OAKTON Community College Electrochemistry

Redox: Transfer of Electrons General College Chemistry I Lecture 4, June 23 2

Redox: Transfer of Electrons General College Chemistry I Lecture 4, June 23 3

Oxidation Numbers What is Chemistry? Oxidation Numbers =Charge Oxidation Numbers What is Chemistry? General College Chemistry I Lecture 4, June 23 4 4

Oxidation Numbers =Charge Redox View Points General College Chemistry I Lecture 4, June 23

Oxidation Numbers =Charge Oxidation Numbers General College Chemistry I Lecture 4, June 23

Recognizing Redox Reactions General College Chemistry I Lecture 4, June 23

Electromagnetic Radiation Types of Electrochemical Cells: Electrolytic Cell Electromagnetic Radiation An electrolytic cell uses electrical energy to drive a nonspontaneous reaction (D G > 0). In the cell reaction, an external source supplies free energy to convert lower energy reactants into higher energy products. Thus, the surroundings do work on the system. Electroplating and the recovery of metals from ores utilize electro- lytic cells.

Electromagnetic Radiation Types of Electrochemical Cells: Voltaic Cell Electromagnetic Radiation A voltaic cell (also called a galvanic cell) uses a spontaneous reaction (D G < 0) to generate electrical energy. In the cell reaction, some of the difference in free energy between higher energy reactants and lower energy products is converted into electrical energy, which operates the load (surroundings)—flashlight, MP3 player, car starter motor, and so forth. Thus, the system does work on the surroundings. All batteries contain voltaic cells.

Electromagnetic Radiation Types of Electrochemical Cells Electromagnetic Radiation Both cells have several similarities: Two electrodes, which conduct the electricity between cell and surroundings, dip into an electrolyte An electrode is identified as either anode or cathode depending on the half-reaction that takes place there: The oxidation half-reaction occurs at the anode; electrons leave the oxidation half-cell at the anode. The reduction half-reaction occurs at the cathode; electrons enter the reduction half-cell at the cathode.

Electromagnetic Radiation Types of Electrochemical Cells Electromagnetic Radiation The spontaneous rxn occurs in the cell e-’s flow from – to + (“get to go where they want to go”) Anode = where ox occurs Cathode = where red occurs Salt bridge prevents charge buildup (which would stop flow)

Electromagnetic Radiation Types of Electrochemical Cells Electromagnetic Radiation Anode and Oxidation start with vowels while the words Cathode and Reduction start with consonants Alphabetically, the A in anode comes before the C in cathode, and the O in oxidation comes before the R in reduction ANode = OXidation AN OX REDuction = CAThode RED CAT

Electromagnetic Radiation A Voltaic Cell Based on the Zinc-Copper Reaction. Electromagnetic Radiation The spontaneous reaction between zinc and copper(II) ion. When zinc metal is placed in a solution of Cu2+ ion (left), zinc is oxidized to Zn2+, and Cu2+ is reduced to copper metal (right). (The very finely divided Cu appears black.)

Electromagnetic Radiation Operation of Voltaic Cell

Electromagnetic Radiation A Voltaic Cell Using Inactive Electrodes Electromagnetic Radiation In the anode half-cell, I- ions are oxidized to solid I2, and the released electrons flow into the graphite electrode (C) and through the wire. From the wire, the electrons enter the graphite electrode in the cathode half-cell and reduce MnO4- ions to Mn2+ ions. A KNO3 salt bridge is used.

Inert Platinum Electrode Used when neither redox species in a half reaction (or electrode) is a neutral metal. (i.e.both are solution species) You used graphite in place of Pt in lab for Fe3+/Fe2+ and I2/I-cells. A cheaper “inert” electrode. Chapter 18, Figure 18.4 Inert Platinum Electrode Neither is a neutral metal

Electromagnetic Radiation the cathode compartment Notation for a Voltaic Cell Electromagnetic Radiation Components of the anode compartment Components of the cathode compartment Phase boundary Half-cells are physically separated half--cell components Half-cell components usually appear in the same order as in the half-reaction. Ions in the salt bridge are not part of the reaction so they are not in the notation.

