Electrochemistry Chapter 20 Fall 2024 Lecture Notes PDF

Electrochemistry Chapter 20 Outline How are redox reactions balanced? 20.1 Oxidation States and Oxidation-Reduction Reactions (REVIEW) 20.2 Balancing Redox Equations (REVIEW-ish) How is electricity produced from a spontaneous redox reaction? 20.3 Voltaic Cells 20.4 Cell Potentials...

Electrochemistry Chapter 20 Outline How are redox reactions balanced? 20.1 Oxidation States and Oxidation-Reduction Reactions (REVIEW) 20.2 Balancing Redox Equations (REVIEW-ish) How is electricity produced from a spontaneous redox reaction? 20.3 Voltaic Cells 20.4 Cell Potentials under Standard Conditions How is the spontaneity of a redox reaction determined? 20.5 Free Energy and Redox Reactions 20.6 Cell Potentials under Nonstandard Conditions What are some applications of electrochemistry? 20.7 Batteries and Fuel Cells 20.8 Corrosion BRIEFLY 20.9 Electrolysis ENERGY Non renewables Vs Renewables? Non Renewables Renewables - Solar, Wind power - Petroleum Oil spills 😞 Eg- Deepwater Horizon 2010 Intermittent in nature. - Energy generated during peak times must be stored. https://response.restoration.noaa.gov/oil-and- - Converted to chemical energy chemical-spills (Check it out! ) https://www.fisheries.noaa.gov/topic/offshore- - Gas, Coal etc.. The Guyandotte River Spill 2011 wind-energy (Check it out! ) 😞 https://www.epa.gov/hfstudy How are redox reactions balanced? Sections 20.1 – 20.2 Electrochemistry Why is it important? The electricity that powers much of modern society cannot be easily stored. Used as it is generated. Electrical energy can be converted to chemical energy which can be stored AND is portable. Electrochemistry is the study of the relationships between electricity and chemical reactions. Electrochemistry Electrochemical processes are oxidation-reduction reactions in which: The energy released by a spontaneous redox reaction is converted to electricity. OR Electrical energy is used to “drive” a nonspontaneous reaction. 𝒁𝒏 𝒔 + 𝟐 𝑯𝑪𝒍 𝒂𝒒 → 𝒁𝒏𝑪𝒍𝟐 𝒂𝒒 + 𝑯𝟐 (𝒈)  Oxidation-Reduction (Redox) Reactions REVIEW Oxidation Reduction Loss of electrons Gain of electrons Increase in oxidation number Decrease in oxidation number “Reducing agent” “Oxidizing agent” 𝒁𝒏 𝒔 → 𝒁𝒏𝟐+ 𝒂𝒒 + 𝟐 𝒆− 𝟐 𝑯+ 𝒂𝒒 + 𝟐 𝒆− → 𝑯𝟐 (𝒈) H+ reduced oxidizing agent 𝒁𝒏 𝒔 + 𝟐 𝑯𝑪𝒍 𝒂𝒒 → 𝒁𝒏𝑪𝒍𝟐 𝒂𝒒 + 𝑯𝟐 (𝒈) Zn oxidized 0 +1 +2 0 reducing agent Review Section 4.4 in your eBook Oxidation States As opposed to the formation of cations and anions, MOST oxidation- reduction reactions do NOT involve the complete transfer of electrons from one atom to another. For covalent compounds, the oxidation number (ON) or oxidation state of each atom in the molecule is simply the “charge” the atom would have if the electrons were completely transferred. With oxidation-reduction of molecular (covalent) compounds, there is a change in the “unequal” sharing of electrons. Oxidation States Rules for Assigning Oxidation Numbers Oxidation Number 1. Atoms of pure elements have an ON of zero. (ON)—a number 2. Monatomic ions have an ON equal to their assigned to each atom common charge. in a compound to 3. Nonmetals usually have negative ONs, but represent the “charge” the ON can be positive. each would have if the F is always -1. electrons were divided H is +1 when bound to a non-metal and -1 when bound to a metal. among the atoms. O is -2 in almost all compounds except in peroxides (-1) and when bound to F (+2) Rules for assigning 4. The sum of the ONs in a neutral compound oxidation numbers can must equal zero. The sum of ONs in a be found in Section 4.4. polyatomic ion must equal the overall charge.  Concept Check Oxidation Numbers What is the oxidation number on nitrogen in each of the following? Enter each oxidation number as a whole number. 