Chapter 20 Electrochemistry

Oxidation and Reduction Oxidation (loss of e-) Na Na+ + e- Reduction (gain of e-) Cl + e- Cl- Oxidation-reduction (redox) reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced. Electron transfer can produce electrical energy spontaneously, but sometimes electrical energy is needed to make them occur (nonspontaneous). CHEM 1033 Chapter 20

CHEM 1033 Chapter 20

Oxidation-Reduction (Redox) Reactions BOTH reduction and oxidation must occur. A substance that gives up electrons is oxidized and is called a reducing agent or reductant (causes another substance to be reduced). A substance that accepts electrons is reduced and is therefore called an oxidizing agent or oxidant (causes another substance to be oxidized). CHEM 1033 Chapter 20

Oxidation-reduction (Redox) Reactions Is it a redox reaction?? Zn(s) + 2H+(aq) Zn2+(aq) + H2(g) Quick hint: CHEM 1033 Chapter 20

Oxidation Number Guidelines Atoms in elemental form, oxidation number is zero. (Cl2, H2, P4, Ne are all zero) Monoatomic ion, the oxidation number is the charge on the ion. (Na+: +1; Al3+: +3; Cl-: -1) O is usually -2. But in peroxides (like H2O2 and Na2O2) it has an oxidation number of -1. H is +1 when bonded to nonmetals and -1 when bonded to metals. (+1 in H2O, NH3 and CH4; -1 in NaH, CaH2 and AlH3) The oxidation number of F is -1. The sum of the oxidation numbers for the molecule is the charge on the molecule (zero for a neutral molecule). CHEM 1033 Chapter 20

Example Determine the oxidation state of all elements in ammonium thiosulfate (NH4)2(S2O3). (NH4)2(S2O3) CHEM 1033 Chapter 20

Determining Oxidation States What is the oxidation state of Mn in MnO4-? Answer: +7 CHEM 1033 Chapter 20

Balancing oxidation-reduction equations We know: Balancing chemical equations follows law of conservation of mass. AND, for redox reactions, gains and losses of electrons must also be balanced. CHEM 1033 Chapter 20

Half-Reactions Separate oxidation and reduction processes in equation, Sn2+(aq) + 2Fe3+(aq) Sn4+(aq) + 2Fe2+(aq) Oxidation: Reduction: Overall, the number of electrons lost in the oxidation half reaction must equal the number gained in the reduction half reaction. CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Acidic Consider : MnO4-(aq) + C2O42-(aq) Mn2+(aq) + CO2(g) Unbalanced half-reactions: MnO4-(aq) Mn2+(aq) C2O42-(aq) CO2(g) First, balance everything EXCEPT hydrogen and oxygen. Deal with half-reactions SEPARATELY. (acidic) CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Acidic MnO4-(aq) + C2O42-(aq) Mn2+(aq) + CO2(g) To balance O: Add 4H2O to products to balance oxygen in reactants. To balance H: Add 8H+ to reactant side to balance the 8H in water. CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Acidic Balance charge: Add up charges on both sides. Add 5 electrons to reactant side. 5e- + 8H+(aq) + MnO4-(aq) Mn2+(aq) + 4H2O(l) Mass balance of C in oxalate half-reaction. C2O42-(aq) 2CO2(g) Balance charge by adding two electrons to the products. Last step: Cancel electrons and add reactions together. CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Acidic 5e- + 8H+(aq) + MnO4-(aq) Mn2+(aq) + 4H2O(l) C2O42-(aq) 2CO2(g) + 2e- Top reaction times 2. Bottom reaction times 5. ALL PARTS!! CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Acidic Summary 1. Divide equation into two incomplete half-reactions. 2. Balance each half-reaction (a) balance elements other than H and O. (b) balance O atoms by adding H2O. (c) balance H atoms by adding H+ (basic conditions will require further work at this step). (d) balance charge by adding e- to the side with greater overall positive charge. CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Summary Multiply each half-reaction by an integer so that the number of electrons lost in one half-reaction equals the number gained in the other half-reaction. 4. Add the two half-reactions and cancel out all species appearing on both sides of the equation. 5. Check equation to make sure there are same number of atoms of each kind and the same total charge on both sides. Errors can be caught!! CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Basic Balancing process is started using H+ and H2O, then adjusting with OH- to uphold reaction conditions. (H+ does not exist in basic solutions.) Balance the following reaction: H2O2(aq) + ClO2(aq) ClO2-(aq) + O2(g) (basic) CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Basic Basic Redox Reactions Split into two half-reactions. H2O2(aq) O2(g) ClO2(aq) ClO2-(aq) Balance elements, then oxygen by adding H2O. Then, add H+ to balance H, just like an acidic redox reaction. H2O2(aq) O2(g) + 2H+ CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Basic Basic Redox Reactions Add OH- to both sides, enough to neutralize all H+ (basic reactions cannot support H+). Combine H+ and OH- to form H2O. CHEM 1033 Chapter 20

