous in the reverse direction. Oxidation Reducing Agent All reaction written as reduction reaction. But in electrochemistry, there can t be just a reduction reaction. It must be coupled with an oxidation reaction. 21 Voltaic Cell

22 Standard Reduction Potential How is E red (Cathode) and E red (Anode) determine. E (EMF) - State Function; there is no absolute scale Absolute E value can t be measured experimentally The method of establishing a scale is to measure the difference in potential between two half-cells. Consider: Zn g Zn e - E =? Can t determine because the reaction must be coupled How can a scale of reduction potential be determine? Reference a pre-selected half rxn and assign it a potential of zero. Electrochemical reaction more spontaneous than this reference will have positive E, and those less spontaneous will have negative E. 22 Voltaic Cell

23 Reference Potential Selected half reaction is: H + / H 2 (g) couple half reaction: 2H + (aq, 1.0M) + 2e - g H 2 (g,1atm) by definition c E = 0.0 V, the reverse is also 0.0 V H + /H 2 couple - Standard Hydrogen Electrode (SHE) To determine E for a another half reaction, the reaction of interest needs to be coupled to this SHE. The potential measured is then assigned to the half-reaction under investigation. E Cell = 0.76 V = E red (Cat) - E red (Anode) 0.0 V - (?) E red (Anode) = V Zn +2 /Zn E = V Reduction rxn 23 Voltaic Cell

24 Example: Half-Cell Potential Example BBL20.19: For the rxn: Tl Cr 2+ g Tl + + 2Cr 3+ E Cell = 1.19 V i) Write both half rxn and balance ii) Calculate the E Cell Tl +3 g Tl + iii) Sketch the voltaic cell and line notation (Cr 2+ g 2Cr e - ) x 2 E = 0.41 V 1.19 V Pt Pt Cr 2+ g Cr 3+ Tl 3+ g Tl + 24 Voltaic Cell

Example: Half-Cell Potential Example BBL20.19: For the rxn: Tl +3 + 2Cr 2+ g Tl + + 2Cr 3+ E Cell = 1.

25 Example: Half-Cell Potential Example BBL20.19: For the rxn: Tl Cr 2+ g Tl + + 2Cr 3+ E Cell = 1.19 V i) Write both half rxn and balance ii) Calculate the E Cell Tl +3 g Tl + iii) Sketch the voltaic cell and line notation (Cr 2+ g 2Cr e - ) x 2 E = 0.41 V i) Tl e - g Tl + (Cr 2+ g 2Cr e - ) x 2 E = 0.41 V ii) E Cell = 1.19 V = E red (Cat) - E red (Anode) 1.19 V= E red (Cat) - (-0.41 V) for Tl e - g Tl + : 1.19 V = E red (Cat) = 0.78 V Pt 1.19 V Pt Cr 2+ g Cr 3+ Tl 3+ g Tl + 25 Voltaic Cell

26 3+ Ga! Ga + 3 e- E = Cd e -! 3 Cd E = Cr + 3 Eu 3+! 2Cr Eu E = Cr! 2Cr + 6 e- E = Eu e - 2+! 3 Eu E = Cr + 3 Eu 3+! 2Cr Eu E = Eu 3+ / Eu 2+ = V In Class work i) Calculate E cell : Ga (s) + Cd2+ (aq) g Ga3+ (aq) + Cd (s) ii) Calculate the E for Eu 3+ / Eu 2+ : Cr (s) Cr 3+ (aq) Eu3+ (aq), Eu2+ (aq) Pt (s) E cell = V 26 Voltaic Cell

27 Voltaic Vs. Electrolytic Cells Voltaic Cell Energy is released from spontaneous redox reaction System does work on load (surroundings) Oxidation Reaction X g X + + e- Reduction Reaction e- + Y + g Y Overall (Cell) Reaction X + Y + g X + + Y, ΔG = 0 Electrolytic Cell Energy is absorbed to drive nonspontaneous redox reaction Surrounding (power supply) do work on system (cell) Oxidation Reaction A - g A + e- Reduction Reaction e- + B + g B Overall (Cell) Reaction A - + B + g A + B, Δ G> 0 General Characteristics of voltaic and electrolytic cells. A voltaic cell generates energy from a spontaneous reaction (ΔG < 0), whereas an electrolytic cell requires energy to drive a nonspontaneous reaction (ΔG > 0). In both types of cell two electrodes dip into electrolyte solution, and an external circuit provides the means for electrons to flow. Oxidation takes place at the anode, and reduction takes place at the cathode, but the relative electrode changes are opposite in the two cells. The anode in the electrolytic cell is now the positive (+) electrode and the cathode in the electrolytic cell is now the negative (-) electrode. 27 Voltaic Cell

A + B C +D ΔG = ΔG + RTlnKp. Me n+ + ne - Me. Me n n

A + B C +D ΔG = ΔG + RTlnKp. Me n+ + ne - Me. Me n n A + B C +D ΔG = ΔG + RTlnKp Me n+ + ne - Me K p a a Me Me n a n e 1 mol madde 6.2 x 1 23 atom elektron yükü 1.62 x 1-19 C FARADAY SABİTİ: 6.2 x 1 23 x 1.62 x 1-19 = 96485 A.sn (= coulomb) 1 Faraday 965