1. Atoms of pure elements have an ON Include the sign of the charge before the number. of zero. N2O5 2. Monatomic ions have an ON equal to their common charge. 3. Nonmetals usually have negative NO2— ONs, but the ON can be positive. F is always -1. H is +1 when bound to a non-metal and - N2 1 when bound to a metal. O is -2 in almost all compounds except in peroxides (-1) and when bound to F (+2) Na3N 4. The sum of the ONs in a neutral compound must equal zero. The sum of ONs in a polyatomic ion must equal the overall charge.  Concept Check Oxidation Numbers What is the oxidation number on nitrogen in each of the following? Enter each oxidation number as a whole number. 1. Atoms of pure elements have an ON Include the sign of the charge before the number. of zero. N2O5 2. Monatomic ions have an ON equal to their common charge. +5 3. Nonmetals usually have negative NO2— ONs, but the ON can be positive. F is always -1. H is +1 when bound to a non-metal and - N2 1 when bound to a metal. O is -2 in almost all compounds except in peroxides (-1) and when bound to F (+2) Na3N 4. The sum of the ONs in a neutral compound must equal zero. The sum of ONs in a polyatomic ion must equal the overall charge.  Concept Check Redox Reaction 𝟐 𝑷𝒃𝑺 𝒔 + 𝟑 𝑶𝟐 𝒈 → 𝟐 𝑷𝒃𝑶 𝒔 + 𝟐 𝑺𝑶𝟐 𝒈 Match. What is being oxidized? Reduced? What is the oxidizing agent? Reducing agent?  Concept Check Redox Reaction 𝑨𝒔𝟐 𝑶𝟑 𝒔 + 𝑵𝑶𝟑 − 𝒂𝒒 → 𝑯𝟑 𝑨𝒔𝑶𝟒 𝒂𝒒 + 𝑵𝑶 𝒈 Match. What is being oxidized? Reduced? What is the oxidizing agent? Reducing agent? Balancing Redox Equations Half-Reactions Although oxidation and reduction take place simultaneously, it is often useful to consider them as separate processes. The equations written for each separate process are called half- reactions. ̶ Electrons are products in an oxidation half-reaction. 𝒁𝒏 𝒔 → 𝒁𝒏𝟐+ 𝒂𝒒 + 𝟐 𝒆− ̶ Electrons are reactants in a reduction half-reaction. 𝑪𝒖𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑪𝒖 (𝒔) Law of conservation of mass must be obeyed when balancing half- reactions. Gains and losses of electrons must also be balanced. Balancing Redox Equations Half-Reactions Metals at the top are more easily oxidized. REVIEW: Consider the following MOST ACTIVE pairs and determine reaction spontaneity. Zn (s) + Cu2+ (aq) → 𝒁𝒏𝟐+ 𝒂𝒒 + 𝑪𝒖 (𝒔) spontaneous Cu (s) + Zn2+ (aq) → 𝒏𝒐 𝒓𝒆𝒂𝒄𝒕𝒊𝒐𝒏 nonspontaneous Metals at the bottom are more The reverse reaction can be made to occur with easily reduced. electric current but they are not spontaneous. LEAST ACTIVE Balancing Redox Equations Electron Transfer Example: Spontaneous redox reaction involving zinc and copper. 𝒁𝒏 𝒔 + 𝑪𝒖𝟐+ 𝒂𝒒 → 𝑪𝒖 𝒔 + 𝒁𝒏𝟐+ (𝒂𝒒) Electrons transfer at the surface of the Zn metal strip. The energy released in this process can be used to perform electrical work.  Concept Check Writing Half-Reactions What are the correct half−reactions when Al (s) reacts with CuCl2 (aq)? Hint: First write the complete redox reaction for Al (s) and CuCl2 (aq). Balancing Redox Equations Balancing Equations by the Method of Half-Reaction 1. Make two half-reactions (oxidation and reduction). Assign ON to determine this. 2. Balance: all atoms other than O and H first O using H2O H using H+ charges using electrons 3. Multiply each half-reaction according to a common factor to make the electrons in each half-reaction equal. 4. Add half-reactions together to get an overall reaction then simplify by removing chemical species that occur on both sides of the equation and/or dividing by a common multiple if coefficients need to be reduced. If solving under acidic conditions, then go on to Step 5. If solving under basic conditions, add OH⎯ to each side of the equation to neutralize the H+ in the equation and create H2O in its place. If this produces water on both sides, simplify the equation by removing water from each side until it only appears on one side of the equation. 5. Double-check atoms and charges balance!  