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S  0 ``  >*  0 `   @*  0 `   @*h  PA޽h ?Parchment ̙33 Default Design 0 zr@ ( )   0j P    P*    0<{     R*  d  c $ ?    0({  @  RClick to edit Master text styles Second level Third level Fourth level Fifth level!     S  64{ `P   P*    6N `   R*  H  0޽h ? ̙3380___PPT10. 0X8(   $+ X X 0D P    >*  X 0     @*  X 6  `P   >*  X 6 `   @* H X 0޽h ? ̙3380___PPT10.1 0L0 P0(   + 0t 'AN INTRODUCTION TO ELECTRODE POTENTIALST(2 82 f0f<fJ , C "A KTRE: . - <̙? x @. 0 0 . <̙? x @* 0 0 / 0x m OKNOCKHARDY PUBLISHING(2 f 0 0"`>B 2008 SPECIFICATIONS>G$G   B  s *޽h ? ̙33y___PPT10Y+D=' = @B +:  0L0 IA` (  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0I   0n@" INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... www.knockhardy.org.uk/sci.htm Navigation is achieved by... either clicking on the grey arrows at the foot of each page or using the left and right arrow keys on the keyboard (2 2(2 2  75, dB   <D?     0$ bELECTRODE POTENTIALS2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   p 6 (    0؎  B CONTENTS Types of half cells Cell potential The standard hydrogen electrode Measuring electrode potentials The electrochemical series Combining half cells Cell diagrams Uses of E valuesJ 202 2dB  <D)P0Q  0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0  03e .@  0 0  63~m !@  0 0  63p] @  0 0  63Km8 @  0 0  63> ] @  0 0  63! m@ @  0 0  63 e0 @ 0 0  63 ez @  0 0   0\$ bELECTRODE POTENTIALS2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 D(    0ȣHx bBefore you start it would be helpful to& Recall the definitions of oxidation and reduction Be able to balance simple ionic equations Have a knowledge of simple circuitry<*(2 2*!dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0X bELECTRODE POTENTIALS2(2$f<fB  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 (ha(  hdB h <D)P0Q h 0Pp@ 0 0dB h@ <D)PXP h 0X0@ 0 0 h 0Դq~ `TYPES OF HALF CELL2(2ff  h 00~  qThese are systems involving oxidation or reduction METALS IN CONTACT WITH SOLUTIONS OF THEIR IONS Reaction Cu2+(aq) + 2e Cu(s) Electrode copper Solution Cu2+(aq) (1M) - 1M copper sulphate solution Potential + 0.34V Reaction Zn2+(aq) + 2e Zn(s) Electrode zinc Solution Zn2+(aq) (1M) - 1M zinc sulphate solution Potential - 0.76V j3H@2H@@@3! /f f   #       "     @    m F ? h yZ I 8N  h fB h 6DjJlB h <BDjJ{>T  h# L)?fB h 6DjJlB h <BDjJ{F ? h   8N  h fB h 6DjJlB h <BDjJ{>T   h# L)?fB !h 6DjJlB "h <BDjJ{P &h C (Acucell1pXHP 'h C (Azncell2 pXg B h s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 )!@(  @dB @ <D)P0Q @ 0Pp@ 0 0dB @@ <D)PXP @ 0X0@ 0 0 @ 0Xq~ `TYPES OF HALF CELL2(2ff3  @ 0?0H These are systems involving oxidation or reduction GASES IN CONTACT WITH SOLUTIONS OF THEIR IONS Reaction 2H+(aq) + 2e H2(g) Electrode platinum Solution H+(aq) (1M) - 1M HCl or 0.5M H2SO4 Gas hydrogen at 100 kPa (1atm) pressure Potential 0.00V Reaction Cl2(aq) + 2e 2Cl(g) Electrode platinum Solution Cl(aq) (1M) - 1M sodium chloride Gas chorine at 100 kPa (1atm) pressure Potential + 1.36V(3H@.H3@@@@3! .f        -    -      -    ;  + F ?  @ u!  8N   @ fB  @ 6DjJlB  @ <BDjJ{>T  @# L)?fB @ 6DjJlB @ <BDjJ{F ? @  8N  @ fB @ 6D8clB @ <BD8c{>T  @# L)?fB @ 6DjJlB @ <BD8c{N @ C &A hcell3bxN @ C &A hcell3b r1B @ s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   H ( w HdB H <D)P0Q H 0Pp@ 0 0dB H@ <D)PXP H 0X0@ 0 0 H 00?q~ `TYPES OF HALF CELL2(2ff$  H 04?0  These are systems involving oxidation or reduction SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES Reaction Fe3+(aq) + e Fe2+(aq) Electrode platinum Solution Fe3+(aq) (1M) and Fe2+(aq) (1M) Potential + 0.77 V 3H@4H4@i@@3! 4f             F ?  H  8N   H fB  H 6DjJlB  H <BDjJ{>T  H# L)?fB H 6DjJlB H <BDjJ{P H C (A fecell4Q0B H s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 0p^(  pdB p <D)P0Q p 0Pp@ 0 0dB p@ <D)PXP p 0X0@ 0 0 p 0Q?q~ `TYPES OF HALF CELL2(2ff p 0W?0 nThese are systems involving oxidation or reduction SOLUTIONS OF IONS IN TWO DIFFERENT OXIDATION STATES Reaction Fe3+(aq) + e Fe2+(aq) Electrode platinum Solution Fe3+(aq) (1M) and Fe2+(aq) (1M) Potential + 0.77 V SOLUTIONS OF OXIDISING AGENTS IN ACID SOLUTION Reaction MnO4 + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) Electrode platinum Solution MnO4(aq) (1M) and Mn2+(aq) (1M) and H+(aq) Potential + 1.52 V&3H@4H4@i@@/H@@3! 4   1& /ff             &   F ? p  8N   p fB  p 6DjJlB  p <BDjJ{>T   p# L)?fB  p 6DjJlB p <BDjJ{F ? p  yhH 8N  p fB p 6DjJlB p <BDjJ{>T  p# L)?fB p 6DjJlB p <BDjJ{P p C (A fecell4Q0P p C (A mncell5 0B p s *޽h ? ̙33y___PPT10Y+D=' = @B ++  0L0 : 2  ( ( w (dB ( <D)P0Q ( 0Pp@ 0 0dB (@ <D)PXP ( 0X0@ 0 0 ( 04?q~ \CELL POTENTIAL2(2ff  ( 0?  LZEach electrode / electrolyte combination has its own half-reaction which sets up a potential difference The value is affected by ... TEMPERATURE PRESSURE OF ANY GASES SOLUTION CONCENTRATION Measurement it is impossible to measure the potential of a single electrode& BUT... you can measure the potential difference between two electrodes it is measured relative to a reference cell under standard conditions The ultimate reference is the STANDARD HYDROGEN ELECTRODE. However, as it is difficult to set up, secondary standards are used.5XZ3XP@P<@h!!fff= f?!!;H!H . B ( s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0   : (  E  04? 0y }The ultimate reference is the STANDARD HYDROGEN ELECTRODE. However as it is difficult to set up secondary standards are used.8~!E ~ dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0?q~ mTHE STANDARD HYDROGEN ELECTRODE2 (2ff   0HYDROGEN (100 kPa) PLATINUM ELECTRODE HYDROCHLORIC ACID (1M)8@,>&  . ^B  6DoXK^B  6Do |^B  6DochB  s *޽h ? ̙33y___PPT10Y+D=' = @B +? 