Electromagnetic Radiation Skillbuilder Draw a diagram, show balanced equations, and write the notation for a voltaic cell that consists of one half-cell with a Cr bar in a Cr(NO3)3 solution, another half-cell with an Ag bar in an AgNO3 solution, and a KNO3 salt bridge. Measurement indicates that the Cr electrode is negative relative to the Ag electrode

Electromagnetic Radiation Skillbuilder: Quick Quiz NOTE: Every electrode compartment has one oxidizing agent and one reducing agent (this pair is called the redox “couple”) If an electrode has Ni(s) and Ni2+ ions in it, which species is the oxidizing agent and which the reducing agent (of the pair)? Oxidizing agent is ___ Ni2+ (b/c it’s “more positive”; it has “room for an electron” Reducing agent is ___ Ni (b/c it’s “more negative”; it has an electron to give)

Cell Potential Why Do Electrons Flow? Electrode in each half-cell have its own individual potential = standard electrode potential The overall standard cell potential Eºcell is the difference between the two standard electrode potentials: Eºcell = Eºfinal – Eºinitial therefore, Eºcell = Eºcathode - Eºanode

Electromagnetic Radiation “Competition for the electrons”. Which oxidizing agent “wants them more”? Electromagnetic Radiation Who is the (possible) oxidizing agent on the left? _____ Who is the (possible) oxidizing agent on the right? _____ Zn2+ Cu2+ Hint: The two “players” are Zn and Zn2+ Hint: The two “players” are Cu and Cu2+. Who wins? (Which one “got” the electrons?) ____ Cu2+  Cu2+ “pulled harder” Cu2+ + 2 e-  Cu(s) So…which of the half reactions shown at the right is more favorable (greater tendency to happen)? Zn2+ + 2 e-  Zn(s) By how much?.....

Electromagnetic Radiation Where reduction takes place Standard Reduction Potential Electromagnetic Radiation Reducing Cu2+ is more favorable than reducing Zn2+ …by 1.10 V! (Measure it w/voltmeter!) We define a “standard reduction potential”, E°red, for every reduction half-reaction such that: E°cell = E°red(cathode) - E°red(anode) Where reduction takes place The more positive the “E” (Ecell, Ered, or Eox), the more favorable the process

Electromagnetic Radiation Standard Reduction Potential Electromagnetic Radiation Reducing Cu2+ is more favorable than reducing Zn2+ …by 1.10 V! (Measure it w/voltmeter!) E°cell = E°red(cathode) - E°red(anode) 1.10 V = E°red(Cu2+/Cu) - E°red(Zn2+/Zn) NOTE: If E°red(Zn2+/Zn) were 0 V, E°red(Cu2+/Cu) would be +1.10 V If E°red(Zn2+/Zn) were -1.0 V, E°red(Cu2+/Cu) would be +0.10 V If E°red(Zn2+/Zn) were +1.0 V, E°red(Cu2+/Cu) would be +2.10 V  The “zero” is arbitrary, but must be chosen / agreed upon!

Electromagnetic Radiation Standard Hydrogen Electrode Electromagnetic Radiation This electrode (SHE) was ultimately the one chosen by the scientific community to be the “zero” of potential. 2 H+ + 2 e-  H2 (g); E°red = 0.0 V Upshot: One can determine any E°red experimentally by just setting up a cell where one of the half cells is SHE!

Electromagnetic Radiation Determning a Standard Reduction Potential Using the SHE Electromagnetic Radiation Determining a Standard Reduction Potential using the SHE 0.76 V = E°red(SHE) - E°red(Zn2+/Zn)  0.76 V = 0 - E°red(Zn2+/Zn)  E°red(Zn2+/Zn) = - 0.76 V Both as reductions Ecell = Ecathode - Eanode The reduction of H+ is more favorable than the reduction of Zn2+…. by 0.76 V! H+ (not Zn2+) gets reduced

Compare Potential Energy Cell Potential Compare Potential Energy The electrode in any half-cell with a lesser tendency to undergo reduction (or greater tendency to undergo oxidation) is negatively charged relative to the SHE and, there- fore, has a negative E°. The electrode in any half-cell with a greater tendency to undergo reduction is positively charged relative to the SHE and therefore has a positive E°.