Balancing Equations by the Method of Half-Reactions: Basic Basic Redox Reactions Balance charge by adding e-. 2OH- + H2O2(aq) O2(g) + 2H2O + + ClO2(aq) ClO2-(aq) Multiply each reaction so both have same e-. Then add them together, cancelling where possible. 2OH- + H2O2(aq) + 2ClO2(aq) O2(g) + 2H2O + 2ClO2-(aq) Double-check your answer! CHEM 1033 Chapter 20

Balance this redox equation Cu(s) + NO3-(aq) Cu2+(aq) + NO2(g) (acidic) Ans: Cu(s) + 2NO3-(aq) + 4H+(aq) Cu2+(aq) + 2NO2(aq) + 2H2O(l) CHEM 1033 Chapter 20

Balance this redox equation NO2-(aq) + Al(s) NH3(aq) + Al(OH)4-(aq) (basic) 2Al(s) + NO2-(aq) +OH-(aq) + 5H2O(aq) 2Al(OH)4-(aq) + NH3(aq) CHEM 1033 Chapter 20

Voltaic Cells Voltaic (aka galvanic) cells: electrochemical reactions in which electron transfer occurs via an external circuit. The energy released in a voltaic cell reaction can be used to perform electrical work. Reactions are spontaneous. CHEM 1033 Chapter 20

Voltaic Cells: Components For example Anode: Zn(s)  Zn2+(aq) + 2e- ( half-reaction) Cathode: Cu2+(aq) + 2e-  Cu(s) ( half reaction) Salt bridge: cations move from anode to cathode, anions move from cathode to anode. External circuit (wire): electrons move from anode to cathode (between two solid metal electrodes). Electron transfer can naturally occur in OTHER forms besides a wire (direct electron transport between solution and metal). CHEM 1033 Chapter 20

Voltaic Cells: Ion Flow Anions and cations move through a porous barrier or salt bridge. Cations move into the cathodic compartment to neutralize the excess negatively charged ions. Anions move into the anodic compartment to neutralize the excess Zn2+ ions formed by oxidation. CHEM 1033 Chapter 20

Simplified Voltaic Cell In any voltaic cell, the electrons flow from the anode through the external circuit to the cathode. Anode labeled with negative sign (-) and cathode with positive sign (+). Anions migrate toward anode. Cations toward the cathode. CHEM 1033 Chapter 20

Voltaic Cells: Electron Flow As oxidation occurs, Zn(s) is converted to Zn2+ and 2e-. And Cu2+ is converted to Cu(s). CHEM 1033 Chapter 20

One-Pot Voltaic Cells If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn dissolves, forming Zn2+. Zn is spontaneously oxidized to Zn2+ by Cu2+ (Cu is the oxidizing agent). The Cu2+ is spontaneously reduced to Cu0 by Zn (Zn is the reducing agent). The entire process is spontaneous. CHEM 1033 Chapter 20

The Atomic Level A Cu2+(aq) ion comes into contact with a Zn(s) atom on the surface of the electrode. Two electrons are directly transferred from the Zn(s), forming Zn2+(aq), to the Cu2+(aq) forming Cu(s). CHEM 1033 Chapter 20

Cell Electromotive Force (EMF) Electromotive force (emf): the force required to push electrons through the external circuit. Potential difference: difference per electrical charge between two electrodes. Units = Volts (V). One volt is the potential required to impart one joule of energy to a charge of one coulomb: CHEM 1033 Chapter 20