0L0 ??{}`>(  `z ` 0D0\  R E / V F2(g) + 2e 2F(aq) +2.87 MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) +1.52 Cl2(g) + 2e 2Cl(aq) +1.36 Cr2O72-(aq) + I4H+(aq) + 6e 2Cr3+(aq) + 7H2O(l) +1.33 Br2(l) + 2e 2Br(aq) +1.07 Ag+(aq) + e Ag(s) +0.80 Fe3+(aq) + e Fe2+(aq) +0.77 O2(g) + 2H+(aq) + 2e H2O2(aq) +0.68 I2(s) + 2e 2I(aq) +0.54 Cu+(aq) + e Cu(s) +0.52 Cu2+(aq) + 2e Cu(s) +0.34 Cu2+(aq) + e Cu+(aq) +0.15 2H+(aq) + 2e H2(g) 0.00 Sn2+(aq) + 2e Sn(s) -0.14 Fe2+(aq) + 2e Fe(s) -0.44 Zn2+ (aq) + 2e Zn(s) -0.76 Layout If species are arranged in order of their standard electrode potentials you get a series that shows how good each substance is at gaining electrons. All equations are written as reduction processes ... i.e. gaining electrons A species with a higher E value oxidise (reverses) one with a lower value}SJ@! ! !- ! !  ! !  ! !  ! ! ! ! !  ! ! ! ! ! ! !  ! ! ! !  ! ! ! !3 ! !. ! !  ! ! ! !  ! ! ! ! ! ! ! !1 ! !, ! !/ ! !  ! ! !           3 3- 3 3, 3 3 3ff3OL+        +  ,  Z  +  !    F dB ` <D)P0Q ` 0Pp@ 0 0dB `@ <D)PXP ` 0X0@ 0 0  ` 04jq~ hTHE ELECTROCHEMICAL SERIES2(2ffF ?  `  8N   ` fB  ` 6D!jJlB  ` <BD!jJ{>T  `# L)?fB ` 6D!jJlB ` <BD!jJ{F ? ` A y8N  ` fB ` 6D!jJlB ` <BD!jJ{>T  `# L)?fB ` 6D!jJlB ` <BD!jJ{F ? `  8N  ` fB ` 6D!jJlB ` <BD!jJ{>T  `# L)?fB ` 6D!jJlB ` <BD!jJ{F ? ` y 8N   ` fB !` 6D!jJlB "` <BD!jJ{>T  #`# L)?fB $` 6D!jJlB %` <BD!jJ{F ? &`  I8N  '` fB (` 6D!jJlB )` <BD!jJ{>T  *`# L)?fB +` 6D!jJlB ,` <BD!jJ{F ? -`  8N  .` fB /` 6D!jJlB 0` <BD!jJ{>T  1`# L)?fB 2` 6D!jJlB 3` <BD!jJ{F ? 4` Y 8N  5` fB 6` 6D!jJlB 7` <BD!jJ{>T  8`# L)?fB 9` 6D!jJlB :` <BD!jJ{F ? ;`  )8N  <` fB =` 6D!jJlB >` <BD!jJ{>T  ?`# L)?fB @` 6D!jJlB A` <BD!jJ{F ? B`  8N  C` fB D` 6D!jJlB E` <BD!jJ{>T  F`# L)?fB G` 6D!jJlB H` <BD!jJ{F ? I` ) a8N  J` fB K` 6D!jJlB L` <BD!jJ{>T  M`# L)?fB N` 6D!jJlB O` <BD!jJ{F ? P`  8N  Q` fB R` 6D!jJlB S` <BD!jJ{>T  T`# L)?fB U` 6D!jJlB V` <BD!jJ{F ? W` Q  8N  X` fB Y` 6D!jJlB Z` <BD!jJ{>T  [`# L)?fB \` 6D!jJlB ]` <BD!jJ{F ? ^`   ) 8N  _` fB `` 6DjJlB a` <BDjJ{>T  b`# L)?fB c` 6DjJlB d` <BDjJ{F ? e`   8N  f` fB g` 6D3jJlB h` <BD3jJ{>T  i`# L)?fB j` 6D3jJlB k` <BD3jJ{F ? l` )  a 8N  m` fB n` 6D3jJlB o` <BD3jJ{>T  p`# L)?fB q` 6D3jJlB r` <BD3jJ{F ? s`    8N  t` fB u` 6D3jJlB v` <BD3jJ{>T  w`# L)?fB x` 6D3jJlB y` <BD3jJ{,8   }`p lB z` <D>PPpfB {`B 6D>PP  |` 0 PREACTION MORE LIKELY TO WORK SPECIES ON LEFT ARE MORE POWERFUL OXIDATION AGENTSQQ  Q B ` s *޽h ? ̙33y___PPT10Y+D=' = @B +0E 0L0 ?D7DydC(  d d 00 , E / V F2(g) + 2e 2F(aq) +2.87 MnO4(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l) +1.52 Cl2(g) + 2e 2Cl(aq) +1.36 Cr2O72-(aq) + I4H+(aq) + 6e 2Cr3+(aq) + 7H2O(l) +1.33 Br2(l) + 2e 2Br(aq) +1.07 Ag+(aq) + e Ag(s) +0.80 Fe3+(aq) + e Fe2+(aq) +0.77 O2(g) + 2H+(aq) + 2e H2O2(aq) +0.68 I2(s) + 2e 2I(aq) +0.54 Cu+(aq) + e