Electromagnetic Radiation Standard Reduction Potentials Electromagnetic Radiation Could use this info to predict that this direct reaction would occur: Use these values to: predict which reactions are spontaneous at standard state and to find any E°cell! E°cell = E°red(cat) - E°red(an) 2 H+(aq) + 2 e- H2(g) 0 E°cell = 0 – (-0.76) = +0.76 V H+ gets reduced; Zn2+ does not (Zn gets oxidized): 2 H+ + Zn → H2(g) + Zn2+ is spontaneous: E°cell > 0 Zn2+(aq) + 2 e- Zn(s) -0.76

Electromagnetic Radiation Types of Electrochemical Cells Electromagnetic Radiation Standard Electrode Potential Refers only to species on the left side of the arrow. E.g., F2, is a better ox agent than H2O2 which is better ox agent than Au3+ (but all of these species are very good oxidizing agents relative to most!) Refers only to species on the right side of the arrow. E.g., F-, is a poorer red agent than H2O which is poorer red agent than Au(s) (but all of these are very poor reducing agents relative to most!) 28

Standard Electrode Potential Table 18.1 (continued) Standard Electrode Potential Click to add notes

Skillbuilder Calculate Ecell for the reaction at 25 C 2 IO3– + 12 H+ + 10 I−→ 6 I2 + 6 H2O separate the reaction into the oxidation and reduction half-reactions ox (anode): 2 I−(s)  I2(aq) + 2 e− red (cathode): IO3−(aq) + 6 H+(aq) + 5 e−  ½ I2(s) + 3 H2O(l) find the E for each half-reaction and Ecell Eanode= −0.54 V Ecathode= +1.20 V Ecell = 1.20 V-(-0.54) = 0.66 V Is the reaction spontaneous?

Calculate Ecell for the reaction at 25 C Skillbuilder Calculate Ecell for the reaction at 25 C separate the reaction into the oxidation and reduction half-reactions find the E for each half-reaction and Ecell

Predicting the Spontaneous Direction of a Redox Rxn Consider the two reduction half-reactions: Since the Mn half-reaction has the more negative electrode potential, it repels electrons and proceeds in the reverse direction (oxidation). Since the Ni half-reaction has the more positive (or least negative) electrode potential, it attracts electrons and proceeds in the forward direction (reduction).

Predicting the Spontaneous Direction of a Redox Rxn Another way to predict the spontaneity of a redox reaction is to note the relative positions of the two half-reactions: Any reduction half-reaction listed will be spontaneous when paired with the reverse of any half-reaction that appears below it in table.

State the better oxidizing agent, determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow Ni2+(aq) + 2 e- Ni(s) -0.23 V Mn2+(aq) + 2 e- Mn(s) -1.18 V The better oxidizing agent is:___ Ni2+ (Because its Ered is more positive) So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode. Ni2+ right Chapter 18, Figure 18.8 Mn/Ni2+ Electrochemical Cell cath So the Ni electrode must be ______ive posit E°cell = _____ - _____ -0.23 V -1.18 V OR E°cell = _____ + _____ -0.23 V -(-1.18 V) = _______ +0.95 V

1 M Fe2+ 1 M Pb2+ e Salt bridge Pb(s) Fe(s) State the better oxidizing agent, determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow Fe2+(aq) + 2 e- Fe(s) -0.45 V Pb2+(aq) + 2 e- Pb(s) -0.13 V e The better oxidizing agent is:___ Pb(s) Fe(s) Pb2+ So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode. Pb2+ Salt bridge Chapter 18, Figure 18.10 Mn/Ni2+ Electrochemical Cell left cath So the Pb electrode must be ______ive 1 M Fe2+ 1 M Pb2+ posit E°cell = _____ - _____ -0.13 V -0.45 V = _______ +0.32 V

Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow Fe2+(aq) + 2 e- Fe(s) -0.45 V Mg2+(aq) + 2 e- Mg(s) -2.37 V The better oxidizing agent is:___ Fe2+ So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode. Fe2+ Chapter 18, Figure 18.8 Mn/Ni2+ Electrochemical Cell right cath So the Fe electrode must be ______ive posit E°cell = _____ - _____ -0.45 V -2.37 V = _______ +1.92 V