Balancing Redox Equations Balancing Equations by the Method of Half-Reaction Example: Consider the reaction between permanganate and oxalate. 1. Make two half-reactions (oxidation and reduction). 2. Balance atoms other than O and H. Then, balance O and H using H2O/H+. Add electrons to balance charges. 3. Multiply by a common factor to make electrons in half-reactions equal. 4. Add half-reactions and simplify by dividing by common factor 𝑴𝒏𝑶𝟒 − 𝒂𝒒 + 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝑴𝒏𝟐+ 𝒂𝒒 + 𝑪𝑶𝟐 (𝒂𝒒) OR converting H+ to OH— if basic. 5. Double-check atoms and charges balance! 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝑪𝑶𝟐 (𝒂𝒒) 𝑴𝒏𝑶𝟒 − 𝒂𝒒 → 𝑴𝒏𝟐+ 𝒂𝒒  Balancing Redox Equations Balancing Equations by the Method of Half-Reaction Example: Consider the reaction between permanganate and oxalate. 1. Make two half-reactions (oxidation and reduction). 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝑪𝑶𝟐 (𝒂𝒒) 𝑴𝒏𝑶𝟒 − 𝒂𝒒 → 𝑴𝒏𝟐+ 𝒂𝒒 2. Balance atoms other than O and 𝟐− The Mn is balanced; to H. Then, balance O and H using 𝑪𝟐 𝑶𝟒 𝒂𝒒 → 𝟐 𝑪𝑶𝟐 (𝒂𝒒) balance the O, add FOUR H2O/H+. Add electrons to waters to the right side. balance charges. 𝑴𝒏𝑶𝟒 − 𝒂𝒒 → 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟒 𝑯𝟐 𝑶 (𝒍) 3. Multiply by a common factor to make electrons in half-reactions To balance H, add 8 H+ to the left side. equal. 4. Add half-reactions and simplify 𝟖 𝑯+ 𝒂𝒒 + 𝑴𝒏𝑶𝟒 − 𝒂𝒒 → 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟒 𝑯𝟐 𝑶 (𝒍) by dividing by common factor Now add ELECTRONS to balance out the charges! + — OR converting H to OH if basic. 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝟐 𝑪𝑶𝟐 𝒂𝒒 + 𝟐 𝒆− 5. Double-check atoms and charges balance! 𝟓 𝒆− + 𝟖 𝑯+ 𝒂𝒒 + 𝑴𝒏𝑶𝟒 − 𝒂𝒒 → 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟒 𝑯𝟐 𝑶 (𝒍)  Balancing Redox Equations Balancing Equations by the Method of Half-Reaction Example: Consider the reaction between permanganate and oxalate. 1. Make two half-reactions (oxidation and reduction). 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝟐 𝑪𝑶𝟐 𝒂𝒒 + 𝟐 𝒆− 2. − Balance atoms other than O and 𝟓 𝒆 + 𝟖 𝑯 + 𝒂𝒒 + 𝑴𝒏𝑶𝟒 − 𝒂𝒒 → 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟒 𝑯𝟐 𝑶 (𝒍) H. Then, balance O and H using H2O/H+. Add electrons to To combine the two half-reactions, the number of electrons must be balance charges. EQUAL on both sides. Multiply the first reaction by 5 and the second by 2: 3. Multiply by a common factor to make electrons in half-reactions 𝟓 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝟏𝟎 𝑪𝑶𝟐 𝒂𝒒 + 𝟏𝟎 𝒆− equal. 𝟏𝟎 𝒆 + 𝟏𝟔 𝑯+ − 𝒂𝒒 + 𝟐 𝑴𝒏𝑶𝟒 − 𝒂𝒒 → 𝟐 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟖 𝑯𝟐 𝑶 (𝒍) 4. Add half-reactions and simplify by dividing by common factor OR converting H+ to OH— if basic. 5. Double-check atoms and charges balance!  Balancing Redox Equations Balancing Equations by the Method of Half-Reaction Example: Consider the reaction between permanganate and oxalate. 1. Make two half-reactions 𝟓 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝟏𝟎 𝑪𝑶𝟐 𝒂𝒒 + 𝟏𝟎 𝒆− (oxidation and reduction). 𝟏𝟎 𝒆− + 𝟏𝟔 𝑯+ 𝒂𝒒 + 𝟐 𝑴𝒏𝑶𝟒 − 𝒂𝒒 → 𝟐 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟖 𝑯𝟐 𝑶 (𝒍) 2. Balance atoms other than O and H. Then, balance O and H using Add together to get: + H2O/H. Add electrons to balance charges. 𝟏𝟎 𝒆− + 𝟏𝟔 𝑯+ 𝒂𝒒 + 𝟐 𝑴𝒏𝑶𝟒 − 𝒂𝒒 + 𝟓 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 3. Multiply by a common factor to make electrons in half-reactions equal. 𝟐 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟖 𝑯𝟐 𝑶 𝒍 + 𝟏𝟎 𝑪𝑶𝟐 𝒂𝒒 + 𝟏𝟎 𝒆− 4. Add half-reactions and simplify Simplify by cancelling out substances that appear on BOTH sides of the by dividing by common factor arrow. Can you reduce the coefficients? + — OR converting H to OH if basic. 