Ecell = Ered(cathode) - Ered(anode) Cell EMF Cell potential (Ecell): the emf of a cell. Aka cell voltage. Positive for spontaneous cell reactions. Ecell strictly depends on reactions that occur at the cathode and anode, the concentration of reactants and products, and the temperature and is described by the equation: Ecell = Ered(cathode) - Ered(anode) For 1M solutions at 25˚C and 1 atm (standard conditions), the standard emf (standard cell potential), Ecell, is written as E˚cell. CHEM 1033 Chapter 20

Standard Reduction Potentials Potential associated with each half-reaction is chosen to be the potential for reduction to occur at that electrode. Hence, Standard Reduction Potentials. Standard reduction potentials, E˚red, are measured relative to the standard hydrogen electrode (SHE). CHEM 1033 Chapter 20

Standard Hydrogen Electrode (SHE) SHE is assumed as the cathode (for consistency). Pt electrode in a tube containing 1M H+ solution. H2(g) is bubbled through the tube and equilibrium is established. CHEM 1033 Chapter 20

Determining Standard Reduction Potentials For the SHE: 2H+(aq, 1M) + 2e-  H2(g, 1 atm) E˚red = . Standard reduction potentials can then be calculated using the SHE (E˚= 0V) as E˚red(cathode). Each calculated E˚ is rewritten as a reduction and tabulated. CHEM 1033 Chapter 20

Finding the Standard Reduction Potential E˚cell is measured, 0V (SHE) is used for E˚(cathode), and the reduction potential for Zn is found. CHEM 1033 Chapter 20

Zn Standard Reduction Potential E˚cell = E˚red(cathode) - E˚red(anode) We measure E˚cell relative to the SHE: E˚cell = E˚red(cathode) - E˚red(anode) 0.76V = 0V - E˚red(anode). Therefore, E˚red(anode) = -0.76V And we find that -0.76V can be assigned to reduction of zinc. CHEM 1033 Chapter 20

Zn Standard Reduction Potential Since E˚red = -0.76 V (negative!) we conclude that the reduction of Zn2+ in the presence of the SHE is not spontaneous. However, the oxidation of Zn with the SHE is spontaneous. Reactions with E˚red > 0 are spontaneous reductions relative to the SHE. E˚red < 0 are spontaneous oxidations. Changing the stoichiometric coefficient does not affect E˚red. 2Zn2+(aq) + 4e-  2Zn(s) E˚red = -0.76 V CHEM 1033 Chapter 20

Full list in Appendix E CHEM 1033 Chapter 20

EMF Trends The larger the difference between E˚red values, the larger E˚cell. Spontaneous voltaic cell: E˚red(cathode) is more positive than E˚red(anode), resulting in a positive E˚cell. Recall CHEM 1033 Chapter 20

Calculating E˚red from E˚cell Zn(s) + Cu2+(aq, 1M) Zn2+(aq, 1M) + Cu(s) E˚cell = 1.10V Given: Zn2+ + 2e- Zn(s) E˚red = -0.76V Calculate the E˚red for the reduction of Cu2+ to Cu. What about E˚red for the oxidation of Cu(s) – reverse reaction? CHEM 1033 Chapter 20

Oxidizing and Reducing Agents We can use E˚red values to understand aqueous reaction chemistry. e.g. Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Since Cu2+ is responsible for the oxidation of Zn(s), Cu2+ is called the . Since Zn(s) is responsible for the reduction of Cu2+, Zn(s) is called the . E˚red for Cu2+ (0.34V) indicates that, compared to E˚red for Zn (-0.76V), it will be reduced. CHEM 1033 Chapter 20

Example Problem: Cell emf Calculate the standard emf for the following: Ni(s) + 2Ce4+(aq) Ni2+(aq) + 2Ce3+(aq) Ask yourself…What are the half-reactions? Which one is oxidized? Reduced? Ni2+(aq) + 2e- Ni(s) E˚red = -0.28V Ce4+(aq) + 1e- Ce3+(aq) E˚red = 1.61V is the oxidizing agent. CHEM 1033 Chapter 20

