Electrochemistry An electrochemical cell produces electricity using a chemical reaction. It consists of two half-cells connected via an external wire with a salt bridge connecting the solutions An electrolytic cell uses an external electricity source to produce a chemical reaction. This is usually called electrolysis.

Historically Historically oxidation involved reaction with O2. i.e., Rusting 4 Fe(s) + 3O2 (g) Fe2O3 (s) Another example Zn(s) + Cu2+(aq) g Zn2+(aq) + Cu(s) In this reaction: Zn(s) g Zn2+(aq) Oxidation Cu2+(aq) g Cu(s) Reduction In a redox reaction, one process can’t occur without the other. Oxidation and Reduction reactions must simultaneously occur.

Redox Between Zn and Cu If Zn(s) and Cu2+(aq) is in the same solution, then the electron is a transferred 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...

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 a very short time the process stops Electron transport stops because of charge build up on both sides Build up of positive charge Build up of negative charge The charge separation will lead to process where it cost too much energy to transfer electron.

Completing the Circuit Electron transfer can occur if the circuit is closed Parts: Two conductors Electrolyte solution Salt Bridge / Porous membrane Process that must happen if e- is to flow. A. e- transport through external circuit B. In the cell, ions must migrate C. Circuit must be closed using a salt bridge (no charge build up) Anode (-) Green Negative electrode generates electron Oxidation Occurs Cathode (+) Red Positive electrode accepts electron Reduction Occurs A C B Anode/Anion (-) Cathode/Cation(+)

Voltaic Cell 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(+)

Completing the Circuit: Salt Bridge 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 neutralize any charge build up.

Sign Convention of Voltaic Cell @ Anode: Negative Terminal (anions). Source of electron then repels electrons. Oxidation occurs. Zn(s) g Zn+2(aq) + 2e- : Electron source @ 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° = -1.10 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)

Another Voltaic Cell Zn(s) + 2H+ (aq) g Zn+2(aq) + H2 (g) E° = 0.76 V @ Anode: Negative Terminal (anions): Zn(s) g Zn+2(aq) + 2e- : Source of electron then repels electrons. Oxidation occurs. @ Cathode: Positive Terminal (cation): 2e- + 2H+(aq) g H2 (g) Attracts electron and then consumes electron. Reduction occurs. Overall: Zn(s) + 2H+ (aq) g Zn2+ (aq) + H2 (g)

Another Voltaic Cell Zn(s) + 2H+ (aq) g Zn+2(aq) + H2 (g) E° = 0.76 V @ Anode: Negative Terminal (anions): Zn(s) g Zn+2(aq) + 2e- : Source of electron then repels electrons. Oxidation occurs. @ Cathode: Positive Terminal (cation): 2e- + 2H+(aq) g H2 (g) Attracts electron and then consumes electron. Reduction occurs. Overall: Zn(s) + 2H+ (aq) g Zn2+ (aq) + H2 (g)

Line Notation Convention basically: metal | ionic solution || ionic solution | metal This is a convenient way of representing cells 1. Anode g Cathode [oxidation (-) ] [reduction (+)] 2. | represents a phase boundary 3. || represents the salt bridge 4. Concentration of component Zn(s) ZnSO4 (aq,1.0M) CuSO4 (aq,1.0M) Cu(s) 1 4 3 2

Line Notation Examples Consider : Zn(s) + Cu+2(aq) g Zn+2(aq) + Cu(s Anode: Zn g Zn+2 + 2e- Cathode: Cu+2 + 2e- 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) + H2(g) Cathode: 2H+ + 2e- g H2 (g) Zn (s) Zn+2 (aq)(1.0M) H+(aq) (1.0M), H2(g, 1atm) Pt(s)

EMF - ElectroMotive Force Potential energy of electron is higher at the anode. This is the driving force for the reaction (e- transfer) e Anode (-) D P.E. = V = J e - C e- flow toward cathode (+) Cathode Larger the gap, the greater the potential (Voltage)

Charge (C) = Current (A) x Time (s) ElectroMotive Force For electrochemical cells Emf – electromotive force Potential energy difference between the two electrodes Units of emf: V - Volts : 1V - 1 Joule / Coulomb 1 Joule of work per coulomb of charge transferred. For electrolysis When a current of one amp flow for 1 second this equals one coulomb of charge passed. Charge (C) = Current (A) x Time (s) 96,400 coulombs = 1 Faraday of charge 1 Faraday is equal to 1 mole of electrons