𝟏𝟔 𝑯+ 𝒂𝒒 + 𝟐 𝑴𝒏𝑶𝟒 − 𝒂𝒒 + 𝟓 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 5. Double-check atoms and charges balance! 𝟐 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟖 𝑯𝟐 𝑶 𝒍 + 𝟏𝟎 𝑪𝑶𝟐 𝒂𝒒  Balancing Redox Equations Balancing Equations by the Method of Half-Reaction Example: Consider the reaction between permanganate and oxalate. 1. Make two half-reactions Verify that the equation is balanced by counting atoms and charges on (oxidation and reduction). each side of the equation. 2. Balance atoms other than O and 𝟏𝟔 𝑯+ 𝒂𝒒 + 𝟐 𝑴𝒏𝑶𝟒 − 𝒂𝒒 + 𝟓 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → H. Then, balance O and H using H2O/H+. Add electrons to balance charges. 𝟐 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟖 𝑯𝟐 𝑶 𝒍 + 𝟏𝟎 𝑪𝑶𝟐 𝒂𝒒 3. Multiply by a common factor to make electrons in half-reactions equal. 4. Add half-reactions and simplify What about balancing in a BASIC solution? by dividing by common factor OR converting H+ to OH— if basic. 5. Double-check atoms and charges balance!  Balancing Redox Equations Balancing Equations by the Method of Half-Reaction A reaction that occurs in a BASIC solution can be balanced as if it occurred in acid. Once the equation is balanced (at Step 4), add OH— TO EACH SIDE to “neutralize” the H+ in the equation and create water in its place. If this produces water on both sides, subtract water from each side so it appears on only one side of the equation. 𝟏𝟔 𝑯+ 𝒂𝒒 + 𝟐 𝑴𝒏𝑶𝟒 − 𝒂𝒒 + 𝟓 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝟐 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟖 𝑯𝟐 𝑶 𝒍 + 𝟏𝟎 𝑪𝑶𝟐 𝒂𝒒 Add 16 OH— 8 𝑯𝟐 𝑶 𝒍 + 𝟐 𝑴𝒏𝑶𝟒 − 𝒂𝒒 + 𝟓 𝑪𝟐 𝑶𝟒 𝟐− 𝒂𝒒 → 𝟏𝟔 𝟐 𝑴𝒏𝟐+ 𝒂𝒒 + 𝟖 𝑯𝟐 𝑶 𝒍 + 𝟏𝟎 𝑪𝑶𝟐 𝒂𝒒 + 𝟏𝟔 𝑶𝑯− (𝒂𝒒)  Concept Check Balancing Redox Reactions (Acidic) 𝑨𝒔𝟐 𝑶𝟑 𝒔 + 𝑵𝑶𝟑 − 𝒂𝒒 → 𝑯𝟑 𝑨𝒔𝑶𝟒 𝒂𝒒 + 𝑵𝑶 𝒈 1. Make two half-reactions (oxidation and reduction). 2. Balance atoms other than O and H. Then, balance O and H using H2O/H+. Add electrons to balance charges. 3. Multiply by a common factor to make electrons in half-reactions equal. 4. Add half-reactions and simplify by dividing by common factor OR converting H+ to OH— if basic. 5. Double-check atoms and charges balance!  Concept Check Balancing Redox Reactions (Acidic) 𝑨𝒔𝟐 𝑶𝟑 𝒔 + 𝑵𝑶𝟑 − 𝒂𝒒 → 𝑯𝟑 𝑨𝒔𝑶𝟒 𝒂𝒒 + 𝑵𝑶 𝒈  Concept Check Balancing Redox Reactions (Acidic) 𝑨𝒔𝟐 𝑶𝟑 𝒔 + 𝑵𝑶𝟑 − 𝒂𝒒 → 𝑯𝟑 𝑨𝒔𝑶𝟒 𝒂𝒒 + 𝑵𝑶 𝒈  Concept Check Balancing Redox Reactions (Acidic) 𝑨𝒔𝟐 𝑶𝟑 𝒔 + 𝑵𝑶𝟑 − 𝒂𝒒 → 𝑯𝟑 𝑨𝒔𝑶𝟒 𝒂𝒒 + 𝑵𝑶 𝒈 How is electricity produced from a spontaneous redox reaction? Sections 20.3 – 20.4 Voltaic Cell Voltaic cell (or galvanic cell) is an electrochemical cell that uses a spontaneous reaction to generate electricity. Two isolated half-reactions connected by a salt bridge or other porous barrier. Conductive solids are used for each electrode. Electrons flow from one electrode to the other via an external wire, forming an external circuit.  Voltaic Cell Cathode Reduction Cations flow toward cathode Anode Oxidation Anions flow toward anode The ions migrate to balance the charge. 𝒁𝒏 𝒔 + 𝑪𝒖𝟐+ 𝒂𝒒 → 𝑪𝒖 𝒔 + 𝒁𝒏𝟐+ (𝒂𝒒) Voltaic Cell 𝒁𝒏 𝒔 + 𝑪𝒖𝟐+ 𝒂𝒒 → 𝑪𝒖 𝒔 + 𝒁𝒏𝟐+ (𝒂𝒒) https://www.youtube.com/watch?v=J1ljxodF9_g Voltaic Cell Components Summary of the Components of a Voltaic Cell: An Ox and a Red Cat Electrons flow spontaneously from anode to cathode. ̶ Oxidation occurs at the anode ( ̶ ). ̶ Reduction occurs at the cathode ( + ). ̶ We expect the anode to lose mass and the cathode to gain mass. The voltage of the cell is positive. The salt bridge provides mobile ions required to maintain electrical neutrality in each half-cell. ̶ Anions flow toward the anode. ̶ Cations flow toward the cathode. ̶ What would happen if the salt bridge was removed? Voltaic Cells Cell Diagram Cell diagram is used to represent the chemical reactions in a voltaic cell. Example: For the redox reaction: 𝒁𝒏 𝒔 + 𝑪𝒖𝟐+ 𝒂𝒒 → 𝑪𝒖 𝒔 + 𝒁𝒏𝟐+ (𝒂𝒒) with: [Cu2+] = 1 M and [Zn2+] = 1 M Single vertical line indicates phase boundary We write: 𝒁𝒏 𝒔 𝒁𝒏𝟐+ 𝟏 𝑴 || 𝑪𝒖𝟐+ 𝟏 𝑴 𝑪𝒖 (𝒔) Electrodes are anode cathode shown on ends Double vertical lines indicate salt bridge  Concept Check Voltaic Cell Consider the following voltaic cell for: 𝑭𝒆 𝒔 𝑭𝒆𝟐+ 𝟏 𝐌 || 𝑪𝒖𝟐+ 𝟏 𝑴 𝑪𝒖 (𝒔) 0.972 where the voltage is +0.972 V. Which of the following statements is true? A. The iron electrode is the cathode. B. Cations migrate through the salt bridge to the compartment with the iron electrode. C. The iron electrode will gain mass. D. Electrons flow from the iron electrode to the copper electrode.  