CH21_elec_notes.W5_6

CH21_elec_notes.W5_6 - ~ Chapter 21 Overview of an...

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Unformatted text preview: ~ Chapter 21 Overview of an Electrochemical Process at Constant T and P I AG = AGO + RTan : welec, max (When < 0) _ QEcell : _ ItEcell (exp) = — nC_FEceu (theory) or rearranging to the N ernst equation RT Ecell = gell — e‘17an 0.0592 = Egell — “IOgQ (at 25°C) 6. 1. n6_ determined by balancing redox equation 2. E3611 determined by E0 of species being oxidized and E0 of species being reduced (stan- dard states) 3. therefore only Q (reaction quotient) can change the voltage, 'Eceu Oxidation - Reduction Reactions Cu(s) + 2Ag+(aq) —-> Cu2+(aq) + 2Ag(s) oxidation reduction Balancing Redox Equations (half-reaction method) oxidation numbers (in order of priority) 1. Sum of oxidation numbers of the atoms in a neutral species is zero; in an ion the sum equals the ion charge. 2. Alkali metals (Group I) ON = 1, alkaline earth metals (Group II) ON 2 2, Group III usually have ON = 3. 3. Fluorine ON = —1 always. Other halogens usually have ON = -1 except in compounds with oxygen or other halogens when the oxidation number can be positive (and follows the trend in electronegativity). 4. Hydrogen has ON 2 1 except in metal hydrides when the oxidation number is negative. 5. Oxygen usually has ON = -2 except in compounds with fluorine when the oxidation number can be positive and in compounds containing the 0-0 bond. For peroxides ON = —1 and for superoxides ON 2 -1/2. EX 1. Balance in acidic solution MnOXaq) + Fez+(aq) —> Mn2+(aq) + Fe3+(aq) EX 2. Balance in basic solution Ag(s) + HS'(aq) + CrOZTaq) —) AgZS(s) + Cr(OH)3(s) disproportionation EX 3. Balance: HNOZ —> NO; + NO Electrochemical Cells FIG I - Terms Used in a Voltaic (Galvanic) Cell half—cells electrodes anode cathode electrolyte salt bridge Copper types of electrochemical cells FIG II - Electrochemical Cells a) Voltaic (Galvanic) (spontaneous) Voltmeter Cathode If Oxidation half-reaction Sn'(51)‘——> Sn2*(aq) + 2e“ Reduction half-reaction Cu2+(aq) + 2e“ —-> Cu(s) Overall (cell) reaction Sn(s) + Cu2+(aq) —> Sn2+(aq) + Cu(s) line notation for cells b) Electrolytic (nonspontaneous) External source greaterthan 0.48 V Cathode (-) Oxidation half-reaction Cu(s)———> Cu2+(aq) + 2e" Reduction half-reaction Sn2+(aq) + 2e‘ —>- Sn(s) Overall (cell) reaction Cu(s) + Sn2+(aq) —> Cu2+(aq) + Sn(s) -4- Stoichiometry in Electrochemical Cells electrical work electric current EX 4. A voltaic electrochemical cell whose overall reaction is Cu(s) + 2 Ag+(aq) —> Cu2+(aq) + 2 Ag(s) oxidizes metallic copper at the anode dissolving the metal Cu(s) ——> Cu2+(aq) + 2 e— and reduces silver ions in solution by plating them out as metallic silver at the cathode. Ag+(aq) + e- -> Ag(S) a) The cell delivers 0.1 A. How long does it take to dissolve 5.00 g of copper at theanode? (MCu = 63.546) b) An external power source is connected to the cell and the cell is run electrolytically. If the cell produces 0.500 A for 101 minutes, how much copper is deposited and how much silver dissolves? (M Ag = 107.87) electrolysis EX 5. An aqueous solution of a palladium salt is electrolyzed by 1.50 A for one hour which produces 2.9779 g of Pd at the cathode. What is the oxidation state of Pd? (Mpd = 106.42 g/mol) Free Energy and Cell Voltage EX 6. A 6 volt battery delivers 1.25 amperes for 1.5 hours. How much work does the battery do? Standard Half-Cell Reduction Potentials standard states - stable form (allotrope) at P = 1 atm and T = 25°C (usually) pure solid pure liquid gas — ideal gas behavior 1 molar solution (1 M) - ideal solution behavior 0 standard cell potential, E6611 standard reduction potential, E o — by international convention, E° for the reduction of H+(aq) to give H2(g) is taken to be zero at all temperatures: 2 H+(aq, 1M) + 2 e‘ —> H2(g, P =1 atm) FIG III - Standard Hydrogen ' FIG IVa - SHE/Zinc Electrode (SHE) ’ I J Vollmater ' Chemically inert ’ l Pl electrode ’ Zn(s) ———> Zn2+(aq) + 2 e' 2 H+(aq) + 2 e' —> H2(g)' Net reaction: Zn(s) + 2 H+(aq) ——->J-lz(g) + Zn2+(aq) FIG IVb - SHE/Silver e: a + 0.20m! H2(8) <=> 2 H+(aq) + 26“ “3:56 ® I “53:2,: OX: H2(g) —> 2H+(aq) + 2e: E0 = 0 "- RED: 2H+(aq) + 26 —> H2(g), E° = 0 OX: Ptl H2(g, 1 atm)[H+(aq, 1 M) l mua. I M “no, RED: H+(a'q’ 1 M)! H2(g’ 1 athPt Deserted WE‘VEQMlC, fate) Made my? a; -6- FIG V - Standard Reduction Potentials EX 7. Will Ag+ oxidize Zn(s) or will Zn2+ oxidize Ag(s)? Table 21.2 Selected Standard Electrode Potentials (298 K) Half-Reaction Efiulhe" (V) +2.87 +1.36 +1.23 +0.96 +0.80 1 +0.77 +0.40 +0.34 Fz(g) + 26‘ :2 2F‘(aq) C]2(g) + 212‘ :: 2C1“(aq) Mn02(.3') + 4H+(aq) + 2e" : Mn2+(aq) + 2H20(l) N03'(aq) + 4H+(aq) + 3c“ : NO(g) + 21-1200) Ag+(aq) + e" : Ag(s) Fe3+(aq) + e' :: F62+(aq) 03( g) + 214200) + 4e” : 40H—(aq) Cu2+(aq) + 2e— :—: Cu(s) E5. 2H+(aq) + 2e_ :2 Hz(g) 0.00 g N2(g) + 5H*(aq) + 4e‘ : N2H5+(aq) —o.23 g» Fe2+(aq) + 2e“ :: Fe(s) —0.44 g .zii-“(aq).+-2e~ :: Zn(s) —0.76 0’ 2H20(l) + 2e_ :2 112(3) + ZOH—(aq) —O.83 Na+(aq) + e' z: Na(s) ~2.71 Li+(aq) + e“ :— Li(x) —3.05 EX 8. Find E ° for the electrode AgZCrO4(s)lAg(s) and determine E261, for the following reaction if E °(Fe3+lFeZ+) = 0.770 V and AG" = —62.5 k] for the reaction: ‘ KzCrO4(aq) + 2 Ag(s) + 2 FeCl3(aq) ———> AgZCrO4(s) + 2 FeClfiaq) + 2 KC1(aq) Disproportionation anlexidri/zing/Reducing Agents 3. 3.: (V) " ' 1V) 0.99 0.94 . ‘ L Basic solution pH 14 0.40 0.17 ‘ N05 + H20 +12 e' + N0; + 2 cm ~ ' " r . . 0.01 —0.28 V ‘ H20 + é‘—+ N0 + 2 OH‘ ‘0'50 I ' —-0.46 ' H20 '+,4 e' —> 820’; + 6 on- “0'58 7' r v ' ’ - _, 2— _ Acidic solution .+‘.H;o + 2 9 $03 + 2 0H 4.98 pH 0 ' —1 12 H20+2e’-> HPo§-+30H- ' 0 + 2 e? —' HzPOZ' + 3 OH“ ' ‘ , —1.57 _7_ Concentration Effects and the Ner’n‘st Equation Han—cans: EX 9. Determine E for the hydrogen electrode: 2 H+ + 2 e‘ —> H2 a) [H+] = 0. 00100 M, PHZ = 0.500 atm b) H+ + e“ —> 1/2 H2: [W] = 0.00100 M, PHz = 0.500 atm C) [W] = 1. 00 ><10‘l4 M, PHz = 1. 00 atm d) rewrite hydrogen electrode equation in base, [OH‘] = 1. 00 M, PH2 = 1. 00 atm EX 10. For the half reaction, Cu” + 2e‘ —> Cu, where E? = 0.340 V what is the reduction potential of the half-cell when [Cu2+] = 0.10 M? Cells: EX 11. If [Cu2+] = 0.10 M and [Ag] = 0.20 M, what is the voltage of an electrochemical cell based upon the couple: E°(Cu2+ICu) = 0.340 V and E°(Ag+lAg) = 0.800 V? EX 12. An electrochemical cell is based upon the couple: E°(Cu2+lCu) = 0.340 V and E° (Ag+lAg) = 0.800 V If the voltage of the cell were 0.448 V when [Cu2+] = 0.10 M, what is the silver ion concentration? EX 13. If the cell voltage is 0.473 V what is the pH _of the cathode compartment for the following cell? E"(Zn2+|Zn) = — 0.7618 V Zn(s) | Zn2+(aq, 1.00 M) II H+(aq, ? M) | H2(g, 1 atm) | Pt(s) -9- Equilibrium Constants from Electrochemistry EX 14. What is the equilibrium constant for the cell of example 11? EX 15. If E°(Cr3+lCr) = -0.74 V and E°(Zn2+lZn) = -0.7628 V what is the equilibrium constant for 2Cr3+(aq) + 3 Zn(s) —> 2Cr(s) + 3Zn2+(aq) Solubility Product, K Sp EX 16. From the following reaction cell where Emu = 0.397 V PtI H2(1 atm) I H+(1. OO M)I|Cl‘(1. 00 X 10'3 M), Ag+(? M) I AgCl(s) I Ag(s) determine the value of K SP(AgCl). Relevant standard reduction potentials are 2 H+(aq) + 26‘ —9 H2(g) E0 = 0.0000V Ag+(aq) + e" —> Ag(s) E° = 0.7996. V -10- EX 17. Determine the value of K Sp(AgSCN) given that ' AgSCN(s) + e‘ —> Ag(s) + SCN(aq) E° = 0.0895 V Ag+(aq) + e‘ —> Ag(s) E° = 0.7996 V Concentration Cells FIG VI - Ni2+lNi Concentration Cell : mam / Salt bridge \ Ni cathode [N12+] = 1.00 x 10‘3M '[Ni2+] = 1.00M [Ni2+] = 0-5M [Ni2+1 = 0:5 M Corrosion ...
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This note was uploaded on 04/13/2010 for the course CHEM 114 taught by Professor Jursich during the Spring '08 term at Ill. Chicago.

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CH21_elec_notes.W5_6 - ~ Chapter 21 Overview of an...

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