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

Zoom of Std. Electrode Potentials Cell Potential is written as a reduction equation. M+ + e- g M E° = F2 (g) + 2e- g 2 F - (aq) 2.87 V Ce4+ + e- g Ce3+ (aq) 1.61 V 2H+ + 2e- g H2 (g) 0.00 V Li+(aq) + e- g Li(s) -3.045 V Most spontaneous Reduction Oxidizing Agent Written as reduction Most non-spontaneous 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.

Eº(cell) = Eº(red) – Eº(ox) E°Cell Evaluation E°Cell Function of the reaction g Oxidation Process (Anode reaction) g Reduction Process (Cathode reaction) E°Cell = E°red (cathode) - E°red (anode) Simple to remember: Eº(cell) = Eº(red) – Eº(ox) E = E redox The potential difference of the cell is equal to the electrode potential of the reduced component minus the electrode potential of the oxidised component.

Standard Electrode 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+2 + 2e- E°=? Can’t determine because the reaction must be coupled Question? - How can a scale of reduction potential be determined ? Answer - Use a half reaction as reference and assign it a potential of zero. Measure all other half cells in comparison to this reaction.

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.

Standard Reference Half Cell Selected half reaction is: H+ / H2 (g) couple half reaction: 2H+ (aq, 1.0M) + 2e- g H2 (g,1atm) by definition c E° = 0.0 V, the reverse is also 0.0 V H+/H2 couple - Standard Hydrogen Electrode (SHE) To determine E° for a another half reaction, the reaction of interest needs to be connected to this SHE. The potential measured is then assigned to the half-reaction under investigation. Example: E°Cell = 0.76 V = E°red (Cat) - E°red (Anode) 0.0 V - (?) E°red (Anode) = - 0.76 V \Zn+2/Zn E° = -0.76 V reduction reaction

Determining Other Half-Cell Potentials Now consider the reaction: Zn(s)|Zn+2 (1.0 M)||Cu+2(1.0 M)|Cu(s) E°Cell = 1.10 V E°Cell = E°red (Cat) - E°red (Anode) recall, E° Zn+2/Zn = - 0.76 V Therefore, E°Cell = E°Cu+2/Cu - E° Zn+2/Zn 1.10 V = (?) - (- 0.76 V) E°Cu+2/Cu = + 0.34 V

Example: Half-Cell Potential For the reaction: Tl+3 + 2Cr2+  Tl+ + 2Cr3+ E°Cell = 1.19 V i) Write both half reaction and balance ii) Calculate the E°Cell Tl+3  Tl+ iii) Sketch the voltaic cell and line notation i) Tl+3 + 2e-  Tl+ (Cr2+  2Cr3+ + 2e- ) x 2 E° = 0.41 V ii) E°Cell = 1.19 V = E°red (Cat) - E°red (Anode) 1.19 V= E°red (Cat) - 0.041 V for Tl+3 + 2e-  Tl+ : 1.19 V - 0.41 = E°red (Cat) = 0.78 V Pt Cr2+ g Cr3+ Tl3+ g Tl+ 1.19 V

Voltaic Vs. Electrolytic Cells Voltaic Cell Energy is released from spontaneous redox reaction Electrolytic Cell Energy is absorbed to drive nonspontaneous redox reaction General characteristics of voltaic and electrolytic cells. A voltaic cell generates energy from a spontaneous reaction (DG<0), whereas an electrolytic cell requires energy to drive a nonspontaneous reaction (DG>0). In both types of cell, two external circuits provides the means or electrons to flow. Oxidation takes place all the anode, and reduction takes place at the cathode, but the relative electrode changes are opposite in the two cells. System does work on load (surroundings) Surrounding (power supply) do work on system (cell) Anode (Oxidation) Oxidation Reaction X g X+ + e- Oxidation Reaction A- g A + e- Reduction Reaction e- + Y+ g Y Reduction Reaction e- + B+ g B Overall (Cell) Reaction A- + B+ g A + B, DG> 0 Overall (Cell) Reaction X + Y+ g X+ + Y, DG = 0