Cell Potential Cell potential (𝑬𝒄𝒆𝒍𝒍 ) is the potential difference between two electrodes. Electromotive force (emf) Anode is higher in potential energy. Cathode is lower in potential energy. Measured in volts (V). Also called voltage 1 V: The potential difference required ANODE to impart 1 Joule of energy to a charge of 1 coulomb. 𝑱 CATHODE 𝟏𝑽=𝟏 𝑪 Cell Potential Cell potential (𝑬𝒄𝒆𝒍𝒍 ) is the potential difference between two electrodes. Magnitude is always positive. Magnitude depends on: 1. half-reactions 2. concentrations of reactants/products 3. temperature Standard cell potential (𝑬°𝒄𝒆𝒍𝒍 ) requires standard conditions. ̶ 298 K ̶ 1 M concentrations OR 1 atm pressure Cell Potential under Standard Conditions Calculating E°cell 𝑬°𝒄𝒆𝒍𝒍 = 𝑬°𝒓𝒆𝒅 𝒄𝒂𝒕𝒉𝒐𝒅𝒆 − 𝑬°𝒓𝒆𝒅 (𝒂𝒏𝒐𝒅𝒆) where 𝑬°𝒓𝒆𝒅 is the standard reduction potential for the electrode Reduction potential is the tendency for a species to be reduced (gain electrons) in a reduction half-reaction. Standard reduction potential (𝑬°𝒓𝒆𝒅 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm at 25°C. Cell Potential under Standard Conditions Reduction Potentials (E°red) Large +𝑬°𝒓𝒆𝒅 values easily reduced, good oxidizing agents Large −𝑬°𝒓𝒆𝒅 values easily oxidized, good reducing agents Because electrical potential measures potential energy per electrical charge, standard reduction potential is an INTENSIVE property. Standard Reduction Potentials (E°red) SHE By convention, standard reduction potentials are listed relative to the: Standard Hydrogen Electrode 𝑬°𝒄𝒆𝒍𝒍 = 𝟎 Reduction reaction (E°red = 0 V): 𝟐 𝑯+ 𝒂𝒒 + 𝟐 𝒆− → 𝑯𝟐 𝒈 1M 1 atm A platinum electrode provides an inert surface for the reaction. SHE can operate either as the anode or the cathode. Standard Reduction Potentials (E°red) 𝑬°𝒄𝒆𝒍𝒍 = +𝟎. 𝟕𝟔 𝑽 SHE Other 𝑬°𝒓𝒆𝒅 values are relative to 𝑬°𝒓𝒆𝒅 for the SHE. Example: The measured cell potential (𝐸°𝑐𝑒𝑙𝑙 ) for: 𝒁𝒏 𝒔 + 𝟐 𝑯+ 𝒂𝒒 → 𝒁𝒏𝟐+ 𝒂𝒒 + 𝑯𝟐 𝒈 is +0.76 V. What is the standard reduction potential for the cathode and the anode? 𝑬°𝒄𝒆𝒍𝒍 = 𝑬°𝒓𝒆𝒅 𝒄𝒂𝒕𝒉𝒐𝒅𝒆 − 𝑬°𝒓𝒆𝒅 (𝒂𝒏𝒐𝒅𝒆) 𝟐 𝑯+ 𝒂𝒒 + 𝟐 𝒆− → 𝑯𝟐 (𝒈) 𝑬°𝒓𝒆𝒅 𝒂𝒏𝒐𝒅𝒆 = 𝑬°𝒓𝒆𝒅 𝒄𝒂𝒕𝒉𝒐𝒅𝒆 − 𝑬°𝒄𝒆𝒍𝒍 SHE (𝑬°𝒓𝒆𝒅 = 𝟎 𝑽) 𝒁𝒏 𝒔 → 𝒁𝒏𝟐+ 𝒂𝒒 + 𝟐 𝒆− 𝑬°𝒓𝒆𝒅 𝒂𝒏𝒐𝒅𝒆 = 𝟎 𝑽 − 𝟎. 𝟕𝟔 𝑽 𝑬°𝒓𝒆𝒅 𝒂𝒏𝒐𝒅𝒆 = −𝟎. 𝟕𝟔 𝑽 Standard Reduction Potentials (E°red) Relative Strengths SHE Standard Reduction Potentials (E°red) Relative Strengths The more positive the value of 𝐸°𝑟𝑒𝑑 , the greater the tendency for reduction under standard conditions. 𝒁𝒏 𝒔 + 𝑪𝒖𝟐+ 𝒂𝒒 → 𝑪𝒖 𝒔 + 𝒁𝒏𝟐+ (𝒂𝒒) 𝑬°𝒄𝒆𝒍𝒍 = 𝑬°𝒓𝒆𝒅 𝒄𝒂𝒕𝒉𝒐𝒅𝒆 − 𝑬°𝒓𝒆𝒅 (𝒂𝒏𝒐𝒅𝒆)  Concept Check Relative Strength of Reduction Potentials Which of the following substances would be oxidized by Ni2+ (aq)? Hint: Refer to the table of standard reduction potentials provided. A. Cu2+ (aq) B. Cu (s) C. Cr (s) D. Cr3+ (aq) E. Pb (s)  Concept Check Relative Strength of Reduction Potentials Which of the following is the strongest oxidizing agent among the following substances? Hint: Refer to the table of standard reduction potentials provided. A. Cu2+ (aq) B. Cu (s) C. Cr (s) D. Cr3+ (aq) E. Pb (s) Cell Potential under Standard Conditions Summary To calculate 𝑬°𝒄𝒆𝒍𝒍 for any combination of two half-reactions: Determine which half-reaction is the cathode (reduction) reaction. ̶ The half-cell with the more positive value (less negative) will form the cathode (reduction) half-cell. Determine which half-reaction is the anode (oxidation) reaction. Subtract the anode half-cell potential from the cathode half-cell potential. 