Example Problem: Cell emf Which reaction will undergo reduction? What reaction occurs at the anode? Sn4+(aq) + 2e- Sn2+(aq) E˚red = 0.154V MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) E˚red = 1.51V CHEM 1033 Chapter 20

Oxidizing and Reducing Agents: EMF The more positive E˚red, the greater the tendency for the reactant in the half-reaction to be reduced. Therefore, the stronger the oxidizing agent. The more negative E˚red, the greater the tendency for the product in the half-reaction to be oxidized. This means the product is tends to be a reducing agent. CHEM 1033 Chapter 20

A species at higher left of the table of standard reduction potentials will spontaneously oxidize a species that is lower right in the table. F2 will oxidize H2 or Li. Ni2+ will oxidize Al(s). CHEM 1033 Chapter 20

Example: Redox Agents Which is the strongest reducing agent? Oxidizing agent? Ce4+, Br2, H2O2, or Zn? E˚red = 1.61V for Ce4+ (aq) 1.065V for Br2 (l) 1.776 for H2O2 (aq) -0.763V for Zn(s) CHEM 1033 Chapter 20

Spontaneity of Redox Reactions In a spontaneous voltaic cell, E˚red(cathode) must be more positive than E˚red(anode). A positive E˚cell indicates a spontaneous voltaic cell process. A negative E˚cell indicates a non-spontaneous process. It all relates back to Gibbs Free Energy. CHEM 1033 Chapter 20

Spontaneity of Redox Reactions Cell emf and free-energy change are related by G is the change in free-energy, n is the number of moles of electrons transferred, F is Faraday’s constant, and E is the emf of the cell. G units are J (assumed per mole). 1F = 96,500 C/mol = 96,500 J/Vmol If standard conditions: G˚ = -nFE˚cell If Ecell > 0 then , both of which indicate spontaneous processes. CHEM 1033 Chapter 20

Example: Cell spontaneity Calculate ΔG˚ for the following reaction. Is it spontaneous? Cl2(g) + 2I-(aq) 2Cl-(aq) + I2(s) E˚cell = 0.823V CHEM 1033 Chapter 20

Batteries Defn: Self-contained electrochemical power source with one (or more) complete voltaic cell. When multiple cells or multiple batteries are connected in series, greater emfs can be achieved. Cathode labeled with a plus sign; the anode with a minus sign. CHEM 1033 Chapter 20

How Batteries Work Spontaneous redox reactions, as in voltaic cells, serve as the basis for battery operation. Specific reactions at anode and cathode determine the voltage of the battery, and the usable life of the battery depends on the quantity of these substances. cannot be recharged. are rechargeable using an external power source after its emf has dropped below a usable level. CHEM 1033 Chapter 20

Lead-Acid Batteries A 12V car battery consists of 6 cathode/anode pairs, connected in series, each producing 2V. Cathode: PbO2(s) on a metal grid in sulfuric acid PbO2(s) + SO42-(aq) + 4H+(aq) + 2e-  PbSO4(s) + 2H2O(l) Anode: Pb(s) in sulfuric acid Pb(s) + SO42-(aq)  PbSO4(s) + 2e- The overall electrochemical reaction is PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l) CHEM 1033 Chapter 20

Lead-Acid Batteries Ecell = Ered(cathode) - Ered(anode) = (+1.685 V) - (-0.356 V) = +2.041 V. Wood or fiberglass spacers prevent electrode contact. H2SO4 consumed during discharge, (voltage may vary with use). Recharging reverses forward reaction: PbO2(s) + Pb(s) + 2SO42-(aq) + 4H+(aq)  2PbSO4(s) + 2H2O(l) CHEM 1033 Chapter 20

Alkaline Batteries Most common nonrechargeable (primary) battery (100 billion produced annually). Anode: Powdered Zn in gel in contact with cKOH solution Zn(s) + 2OH-(aq)  Zn(OH)2(s) + 2e- Cathode: MnO2 and C paste 2MnO2(s) + 2H2O(l) + 2e-  2MnO(OH)(s) + 2OH-(aq) CHEM 1033 Chapter 20