Cu(s) +0.52 Cu2+(aq) + 2e Cu(s) +0.34 Cu2+(aq) + e Cu+(aq) +0.15 2H+(aq) + 2e H2(g) 0.00 Sn2+(aq) + 2e Sn(s) -0.14 Fe2+(aq) + 2e Fe(s) -0.44 Zn2+ (aq) + 2e Zn(s) -0.76 Application Chlorine is a more powerful oxidising agent - it has a higher E Chlorine will get its electrons by reversing the iodine equation Cl2(g) + 2e   > 2Cl(aq) and 2I(aq)   > I2(s) + 2e Overall equation is Cl2(g) + 2I(aq)   > I2(s) + 2Cl(aq)RT}  -               f f f  f f f                 3  .                 /   ,  /                 -  ,    3 ffff ff fff    +        +  ,  Z  +  !    H dB d <D)P0Q d 0Pp@ 0 0dB d@ <D)PXP d 0X0@ 0 0  d 0q~ hTHE ELECTROCHEMICAL SERIES2(2ffF ?  d  8N   d fB  d 6DjJlB  d <BDjJ{>T  d# L)?fB d 6DjJlB d <BDjJ{F ? d A y8N  d fB d 6DjJlB d <BDjJ{>T  d# L)?fB d 6DjJlB d <BDjJ{F ? d  8N  d fB d 6DfjJlB d <BDfjJ{>T  d# L)?fB d 6DfjJlB d <BDfjJ{F ? d y 8N   d fB !d 6DjJlB "d <BDjJ{>T  #d# L)?fB $d 6DjJlB %d <BDjJ{F ? &d  I8N  'd fB (d 6DjJlB )d <BDjJ{>T  *d# L)?fB +d 6DjJlB ,d <BDjJ{F ? -d  8N  .d fB /d 6DjJlB 0d <BDjJ{>T  1d# L)?fB 2d 6DjJlB 3d <BDjJ{F ? 4d Y 8N  5d fB 6d 6DjJlB 7d <BDjJ{>T  8d# L)?fB 9d 6DjJlB :d <BDjJ{F ? ;d  8N  d <BDjJ{>T  ?d# L)?fB @d 6DjJlB Ad <BDjJ{F ? Bd  (8N  Cd fB Dd 6DjJlB Ed <BDjJ{>T  Fd# L)?fB Gd 6DjJlB Hd <BDjJ{F ? Id ) a8N  Jd fB Kd 6DjJlB Ld <BDjJ{>T  Md# L)?fB Nd 6DjJlB Od <BDjJ{F ? Pd  8N  Qd fB Rd 6DjJlB Sd <BDjJ{>T  Td# L)?fB Ud 6DjJlB Vd <BDjJ{F ? Wd Q  8N  Xd fB Yd 6DjJlB Zd <BDjJ{>T  [d# L)?fB \d 6DjJlB ]d <BDjJ{F ? ^d   ) 8N  _d fB `d 6DjJlB ad <BDjJ{>T  bd# L)?fB cd 6DjJlB dd <BDjJ{F ? ed   8N  fd fB gd 6DjJlB hd <BDjJ{>T  id# L)?fB jd 6DjJlB kd <BDjJ{F ? ld )  a 8N  md fB nd 6DjJlB od <BDjJ{>T  pd# L)?fB qd 6DjJlB rd <BDjJ{F ? sd    8N  td fB ud 6DjJlB vd <BDjJ{>T  wd# L)?fB xd 6DjJlB yd <BDjJ{Rr d s *̙x d 0D0 ' OAN EQUATION WITH A HIGHER E VALUE WILL REVERSE AN EQUATION WITH A LOWER VALUEPP P B d s *޽h ? ̙33y___PPT10Y+D=' = @B +V  0L0 e] (    0K 6  rtWhy? The standard hydrogen electrode (SHE) is difficult to set up it is easier to choose a more convenient secondary standard the secondary standard has been calibrated against the SHE Calomel " the calomel electrode contains Hg2Cl2 " it has a standard electrode potential of +0.27V " is used as the LH electrode to determine the potential of an unknown " to get the value of the other cell ADD 0.27V to the measured cell potentialPsff!  %  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0_q~ aSECONDARY STANDARDS2(2ffB  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 q i  (     0f 0  |CELLS " electrochemical cells contain two electrodes " each electrode / electrolyte combination has its own half-reaction " the electrons produced by one half reaction are available for the other " oxidation occurs at the anode " reduction occurs at the cathode. ANODE Zn(s)   > Zn2+(aq) + 2e