𝑬°𝒄𝒆𝒍𝒍 = 𝑬°𝒓𝒆𝒅 𝒄𝒂𝒕𝒉𝒐𝒅𝒆 − 𝑬°𝒓𝒆𝒅 (𝒂𝒏𝒐𝒅𝒆) ̶ If set up correctly, the 𝑬°𝒄𝒆𝒍𝒍 value will be positive, indicating the redox reaction will be spontaneous. ̶ Stoichiometric coefficients DO NOT change 𝑬°𝒄𝒆𝒍𝒍.  Concept Check Calculating E°cell A voltaic cell is based on the two standard half-reactions: 𝑪𝒅𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑪𝒅 𝒔 𝑬°𝒓𝒆𝒅 = −𝟎. 𝟒𝟎 𝑽 𝑺𝒏𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑺𝒏 𝒔 𝑬°𝒓𝒆𝒅 = −𝟎. 𝟏𝟒 𝑽 Which reaction occurs at the cathode? Which reaction occurs at the anode? What is the standard cell potential (in V)? How is the spontaneity of a redox reaction determined? Sections 20.5 – 20.6 Free Energy and Redox Reactions Determining Spontaneity Consider the equation for standard cell potential: 𝑬°𝒄𝒆𝒍𝒍 = 𝑬°𝒓𝒆𝒅 𝒄𝒂𝒕𝒉𝒐𝒅𝒆 − 𝑬°𝒓𝒆𝒅 (𝒂𝒏𝒐𝒅𝒆) Since the standard cell potential for a voltaic cell is positive, then: 𝑬°𝒄𝒆𝒍𝒍 > 0 spontaneous 𝑬°𝒄𝒆𝒍𝒍 < 0 nonspontaneous If E°cell is a negative value, the reaction is spontaneous in the reverse direction (as written). Free Energy and Redox Reactions Determining Spontaneity Example: Determine the standard cell potential (𝑬°𝒄𝒆𝒍𝒍 ) of the spontaneous reaction given the information below. 𝑭𝒆𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑭𝒆 𝒔 𝑬°𝒓𝒆𝒅 = −𝟎. 𝟒𝟒 𝑽 𝑪𝒖𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑪𝒖 𝒔 𝑬°𝒓𝒆𝒅 = +𝟎. 𝟑𝟒𝑽 𝑬°𝒄𝒆𝒍𝒍 = 𝑬°𝒓𝒆𝒅 𝒄𝒂𝒕𝒉𝒐𝒅𝒆 − 𝑬°𝒓𝒆𝒅 (𝒂𝒏𝒐𝒅𝒆) = (+𝟎. 𝟑𝟒 𝑽) − (−𝟎. 𝟒𝟒 𝑽) Which is the reduction half-reaction? = +𝟎. 𝟕𝟖 𝑽 Which is the oxidation half-reaction? Since the 𝑬°𝒄𝒆𝒍𝒍 is positive, this reaction 𝑪𝒖𝟐+ 𝒂𝒒 + 𝑭𝒆 𝒔 → 𝑪𝒖 𝒔 + 𝑭𝒆𝟐+ 𝒂𝒒 is spontaneous. If the values were reversed, then the reaction would be nonspontaneous. Free Energy and Redox Reactions We have already seen three criteria for spontaneity: 1. ∆𝑮° < 𝟎 2. 𝑲 ≫ 𝟏 3. ∆𝑬° > 𝟎 We have already been introduced to the relationship between free energy and the equilibrium constant: ∆𝑮° = −𝑹𝑻 ln 𝑲 Now, we can numerically equate cell potential of a voltaic cell with the free energy for that redox reaction: ∆𝑮° = −𝒏𝑭𝑬°𝒄𝒆𝒍𝒍 Free Energy and Redox Reactions ∆𝑮° = −𝒏𝑭𝑬°𝒄𝒆𝒍𝒍 n = number of moles of electrons transferred in the redox rxn 𝐽 F = 96,500 (Faraday’s constant) 𝑉∙𝑚𝑜𝑙 Example: What is the standard free energy change (in kJ) for this reaction? 𝒁𝒏 𝒔 + 𝑷𝒃𝟐+ 𝒂𝒒 → 𝒁𝒏𝟐+ 𝒂𝒒 + 𝑷𝒃 𝒔 𝑬°𝒄𝒆𝒍𝒍 = +𝟎. 𝟔𝟑 𝑽 𝐽 ∆𝑮° = −𝒏𝑭𝑬° = −(𝟐 𝒎𝒐𝒍) 96,500 (𝟎. 𝟔𝟑 𝑽) = −𝟏𝟐𝟐, 𝟎𝟎𝟎 𝑱 𝑉 ∙ 𝑚𝑜𝑙 = −𝟏𝟐𝟐 𝒌𝑱  Concept Check Standard Free Energy Calculate ΔG° (in kJ) for the following reaction: 𝑷𝒃 𝒔 + 𝟐 𝑯+ 𝒂𝒒 → 𝑷𝒃𝟐+ 𝒂𝒒 + 𝑯𝟐 (𝒈) Hint: Refer to the table of standard reduction potentials provided. Round your answer to ONE place past the decimal. If your answer is negative, include the sign. Free Energy and Redox Reactions Equilibrium Constant Recall the two equations for spontaneity: ∆𝑮° = −𝑹𝑻 ln 𝑲 ∆𝑮° = −𝒏𝑭𝑬°𝒄𝒆𝒍𝒍 Combine them to relate standard cell potential to the equilibrium constant: 𝑹𝑻 𝑬°𝒄𝒆𝒍𝒍 = ln 𝑲 𝒏𝑭 𝑱 n = number of moles of electrons transferred 𝑹 = 𝟖. 𝟑𝟏𝟒 𝒎𝒐𝒍 ∙ 𝑲 in the redox rxn 𝐽 𝑻 = 𝒂𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆 (𝒊𝒏 𝑲) F = 96,500 (Faraday’s constant) 𝑉∙𝑚𝑜𝑙 Free Energy and Redox Reactions Equilibrium Constant 𝟎. 𝟎𝟐𝟓𝟕 𝑽 𝑬°𝒄𝒆𝒍𝒍 = ln 𝑲 𝒏 𝑹𝑻 𝑬°𝒄𝒆𝒍𝒍 = ln 𝑲 At 25°C 𝒏𝑭 𝟎. 𝟎𝟓𝟗𝟐 𝑽 𝑬°𝒄𝒆𝒍𝒍 = log 𝑲 𝒏  Concept Check E°cell and K What is K for the following reaction at 25°C? 𝒁𝒏 𝒔 + 𝑪𝒖𝟐+ 𝒂𝒒 → 𝑪𝒖 𝒔 + 𝒁𝒏𝟐+ 𝒂𝒒 𝑬°𝒄𝒆𝒍𝒍 = +𝟏. 𝟏𝟎 𝑽 𝟎. 𝟎𝟓𝟗𝟐 𝑽 𝑬°𝒄𝒆𝒍𝒍 = log 𝑲 𝒏 Free Energy and Redox Reactions Relationships between ΔG°, K and E°cell 𝑬°𝒄𝒆𝒍𝒍 𝑲 ∆𝑮° Reaction 𝑬°𝒄𝒆𝒍𝒍 > 𝟎 K>1 ∆𝑮° < 0 spontaneous 𝑬°𝒄𝒆𝒍𝒍 < 𝟎 K 0 nonspontaneous 𝑬°𝒄𝒆𝒍𝒍 = 𝟎 K=1 ∆𝑮° = 0 equilibrium Keep in mind that 𝐸°𝑐𝑒𝑙𝑙 and ΔG° are determined/calculated before equilibrium is established. K represents the ratio of products-to-reactants at equilibrium.  