Rechargeable Batteries Lightweight batteries for use in cell phones, notebook computers, etc. Nickel-cadmium (NiCad) battery still common. Cadmium oxidized at anode, nickel oxyhydroxide [NiO(OH)(s)] reduced at cathode. Cathode: 2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq) Anode: Cd(s) + 2OH-(aq)  Cd(OH)2(s) + 2e- CHEM 1033 Chapter 20

Rechargeable Batteries Solid reaction products adhere to the electrodes, permitting electrode reactions to be reversed during recharging. Drawbacks of NiCad battery: Toxicity of Cd (disposal problems), relatively heavy. CHEM 1033 Chapter 20

NiMH and Li-ion Batteries Cathode reaction in nickel metal hydride (NiMH) battery is same as for Ni-Cd battery. 2NiO(OH)(s) + 2H2O(l) + 2e-  2Ni(OH)2(s) + 2OH-(aq) Anode consists of a metal alloy, e.g. ZrNi2, capable of absorbing hydrogen atoms. During oxidation, the hydrogen atoms lose electrons (used by cathode), and the resulting H+ ions react with OH- to form water (drives cathode reaction to the right). Another type: Li-ion battery - lightweight with very high energy density. CHEM 1033 Chapter 20

Fuel Cells Direct, efficient production of electricity. NOT batteries as they are not self-contained (some products must be released). On Apollo moon flights, the H2-O2 fuel cell was the primary source of electricity. Cathode: 2H2O(l) + O2(g) + 4e-  4OH-(aq) Anode: 2H2(g) + 4OH-(aq)  4H2O(l) + 4e- What’s the overall reaction? CHEM 1033 Chapter 20

Corrosion Generally the result of an undesirable redox reaction. Spontaneous! A metal is attacked by some substance in its environment to form an unwanted compound. For many metals oxidation is thermodynamically favored at RT. Corrosion can form an insulating protective coating, preventing further oxidation. (e.g., Al forms a hydrated form of Al2O3) ~20% of iron produced annually in USA is used to replace rusted items!! CHEM 1033 Chapter 20

Corrosion of Iron Rusting of iron requires both oxygen and water. Other factors that can accelerate rusting of iron: - Rusting of iron is electrochemical and iron itself conducts electricity. CHEM 1033 Chapter 20

Corrosion of Iron Recall: E˚red Fe2+ (-0.44V) < E˚red O2 (1.23V), so iron can be oxidized by oxygen. Cathode: O2(g) + 4H+(aq) + 4e-  2H2O(l) E˚red= 1.23V Anode: Fe(s)  Fe2+(aq) + 2e- E˚red= -0.44V Dissolved oxygen in water usually causes Fe oxidation. Fe2+ initially formed can be further oxidized to Fe3+ which forms rust, Fe2O3 . xH2O(s). CHEM 1033 Chapter 20

Corrosion of Iron CHEM 1033 Chapter 20

Corrosion of Iron Oxidation occurs more readily at the site with the greatest concentration of O2 (water-air junction). Salts produce the necessary electrolyte required to complete the electrical circuit. Preventing Corrosion of Iron Corrosion can be prevented by coating the iron with paint or another metal. Galvanized iron (commonly used for nails) is coated with a thin layer of zinc. CHEM 1033 Chapter 20

Corrosion Prevention: Galvanization Zinc protects iron since Zn is more easily oxidized (E˚redZn is more negative than E˚redFe). Anode: Zn2+(aq) +2e-  Zn(s) E˚red = -0.76 V Cathode: Fe2+(aq) + 2e-  Fe(s) E˚red = -0.44 V Zn is oxidized INSTEAD OF IRON, preventing corrosion. Protects even if the surface coat is broken. CHEM 1033 Chapter 20

Corrosion Prevention: Galvanization Preventing Corrosion of Iron CHEM 1033 Chapter 20

Corrosion Prevention: Sacrificial Anodes To protect underground pipelines. The water pipe is the cathode and a more active metal is used as the anode (cathodic protection). 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 Since E˚red of Mg is more negative, it will oxidize first. CHEM 1033 Chapter 20

Corrosion Prevention: Sacrificial Anodes CHEM 1033 Chapter 20

Electrolysis Defn: Use of electrical energy to drive a non-spontaneous redox reaction. In voltaic and electrolytic cells reduction occurs at the cathode, and oxidation occurs at the anode. However, in electrolytic cells, electrons are forced to flow from the anode to cathode. Anode is now designated as POSITIVE(+). CHEM 1033 Chapter 20