OXIDATION CATHODE Cu2+(aq) + 2e   > Cu(s) REDUCTION The resulting cell has a potential difference (voltage) called the cell potential which depends on the difference between the two potentials It is affected by ... " current " temperature " pressure of any gases " solution concentrations&6}u}  #t@  D   dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0q~ cELECTROCHEMICAL CELLS2(2ffB  s *޽h ? ̙33u 0L0 |0,(  ^  0h( 0 zinc is more reactive - it dissolves to give ions Zn(s)   > Zn2+(aq) + 2e the electrons produced go round the external circuit to the copper electrode electrons are picked up by copper ions Cu2+(aq) + 2e   > Cu(s) As a result, copper is deposited overall reaction Zn(s) + Cu2+(aq)   > Zn2+(aq) + Cu(s)[@MHo@D@' Mf&  fZ<    <  * dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0hq~ q#A TYPICAL COMBINATION OF HALF CELLS2$(2#ffL   C $A cuzng ! 0{(` 2ZINC ZINC SULPHATE (1M)D2 3 ^B " 6Doha # 0`h yCOPPER COPPER SULPHATE (1M),  ^B $ 6Doh@U^B % 6Do(`a^B & 6Do`a ' 0p` W E = - 0.76V   ( 0`P U E = + 0.34V   * 0,Ĉ.A 2  Z 1.10V  + 0Ȉ ;; M+ $   , 0\̈ M_ ,  B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 @Kpa(  r  0\ӈ0}  These give a diagrammatic representation of what is happening in a cell. " Place the cell with the more positive E value on the RHS of the diagram. Cu2+(aq) + 2e Cu(s) E = + 0.34V put on the RHS Zn2+(aq) + 2e Zn(s) E = - 0.76V put on the LHS IHM333  333  @  7  # dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0q~ [ CELL DIAGRAMS2(2 ff L F >  # ]`B  0DoF  `B  0DoBF B fB  6DoF  fB  6Do F   6s >  Zn Zn2+ Cu2+ CuL 2    F ? >  m\A8N  ? fB @ 6DfjJlB A <BDfjJ{>T  B# L)?fB C 6DfjJlB D <BDfjJ{F ? E eT 8N  F fB G 6DfjJlB H <BDfjJ{>T  I# L)?fB J 6DfjJlB K <BDfjJ{B  s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 P!?T(  T  T 0w0t  These give a diagrammatic representation of what is happening in a cell. " Place the cell with the more positive E value on the RHS of the diagram. Cu2+(aq) + 2e Cu(s) E = + 0.34V put on the RHS Zn2+(aq) + 2e Zn(s) E = - 0.76V put on the LHS ZINC IS IN CONTACT THE SOLUTIONS A SOLUTION OF WITH A SOLUTION ARE JOINED VIA A COPPER IONS IN OF ZINC IONS SALT BRIDGE TO CONTACT WITH COPPER IHM333 333   @  7   dB T <D)P0Q T 0Pp@ 0 0dB T@ <D)PXP T 0X0@ 0 0  T 0wq~ [ CELL DIAGRAMS2(2 ffL F >   T# ]ZB  T s *DoF  ZB  T s *DoBF B `B  T 0DoF  `B T 0Do F  T 0%ws >  Zn Zn2+ Cu2+ CuL 2    RB T s *D3jJE  X RB T s *D3jJMZ hX RB  T@ s *D3jJEJX L !T C $A cuzng4  UF ? 0T  m\A8N  1T fB 2T 6DfjJlB 3T <BDfjJ{>T  4T# L)?fB 5T 6DfjJlB 6T <BDfjJ{F ? 7T eT 8N  8T fB 9T 6DfjJlB :T <BDfjJ{>T  ;T# L)?fB T 0X2whE Q+ 0   ?T 06w]8 Q_ ,  B T s *޽h ? ̙33y___PPT10Y+D=' = @B +8 0L0 G?