Skillbuilder A spontaneous reaction will take place when a reduction half-reaction is paired with an oxidation half-reaction lower on the table If paired the other way, the reverse reaction is spontaneous Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) spontaneous Cu2+(aq) + 2 e−  Cu(s) Ered = +0.34 V higher on a table Zn2+(aq) + 2 e−  Zn(s) Ered = −0.76 V lower on the table should go reverse! Cu(s) + Zn2+(aq)  Cu2+(aq) + Zn(s) nonspontaneous

Skillbuilder Decide whether each of the following will be spontaneous as written or in the reverse direction F2(g) + 2 I−(aq)  I2(s) + 2 F−(aq) Reduction Half-Reaction F2(g) + 2e− 2 F−(aq) IO3−(aq) + 6 H++ 5e− I2(s) + 3H2O(l) Ag+(aq) + 1e− Ag(s) I2(s) + 2e− 2 I−(aq) Cu2+(aq) + 2e− Cu(s) Cr3+(aq) + 1e− Cr2+(aq) Mg2+(aq) + 2e− Mg(s) spontaneous as written Mg(s) + 2 Ag+(aq)  Mg2+(aq) + 2 Ag(s) spontaneous as written Cu2+(aq) + 2 I−(aq)  I2(s) + Cu(s) spontaneous in reverse Cu2+(aq) + 2 Cr2+(aq)  Cu(s) + 2 Cr3+(aq) spontaneous as written

Skillbuilder F− I− I2 Cr3+ F− I− I2 Cr3+ Which of the following materials can be used to oxidize Cu without oxidizing Ag? F− I− I2 Cr3+ F− I− I2 Cr3+ Reduction Half-Reaction F2(g) + 2e− 2 F−(aq) IO3−(aq) + 6 H++ 5e− I2(s) + 3H2O(l) Ag+(aq) + 1e− Ag(s) I2(s) + 2e− 2 I−(aq) Cu2+(aq) + 2e− Cu(s) Cr3+(aq) + 1e− Cr2+(aq) Mg2+(aq) + 2e− Mg(s)

Predicting whether a Metal Will Dissolve in an Acid Metals dissolve in acids if the reduction of the metal ion is easier than the reduction of H+(aq) Metals whose ion reduction reaction lies below H+ reduction on the table will dissolve in acid as a single displacement reaction Almost all metals will dissolve in HNO3 having N reduced rather than H Au and Pt dissolve in HNO3 + HCl NO3−(aq) + 4H+(aq) + 3e− → NO(g) + 2H2O(l)

Skillbuilder Which of the following metals will dissolve in HC2H3O2(aq)? Write the reaction. Ag Cu 2 Fe(s) + 6 HC2H3O2(aq) → 2 Fe(C2H3O2)3(aq) + 3 H2(g) 2 Cr(s) + 6 HC2H3O2(aq) → 2 Cr(C2H3O2)3(aq) + 3 H2(g) Ag Cu Fe Cr Reduction Half-Reaction Au3+(aq) + 3e− Au(s) Ag+(aq) + 1e− Ag(s) Cu2+(aq) + 2e− Cu(s) 2H+(aq) + 2e− H2(g) Fe3+(aq) + 3e− Fe(s) Cr3+(aq) + 3e− Cr(s) Mg2+(aq) + 2e− Mg(s)

E°cell, DG° and K For a spontaneous reaction one that proceeds in the forward direction with the chemicals in their standard states DG° < 1 (negative) E° > 1 (positive) K > 1 DG° = −RTlnK = −nFE°cell n is the number of mole of electrons F = Faraday’s Constant = 96,485 C/mol e−

Calculate DG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) Skillbuilder Calculate DG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) Given: Find: I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) DG°, (J) Conceptual Plan: Relationships: E°ox, E°red E°cell DG° Solve: ox: 2 Br−(aq) → Br2(l) + 2 e− E° = 1.09 V red: I2(l) + 2 e− → 2 I−(aq) E° = +0.54 V tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 V Answer: because DG° is +, the reaction is not spontaneous

Skillbuilder Calculate DG for the reaction at 25C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l) Reduction Half-Reaction Ered, V F2(g) + 2e− 2 F−(aq) +2.87 IO3−(aq) + 6 H++ 5e− ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e− Ag(s) +0.80 I2(s) + 2e− 2 I−(aq) +0.54 Cu2+(aq) + 2e− Cu(s) +0.34 Cr3+(aq) + 1e− Cr2+(aq) −0.50 Mg2+(aq) + 2e− Mg(s) −2.37