Cell Potential Under Nonstandard Conditions Recall the following three equations: ∆𝑮 = ∆𝑮° + 𝑹𝑻 ln 𝑸 ∆𝑮° = −𝒏𝑭𝑬°𝒄𝒆𝒍𝒍 ∆𝑮 = −𝒏𝑭𝑬𝒄𝒆𝒍𝒍 Combine them to obtain the Nernst Equation: 𝑹𝑻 𝑬𝒄𝒆𝒍𝒍 = 𝑬°𝒄𝒆𝒍𝒍 − ln 𝑸 𝒏𝑭 𝟎. 𝟎𝟐𝟓𝟕 𝟎. 𝟎𝟓𝟗𝟐 At 25°C 𝑬𝒄𝒆𝒍𝒍 = 𝑬°𝒄𝒆𝒍𝒍 − ln 𝑸 𝑬𝒄𝒆𝒍𝒍 = 𝑬°𝒄𝒆𝒍𝒍 − log 𝑸 𝒏 𝒏 Cell Potential Under Nonstandard Conditions Nernst Equation At 25°C, 𝑹𝑻 𝑬𝒄𝒆𝒍𝒍 = 𝑬°𝒄𝒆𝒍𝒍 − ln 𝑸 𝒏𝑭 Before the cell starts (standard state) Q=1 𝐸𝑐𝑒𝑙𝑙 = 𝐸°𝑐𝑒𝑙𝑙 After the cell starts (spontaneous forward) Q>1 𝐸𝑐𝑒𝑙𝑙 < 𝐸°𝑐𝑒𝑙𝑙 When the system reaches equilibrium Q=K 𝐸𝑐𝑒𝑙𝑙 = 0 (The cell has “run down”) 𝑹𝑻 𝑬°𝒄𝒆𝒍𝒍 = ln 𝑲 𝒏𝑭  Concept Check Nernst Equation The standard cell potential (E°cell) for the reaction below is +1.10 V. 𝒁𝒏 𝒔 + 𝑪𝒖𝟐+ 𝒂𝒒 → 𝑪𝒖 𝒔 + 𝒁𝒏𝟐+ 𝒂𝒒 What is the cell potential (in V) for this reaction when the concentration of [Cu2+] = 1.0 × 10-5 M and [Zn2+] = 3.5 M? Round your answer to TWO places past the decimal. If your answer is negative, include the sign. Cell Potential Under Nonstandard Conditions Nernst Equation 𝑭𝒆 𝒔 + 𝑪𝒅𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑪𝒅 𝒔 + 𝑭𝒆𝟐+ 𝒂𝒒 + 𝟐 𝒆− Oxidation: 𝑭𝒆 𝒔 → 𝑭𝒆𝟐+ 𝒂𝒒 + 𝟐 𝒆− Reduction: 𝑪𝒅𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑪𝒅 𝒔 𝑹𝑻 [𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔] [𝑭𝒆𝟐+ ] 𝑬𝒄𝒆𝒍𝒍 = 𝑬°𝒄𝒆𝒍𝒍 − ln 𝑸 𝑸= = 𝒏𝑭 [𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔] [𝑪𝒅𝟐+ ] IF [products] > [reactants], ln 𝑸 is positive and 𝑬𝒄𝒆𝒍𝒍 < 𝑬°𝒄𝒆𝒍𝒍 IF [products] < [reactants], ln 𝑸 is negative and 𝑬𝒄𝒆𝒍𝒍 > 𝑬°𝒄𝒆𝒍𝒍  Concept Check Nernst Equation and Q Predict how each of the following will change the value of 𝑬𝒄𝒆𝒍𝒍. + 𝟐+ 𝑹𝑻 𝒁𝒏 𝒔 + 𝟐 𝑯 (𝒂𝒒) → 𝒁𝒏 𝒂𝒒 + 𝑯𝟐 (𝒈) 𝑬𝒄𝒆𝒍𝒍 = 𝑬°𝒄𝒆𝒍𝒍 − 𝒏𝑭 ln 𝑸 1. Increasing [Zn2+] 2. Reducing partial pressure of H2 3. Increasing pH 4. Increasing Zn (s) mass Cell Potential Under Nonstandard Conditions Concentration Cells Concentration cell is a voltaic cell in which both half-cells contain the same reactions, but at different concentrations. Dilute cell: 𝑵𝒊 𝒔 → 𝑵𝒊𝟐+ 𝒂𝒒 + 𝟐 𝒆− Concentrated cell: 𝑵𝒊𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑵𝒊 𝒔 Electrons will leave the LOWER concentration cell to go to the HIGHER concentration cell.  Cell Potential Under Nonstandard Conditions Concentration Cells 𝑵𝒊 𝒔 → 𝑵𝒊𝟐+ 𝒂𝒒 + 𝟐 𝒆− 𝑵𝒊𝟐+ 𝒂𝒒 + 𝟐 𝒆− → 𝑵𝒊 𝒔 Too dilute! Too concentrated! Sponteous increase [Ni2+] Sponteneous decrease to reach equilibrium [Ni2+] to reach equilibrium  Concept Check Concentration Cell A voltaic cell was constructed with two Ni2+ - Ni electrodes. The two cell compartments have [Ni2+] = 1.00 M and [Ni2+] = 1.00 × 10−3 M, respectively, where 𝑬°𝒓𝒆𝒅 = −𝟎. 𝟐𝟖 𝑽. Which electrode is the anode of the cell? What is the standard emf of this concentration cell? What is the cell emf for the concentrations given? For the anode compartment, predict whether [Ni2+] will increase, decrease, or stay the same as the cell operates. What are some applications of electrochemistry? Sections 20.7 – 20.9 Applications of Electrochemistry Batteries A battery is a portable, self-contained electrochemical power source that consists of one or more voltaic cells. Batteries are marked with: plus sign (+): for cathode negative sign (─): for anode Voltage is the sum of voltages of individual cells. Primary cells (cannot be recharged when “dead”—the reaction is complete) or secondary cells (can be recharged) Applications of Electrochemistry Non-Rechargeable Batteries Alkaline Battery (~1.5 V) Cathode: 2 𝑀𝑛𝑂2 𝑠 + 2 𝐻2 𝑂 𝑙 + 2 𝑒 − → 2 𝑀𝑛𝑂 𝑂𝐻 𝑠 + 2 𝑂𝐻 − 𝑎𝑞 Anode: 𝑍𝑛 𝑠 + 2 𝑂𝐻 − 𝑎𝑞 → 𝑍𝑛 𝑂𝐻 2 𝑠 + 2 𝑒− The cathode is separated from the anode by a porous fabric. Must be discarded/recycled after voltage drops to zero.  