`I(  :  0=w0! `These give a diagrammatic representation of what is happening in a cell. " Place the cell with the more positive E value on the RHS of the diagram. Cu2+(aq) + 2e Cu(s) E = + 0.34V put on the RHS Zn2+(aq) + 2e Zn(s) E = - 0.76V put on the LHS " Draw as shown& the cell reaction goes from left to right " the zinc metal dissolves Zn(s)   > Zn2+(aq) + 2e OXIDATION " copper is deposited Cu2+(aq) + 2e   > Cu(s) REDUCTION " oxidation takes place at the anode " reduction at the cathodeI$HM  +3   %t  7    _  Q dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0Xjwq~ [ CELL DIAGRAMS2(2 ffL F >   # ]ZB   s *DoF  ZB   s *DoBF B `B   0DoF  `B  0Do F   0`qws >  Zn Zn2+ Cu2+ CuL 2    XB  0D3ԔUY U , 0xwhE Q+ 0   - 0}w]8 Q_ ,  F ? <  m\A8N  = fB > 6DjJlB ? <BDjJ{>T  @# L)?fB A 6DjJlB B <BDjJ{F ? C eT 8N  D fB E 6DjJlB F <BDjJ{>T  G# L)?fB H 6DjJlB I <BDjJ{B  s *޽h ? ̙33y___PPT10Y+D=' = @B +W 0L0 f^p*J(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0   0Ĉwq~ [ CELL DIAGRAMS2(2 ff , 0w0 VThese give a diagrammatic representation of what is happening in a cell. " Place the cell with the more positive E value on the RHS of the diagram. Cu2+(aq) + 2e Cu(s) E = + 0.34V put on the RHS Zn2+(aq) + 2e Zn(s) E = - 0.76V put on the LHS " Draw as shown& the electrons go round the external circuit from left to right " electrons are released when zinc turns into zinc ions " the electrons produced go round the external circuit to the copper " electrons are picked up by copper ions and copper is depositedIIM  3?@  7  > L F >   # ]ZB   s *DoF  ZB   s *DoBF B `B   0DoF  `B  0Do F   0ws >  Zn Zn2+ Cu2+ CuL 2    F z 5   > `2  0om   B,woC"?  GV  ZB  s *Doz ZB  s *DofB  6Do fB  6Do&N  ] K   5 #ZB  s *Do ] K fB  6Do e  &N z] zK   z 5 z #ZB  s *Doz] zK fB  6Doz z  - 0нwhE Q+ 0   . 0w]8 Q_ ,  F ? =  m\A8N  > fB ? 6DjJlB @ <BDjJ{>T  A# L)?fB B 6DjJlB C <BDjJ{F ? D eT 8N  E fB F 6DjJlB G <BDjJ{>T  H# L)?fB I 6DjJlB J <BDjJ{B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 *H4(  dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0  0wq~ [ CELL DIAGRAMS2(2 ffN  0w"`0 2These give a diagrammatic representation of what is happening in a cell. " Place the cell with the more positive E value on the RHS of the diagram. Cu2+(aq) + 2e Cu(s) E = + 0.34V put on the RHS Zn2+(aq) + 2e Zn(s) E = - 0.76V put on the LHS " Draw as shown& the cell voltage is E(RHS) - E(LHS) - it must be positive cell voltage = +0.34V - (-0.76V) = +1.10VJIHP:53+f3*@  7   L F >  # ]ZB  s *DoF  ZB  s *DoBF B `B  0DoF  `B  0Do F   0ws >  Zn Zn2+ Cu2+ CuL 2    F z 5   > `2  0om    BwoC"?  GV  ZB ! s *Doz ZB " s *DofB # 6Do fB $ 6Do&N  ] K  % 5 #ZB & s *Do ] K fB ' 6Do e  &N z] zK  ( z 5 z #ZB ) s *Doz] zK fB * 6Doz z  + 0 whE Q+ 0   , 0x]8 Q_ ,  F ? ;  m\A8N  < fB = 6DjJlB > <BDjJ{>T  ?# L)?fB @ 6DjJlB A <BDjJ{F ? B eT 8N  C fB D 6DjJlB E <BDjJ{>T  F# L)?fB G 6DjJlB H <BDjJ{B  s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0  V( w  dB   <D)P0Q   0Pp@ 0 0dB  @ <D)PXP   0X0@ 0 0   0x xq~  USE OF Eo VALUES - WILL IT WORK?X!