Skillbuilder Calculate DG for the reaction at 25C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l) Given: Find: 2IO3−(aq)+12H+(aq)+10I−(aq) → 6 I2(s) + 6 H2O(l) G°, (J) Conceptual Plan: Relationships: E°ox, E°red E°cell G° Solve: ox : 2 I−(s) → I2(aq) + 2 e− Eº = 0.54 V red: IO3−(aq) + 6 H+(aq) + 5 e− → ½ I2(s) + 3 H2O(l) Eº = 1.20 V tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V Answer: because DG° is −, the reaction is spontaneous in the forward direction under standard conditions

Skillbuilder Calculate K at 25 °C for the reaction Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) Given: Find: Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) K Conceptual Plan: Relationships: E°ox, E°red E°cell K Solve: ox: Cu(s) → Cu2+(aq) + 2 e− E° = 0.34 V red: 2 H+(aq) + 2 e− → H2(aq) E° = +0.00 V tot: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) E° = −0.34 V Answer: because K < 1, the position of equilibrium lies far to the left under standard conditions

Skillbuilder Calculate K for the reaction at 25C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l) Reduction Half-Reaction Ered, V F2(g) + 2e− 2 F−(aq) +2.87 IO3−(aq) + 6 H++ 5e− ½I2(s) + 3H2O(l) +1.20 Ag+(aq) + 1e− Ag(s) +0.80 I2(s) + 2e− 2 I−(aq) +0.54 Cu2+(aq) + 2e− Cu(s) +0.34 Cr3+(aq) + 1e− Cr2+(aq) −0.50 Mg2+(aq) + 2e− Mg(s) −2.37

Skillbuilder Calculate K for the reaction at 25C 2IO3–(aq) + 12H+(aq) + 10 I−(aq) → 6I2(s) + 6H2O(l) Given: Find: 2 IO3−(aq)+12 H+(aq)+10 I−(aq) → 6 I2(s) + 6 H2O(l) K Conceptual Plan: Relationships: E°ox, E°red E°cell K Solve: ox : 2 I−(s) → I2(aq) + 2e− Eº = 0.54 V red: IO3−(aq) + 6 H+(aq) + 5e− → ½ I2(s) + 3H2O(l) Eº = 1.20 V tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V Answer: because K >> 1, the position of equilibrium lies far to the right (“product favored [at equilibrium]”)

Standard vs. Nonstandard Cell Cell Potential under Nonstandard Conditions Zn + Cu2+  Zn2+ + Cu; Q = ?? Chapter 18, Figure 18.11 Cell Potential and Concentration (T = 25C) **Always write out the balanced redox equation before using the Nernst Equation or predicting whether Ecell should increase or decrease!!**

Calculate Ecell at 25 °C for the reaction Skillbuilder Calculate Ecell at 25 °C for the reaction Given: Find: 3Cu(s) + 2MnO4−(aq) + 8H+(aq) → 2MnO2(s) + Cu2+(aq) + 4H2O(l) [Cu2+] = 0.010 M, [MnO4−] = 2.0 M, [H+] = 1.0 M Ecell Conceptual Plan: Relationships: E°ox, E°red E°cell Ecell Solve: ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = 0.34 V red: MnO4−(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = +1.68 V tot: 3Cu(s) + 2MnO4−(aq) + 8H+(aq) → 2MnO2(s) + Cu2+(aq) + 4H2O(l)) E° = +1.34 V Check: units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M and [Cu2+] < 1 M

Calculate Ecell at 25 °C for the reaction Skillbuilder Calculate Ecell at 25 °C for the reaction Given: Find: 2 IO3−(aq)+12 H+(aq)+10 I−(aq) → 6 I2(s) + 6 H2O(l) [H+] = 0.10 M, [IO3−] = 0.10 M, [I−] = 0.10 M Ecell Conceptual Plan: Relationships: E°ox, E°red E°cell Ecell Solve: ox : 2 I−(s) → I2(aq) + 2e− Eº = −0.54 V red: IO3−(aq) + 6 H+(aq) + 5e− → ½ I2(s) + 3H2O(l) Eº = 1.20 V tot: 2IO3−(aq) + 12H+(aq) + 10I−(aq) → 6I2(s) + 6H2O(l) Eº = 0.66 V Check: units are correct, Ecell < E°cell as expected because all the ions are reactants and < 1M