Applications of Electrochemistry Rechargeable Batteries 12-V Lead-Acid (2.05 V/cell) Combustion engine (6 cells) 𝑷𝒃𝑶𝟐 𝒔 + 𝑷𝒃 𝒔 + 𝟐 𝑯𝑺𝑶𝟒 − 𝒂𝒒 + 𝟐 𝑯+ → 𝟐 𝑷𝒃𝑺𝑶𝟒 𝒔 + 𝑯𝟐 𝑶 (𝒍) Alternator recharges the battery Ni-Cd (1.30 V/cell) Used in a series for electronic devices Cd is toxic (recycle the battery) NiMH (similar V to Ni-Cd) Developed to replace Ni-Cd batteries Hybrid (gas/electric) cars Applications of Electrochemistry Rechargeable Batteries Lithium Ion (3.70 V/cell) Cell phones, laptops, electric cars High specific energy density (amount of energy stored per unit mass) ─ Li is a very light element High volumetric energy density (amount of energy stored per unit volume) ─ Li+ has a very large E°red Li ions can be inserted/removed from certain layered solids. Applications of Electrochemistry Hydrogen Fuel Cell 𝟐 𝑯𝟐 𝒈 + 𝑶𝟐 𝒈 → 𝟐 𝑯𝟐 𝑶 (𝒍) Hydrogen is fed to the anode, and oxygen from air is fed to the cathode. A catalyst (often Pt or In) at the anode separates hydrogen molecules into protons and electrons, which take different paths to the cathode. The electrons go through an external circuit, creating a flow of electricity. Applications of Electrochemistry Corrosion Corrosion (or rusting) is a spontaneous redox reaction in which a metal is attacked by some substance in its environment and converted to an unwanted compound. Example: corrosion of iron Reduction Half-Reaction E°red (V) 𝑭𝒆𝟐+ + 2 𝑒 − → 𝐹𝑒 ─0.45 𝑶𝟐 + 4 𝐻+ + 4 𝑒 − → 2 𝐻2 𝑂 +1.23 𝑭𝒆𝟑+ + 𝑒 − → 𝐹𝑒 2+ +0.77 Applications of Electrochemistry Corrosion A more active metal (like Zn in this case) in contact with system acts as a sacrificial anode. The more active metal will be oxidized in place of the metal that needs to be preserved in a process known as cathodic protection. Reduction Half-Reaction E°red (V) 𝒁𝒏𝟐+ + 2 𝑒 − → 𝑍𝑛 ─0.77 𝑭𝒆𝟐+ + 2 𝑒 − → 𝐹𝑒 ─0.45 𝑶𝟐 + 4 𝐻+ + 4 𝑒 − → 2 𝐻2 𝑂 +1.23 Applications of Electrochemistry Electrolysis Electrolysis is the process in which electrical energy is used to drive a nonspontaneous chemical reaction. Processes in an electrolytic cell (an apparatus for carrying out electrolysis) are the reverse of those in a galvanic cell. Applications of Electrochemistry Electrolysis Electrolysis is the basis for the following processes: Recharging batteries Forming metals from natural-occurring compounds Purifying metals Electroplating (plating one metal on another) Examples: Electrolysis of water 2 𝐻2 𝑂 𝑙 → 2 𝐻2 𝑔 + 𝑂2 (𝑔) Producing active metals 2 𝑁𝑎𝐶𝑙 𝑙 → 2 𝑁𝑎 𝑙 + 𝐶𝑙2 (𝑔) Applications of Electrochemistry Electrolysis of Molten Salts 𝟐 𝑵𝒂𝑪𝒍 𝒍 → 𝟐 𝑵𝒂 𝒍 + 𝑪𝒍𝟐 (𝒈) When electricity is applied to a molten binary salt during electrolysis, the cation will be reduced and the anion will be oxidized. If more than one cation is present, only the one with highest reduction potential will be reduced. Similarly, if more than one anion is present, only the one with the lowest reduction potential (or highest oxidation potential) will be oxidized. Electrolysis Quantitative Aspects of Electrolysis Quantitative electrolysis—the amount of a substance produced at an electrode by electrolysis depends on the moles of electrons passed through the cell. To determine the moles of electrons passed, we need to calculate charge. To calculate the charge, multiply the applied 𝑪 𝟏𝑨=𝟏 current and the time that current flows: 𝒔 𝒄𝒉𝒂𝒓𝒈𝒆 𝐶𝑜𝑢𝑙𝑜𝑚𝑏𝑠 = 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝑎𝑚𝑝𝑠 × 𝒕𝒊𝒎𝒆 (𝑠) Because 1 mole of e— is 96,500 C, the number of moles of e— passed through a cell is: 1 𝑚𝑜𝑙 𝑒 − 𝑚𝑜𝑙 𝑒 − = 𝒄𝒉𝒂𝒓𝒈𝒆 (𝐶) × 96,500 𝐶  Concept Check Quantitative Aspects of Electrolysis How much Ca (s) (in g) will be produced in an electrolytic cell of molten CaCl2 if a current of 0.452 A is passed through the cell for 1.5 hours? Round your answer to TWO places past the decimal. 2 𝑚𝑜𝑙 𝑒 − = 1 𝑚𝑜𝑙 𝐶𝑎 𝟐 𝑪𝒍− 𝒍 → 𝑪𝒍𝟐 𝒈 + 𝟐 𝒆− 𝐶 𝑪𝒂𝟐+ 𝒍 + 𝟐 𝒆− → 𝑪𝒂 (𝒔) 0.452 𝐴 = 0.452 𝑠 𝑪𝒂𝟐+ 𝒍 + 𝟐 𝑪𝒍− 𝒍 → 𝑪𝒂 𝒔 + 𝑪𝒍𝟐 (𝒈) 1.5 ℎ𝑟 3600 𝑠 × = 5400 𝑠 1 1 ℎ𝑟 5400 𝑠 0.452 𝐶 1 𝑚𝑜𝑙 𝑒 − 1 𝑚𝑜𝑙 𝐶𝑎 × × × − = 𝟎. 𝟎𝟏𝟐𝟔 𝒎𝒐𝒍 𝑪𝒂 1 1𝑠 96,500 𝐶 2 𝑚𝑜𝑙 𝑒 = 𝟎. 𝟓𝟎 𝒈 𝑪𝒂 END OF CHAPTER 20 SLIDES