(2ffff   0x"` j:E values Can be used to predict the feasibility of redox and cell reactions In theory ANY REDOX REACTION WITH A POSITIVE E VALUE WILL WORK In practice, it proceeds if the E value of the reaction is greater than + 0.40V An equation with a more positive E value reverse a less positive one;@ fD!!!> !E!E!!&8   B   s *޽h ? ̙33y___PPT10Y+D=' = @B +x 0L0   h (  hdB h <D)P0Q h 0Pp@ 0 0dB h@ <D)PXP h 0X0@ 0 0 h 0'xq~  USE OF Eo VALUES - WILL IT WORK?X!(2ffff  h 0\3x"`&| rWhat happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected? Write out the equations Cu2+(aq) + 2e Cu(s) ; E = +0.34V Sn2+(aq) + 2e Sn(s) ; E = -0.14V  PH@K@9@ff4ff.f Z  i  2   F ?  h ~ m8N   h fB  h 6D3jJlB  h <BD3jJ{>T  h# L)?fB h 6D3jJlB h <BD3jJ{F ? h } l8N  h fB h 6D3jJlB h <BD3jJ{>T  h# L)?fB h 6D3jJlB h <BD3jJ{ h 0Gx"`~ EAn equation with a more positive E value reverse a less positive oneFHF F B h s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 Px(  xdB x <D)P0Q x 0Pp@ 0 0dB x@ <D)PXP x 0X0@ 0 0 x 0Oxq~  USE OF Eo VALUES - WILL IT WORK?X!(2ffffZ x 0Xx"`&  zWhat happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected? Write out the equations Cu2+(aq) + 2e Cu(s) ; E = +0.34V Sn2+(aq) + 2e Sn(s) ; E = -0.14V the half reaction with the more positive E value is more likely to work it gets the electrons by reversing the half reaction with the lower E value therefore Cu2+(aq)   > Cu(s) and Sn(s)   > Sn2+(aq) PH@K@@,@&@ff4ff.fJLfffffft  i  2     F ? x ~ m8N   x fB  x 6D3jJlB  x <BD3jJ{>T   x# L)?fB  x 6D3jJlB x <BD3jJ{F ? x } l8N  x fB x 6D3jJlB x <BD3jJ{>T  x# L)?fB x 6D3jJlB x <BD3jJ{ x 0}x"`~ EAn equation with a more positive E value reverse a less positive oneFHF F B x s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 `|(  |dB | <D)P0Q | 0Pp@ 0 0dB |@ <D)PXP | 0X0@ 0 0 | 0`xq~  USE OF Eo VALUES - WILL IT WORK?X!(2ffffX | 0x"`&  What happens if an Sn(s) / Sn2+(aq) and a Cu(s) / Cu2+(aq) cell are connected? Write out the equations Cu2+(aq) + 2e Cu(s) ; E = +0.34V Sn2+(aq) + 2e Sn(s) ; E = -0.14V the half reaction with the more positive E value is more likely to work it gets the electrons by reversing the half reaction with the lower E value therefore Cu2+(aq)   > Cu(s) and Sn(s)   > Sn2+(aq) the overall reaction is Cu2+(aq) + Sn(s)   > Sn2+(aq) + Cu(s) the cell voltage is the difference in E values... (+0.34) - (-0.14) = + 0.48V|PH@K@@,@@ff4ff.fJLfffffffffff3f   i  2    ?  u F ? | ~ m8N   | fB  | 6D3jJlB  | <BD3jJ{>T   |# L)?fB  | 6D3jJlB | <BD3jJ{F ? | } l8N  | fB | 6D3jJlB | <BD3jJ{>T  |# L)?fB | 6D3jJlB | <BD3jJ{ | 0x"`~ EAn equation with a more positive E value reverse a less positive oneFHF F B | s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 l)(  ldB l <D)P0Q l 0Pp@ 0 0dB l@ <D)PXP l 0X0@ 0 0 l 0 xq~  USE OF Eo VALUES - WILL IT WORK?X!