Cell Potential under Nonstandard Conditions Concentration Cells Cell Potential under Nonstandard Conditions It is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different. Electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution: oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anode reduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode

Concentration Cell Concentration Cell When the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode) When the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow Cu(s) + Cu2+(aq) (2.0 M) Cu2+(aq) (0.010 M) + Cu(s)

Electrochemical Cell In voltaic cells anode is the source of electrons and has a (−) charge cathode draws electrons and has a (+) charge In electrolytic cells electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery electrons are forced toward the anode, so it must have a source of electrons, the − terminal of the battery

Electrolysis Electrolysis is the process of using electrical energy to break a compound apart. Electrolysis is done in an electrolytic cell (opposite of the spontaneous process); electrolysis use electrical NRG to overcome the NRG barrier of a non-spontaneous reaction. Electrolytic cells can be used to separate elements from their compounds: 2 H2(g) + O2(g)  2 H2O(l) spontaneous 2 H2O(l)  2 H2(g) + O2(g) electrolysis Some applications are (1) metal extraction from minerals and purification, (2) production of H2 for fuel cells, (3) metal plating

Electroplating In electroplating, the work piece is the cathode Cations are reduced at cathode and plate to the surface of the work piece The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution.

Stoichiometry of Electrolysis In an electrolytic cell, the amount of product made is related to the number of electrons transferred essentially, the electrons are a reactant The number of moles of electrons that flow through the electrolytic cell depends on the current and length of time 1 Amp = 1 Coulomb of charge/second 1 mole of e− = 96,485 Coulombs of charge Faraday’s constant

Skillbuilder Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-rxn Au3+(aq) + 3 e− → Au(s) Given: Find: 3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g Conceptual Plan: Relationship s: t(s), amp charge (C) mol e− mol Au g Au Solve: Check: units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−

Skillbuilder Calculate the amperage required to plate 2.5 g of gold in 1 hour (3600 sec) Au3+(aq) + 3 e− → Au(s)

Skillbuilder Calculate the amperage required to plate 2.5 g of gold in 1 hour (3600 sec) Au3+(aq) + 3 e− → Au(s) Given: Find: 3 mol e− : 1 mol Au, mass = 2.5 g, time = 3600 s current, A Conceptual Plan: Relationships: g Au mol Au mol e− charge amps Solve: Check: units are correct, answer is reasonable because 10 A running for 1 hr ~ 1/3 mol e−

Cu2+/Cu & Fe3+/Fe2+ 0.32 V Cu2+/Cu +0.45 V -0.13 V Zn2+/Zn & Ag+/Ag Determined by the color of the leads when voltmeter gave positive reading. Electrons flow away from (-) electrode  Oxidation occurs there. RA loses e-s Cu2+/Cu & Fe3+/Fe2+ 0.32 V Cu2+/Cu Cu  Cu2+ + 2e- +0.45 V Fe3+ + e- Fe2+ -0.13 V Zn2+/Zn & Ag+/Ag 1.50 V Zn2+/Zn Zn  Zn2+ + 2e- 1.50 V Ag+ + e- Ag 0.0 V Cu2+/Cu & Zn2+/Zn 1.05 V Zn2+/Zn Zn  Zn2+ + 2e- 1.50 V Cu2+ + 2e- Cu -0.45 V **Circle the species (not the “electrode”!) that is the better oxidizing agent**

From Text Table 0.0 V 1.50 V -0.45 V -0.13 V 0.0 V 0.80 V 0.80 V (See next slide) Cu2+ + 2e- Cu Zn  Zn2+ + 2e- Fe3+ + e- Fe2+ Ag+ + e- Ag 0.0 V 1.50 V -0.45 V -0.13 V From Table A.1: 0.0 V Ag+ + e- Ag 0.80 V 0.80 V -0.13 V Fe3+ + e- Fe2+ 0.67 V 0.77 V -0.45 V Cu2+ + 2e- Cu 0.35 V 0.34 V -1.50 V Zn2+ + 2e-  Zn -0.70 V -0.76 V Flip Zn2+ + 2e-  Zn -1.50 V