Electrolysis of Molten Salts Used industrially to produce metals such as Na and Al. Requires high temperatures. Example: Electrolysis of molten NaCl: NaCl(s)  Na+(l) + Cl-(l) Two reactants: Na+ and Cl- Cathode: 2Na+(l) + 2e-  2Na(l) E˚red = -2.71V Anode: 2Cl-(l)  Cl2(g) + 2e- E˚red = -1.359V This is clearly a non-spontaneous reaction. E˚cell = CHEM 1033 Chapter 20

Electrolysis of Molten Solutions Electrons still flow from anode to cathode, but are forced. CHEM 1033 Chapter 20

Electrolysis of Aqueous Solutions Aqueous solution electrolysis is complicated by water. Ask: Is water oxidized (to form O2) or reduced (to form H2) in reference to the other components? e.g. For an aqueous solution of NaF in an electrolytic cell, possible reactants are Na+, F- and H2O. Both Na+ and H2O can be reduced but not F-. Thus possible reactions at cathode are: Na+(aq) + e- Na(s) E˚red = -2.71 V 2H2O(l) + 2e- H2(g) + 2OH-(aq) E˚red = -0.83 V CHEM 1033 Chapter 20

Electrolysis of Aqueous Solutions More positive the E˚red favors reduction reactions. Reduction of H2O (E˚red = -0.83V) occurs at cathode with H2 gas produced. At anode, either F- or H2O must be oxidized since Na+ cannot lose additional electrons. 2F-(aq) F2(g) + 2e- E˚red = +2.87 V 4OH-(aq) O2(g) + 2H2O(l) + 4e- E˚red = +0.40 V Thus oxidation of H2O occurs at anode (more negative E˚red favors oxidation reactions). CHEM 1033 Chapter 20

Electrolysis of Aqueous Solutions Cathode 4H2O(l) + 4e- 2H2(g) + 4OH-(aq) E˚red = -0.83 V Anode 4OH-(aq) O2(g) + 2H2O(l) + 4e- E˚red = -0.40 V Process is non-spontaneous (as expected). CHEM 1033 Chapter 20

Electroplating Active electrodes - take part in electrolysis. Example: electrolytic plating. Defn: Electrolysis used to deposit a thin layer of one metal on another in order to improve beauty or resistance to corrosion. e.g. electroplating nickel on a piece of steel. CHEM 1033 Chapter 20

Electroplating Consider an active Ni electrode and another metallic electrode placed in an aqueous solution of NiSO4. Anode: Ni(s)  Ni2+(aq) + 2e- Cathode: Ni2+(aq) + 2e-  Ni(s) With voltage supplied, Ni plates on the steel electrode. Electroplating is important in protecting objects from corrosion (chrome). E˚cell = 0V, so outside source of voltage needed! CHEM 1033 Chapter 20

Quantitative Aspects of Electrolysis How much material can we obtain with electrolysis? Consider the reduction of Cu2+ to Cu(s). Cu2+(aq) + 2e-  Cu(s). 2 mol of electrons will plate 1 mol of Cu The charge of 1 mol of electrons is 96,500 C (=1F). Using Q = It The amount of Cu can be calculated from the current used (I) and time (t) allowed to plate. CHEM 1033 Chapter 20

Calculate the number of grams of aluminum produced in 1 Calculate the number of grams of aluminum produced in 1.00h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A. Example Ans. 3.36g CHEM 1033 Chapter 20

Electrical Work Free-energy is a measure of the maximum amount of useful work that can be obtained from a system. We know If work is negative, then work is performed by the system and E is positive (which means it’s spontaneous) CHEM 1033 Chapter 20

Electrical Work The emf can be thought of as a measure of the driving force for a redox reaction to proceed. In order to drive non-spontaneous, electrolytic reactions, the external, applied emf must be greater than Ecell. If Ecell = -1.35 V, then external emf must be at least +1.36 V. Note: work can be expressed in Watts (W) 1 W = 1 J/s. CHEM 1033 Chapter 20