(2ffff   l 0x"`~ EAn equation with a more positive E value reverse a less positive one(FHE F  l 0{"`&N r|Will this reaction be spontaneous? Sn(s) + Cu2+(aq)   > Sn2+(aq) + Cu(s) Write out the appropriate equations Cu2+(aq) + 2e Cu(s) ; E = +0.34V as reductions with their E values Sn2+(aq) + 2e Sn(s) ; E = - 0.14V The reaction which takes place will involve the more positive one reversing the other i.e. Cu2+(aq)   > Cu(s) and Sn(s)   > Sn2+(aq) The cell voltage will be the difference in E values and will be positive... (+0.34) - (- 0.14) = + 0.48V If this is the equation you want then it will be spontaneous If it is the opposite equation (going the other way) it will not be spontaneousxW@]@@IP@)@"f -ff/f,ff/fUffffffM!fZ(      & F ?  l  n8N   l fB  l 6D3jJlB  l <BD3jJ{>T  l# L)?fB l 6D3jJlB l <BD3jJ{F ? l w f8N  l fB l 6D3jJlB l <BD3jJ{>T  l# L)?fB l 6D3jJlB l <BD3jJ{B l s *޽h ? ̙33y___PPT10Y+D=' = @B + 0L0 XI(  XdB X <D)P0Q X 0Pp@ 0 0dB X@ <D)PXP X 0X0@ 0 0 X 0{q~  USE OF Eo VALUES - WILL IT WORK?X!(2ffff  X 0{"`~ EAn equation with a more positive E value reverse a less positive one(FHE F  X 0<{"`&h Will this reaction be spontaneous? Sn(s) + Cu2+(aq)   > Sn2+(aq) + Cu(s) Split equation into two half equations Cu2+(aq) + 2e   > Cu(s) Sn(s)   > Sn2+(aq) + 2e Find the electrode potentials Cu2+(aq) + 2e Cu(s) ; E = +0.34V and the usual equations Sn2+(aq) + 2e Sn(s) ; E = - 0.14V Reverse one equation and its sign Sn(s)   > Sn2+(aq) + 2e ; E = +0.14V Combine the two half equations Sn(s) + Cu2+(aq)   > Sn2+(aq) + Cu(s) Add the two numerical values (+0.34V) + (+ 0.14V) = +0.48V If the value is positive the reaction will be spontaneousW@L@)@O@5@;P"f &ffffffff,fff-f!ff ff fffff (  |  V  F  9  P   F ?  X  ~8N   X fB  X 6D3jJlB  X <BD3jJ{>T  X# L)?fB X 6D3jJlB X <BD3jJ{F ? X  y8N  X fB X 6D3jJlB X <BD3jJ{>T  X# L)?fB X 6D3jJlB X <BD3jJ{B X s *޽h ? ̙33y___PPT10Y+D=' = @B +  0L0 $    (    0V{ JREVISION CHECK (2f  0P[{ ZWhat should you be able to do? (2   0x`{w JRecall the different types of half cells Recall the structure of the standard hydrogen electrode Recall the methods used to calculate standard electrode electrode potentials Write balanced full and half equations representing electrochemical processes Know that a reaction can be spontaneous if it has a positive E value Calculate if a reaction is feasible by finding its E value  2f#ff1ffFffIff:ff f f+ff    0Xq{   ~,CAN YOU DO ALL OF THESE? YES NO6- 2!    0 H @P 0 0   0f h @U 0 0dB   <D)P0Q  0Pp@ 0 0dB @ <D)PXP  0X0@ 0 0H  0޽h ? ̙33 0L0  ;(  9  0{{  YYou need to go over the relevant topic(s) again Click on the button to return to the menudG(282/$f f) f  s * x  P @S 0 0dB  <D)P0Q  0Pp@ 0 0dB @ <D)PXP   0X0@ 0 0H  0޽h ? ̙33 0L0  D(    04{xK (WELL DONE! 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