Chapter 18 Review - Electrochemistry
(sections 1-6, 8, 11, 13)

 Electrochemistry involves the interconversion of electrical and chemical energy 

  1. Redox Reactions
    1. Reactions where oxidation and reduction occur.  Note you can't have one without the other, oxidation and reduction always occur together.
    2. Oxidation - lose electrons, charge goes up.  Reduction - gain electrons, charge goes down
    3. Oxidizing agent - the chemical species that helps another chemical be oxidized, so itself is reduced
    4. Reducing agent - the chemical species that helps another chemical be reduced, so itself is oxidized
    5. Redox animation
    6. Example of a redox reaction:  Co (s) + Pb2+ (aq) g Pb (s) + Co2+ (aq)
      1. Co goes from zero to +2 and Pb2+ goes from +2 to zero
      2. What is oxidized?  Co (s).  What is reduced?  Pb2+ (aq)
      3. What is the oxidizing agent?  Pb2+ (aq).  What is the reducing agent? Co (s)
    7. Example:  CH4 (g) + 2 O2 (g) g CO2 (g) + 2 H2O (g)  a combustion reaction
      1. C in CH4 goes from -4 to a +4 in CO2 and O in O2 goes from zero to -2 in CO2 and H2O
      2. What is oxidized?  C in CH4.  What is reduced?  O in O2
      3. What is the oxidizing agent?  O2.  What is the reducing agent? CH4
      4. To answer the four questions above, you must name the entire molecule (like CH4) and for what is oxidized and reduced identify the element in that chemical that is changing. Note also that the answer to those four questions is never a product since they are being formed.
    8. Example:  2 Na (s) + 2 HCl (aq) g 2 NaCl (aq) + H2 (g)
      1. Na goes from zero in the solid to +1 in NaCl and H goes from +1 in HCl to zero in hydrogen gas
      2. What is oxidized?  Na (s).  What is reduced?  H in HCl (aq)
      3. What is the oxidizing agent?  HCl (aq).  What is the reducing agent? Na (s)
    9. Types of Redox reactions
      1. combustion (where O2 is the oxidizing agent)
      2. breathing (where O2 is the oxidizing agent)
      3. rusting (where O2 is the oxidizing agent)
      4. bleaching
      5. batteries
      6. single displacement reactions
    10. In order to do these problems you must be able to assign oxidation numbers (aka oxidation states, charge).  Remember that elements in their natural state have a charge of zero.  Alkali metals are always +1 in a compound. Alkaline Earth metals are always +2 in a compound.  H is +1 in a compound, except when bonded to a metal and then it is -1.  Fluorine is always -1 in a compound. Oxygen is -2 in a compound, except peroxides where it is -1 like in H2O2.  The other Halogens are usually -1.
      1. Example:  H2SO4   -  H is +1, S is +6, O is -2
      2. Example:  CO2    -  C is +4, O is -2
    11. Consider this reaction:  Cr3+(aq) + Be(s)  g Cr(s) + Be2+(aq)
      1. Is it balanced?  Actually no because the total charge on each side of the arrow is not equal - we can't create or destroy electrons! 
      2. To balance the reaction break it into oxidation and reduction 1/2 reactions.
      3. ox 1/2 rxn:  Be  g  Be2+ + 2e-.  This is oxidation since it loses electrons and the charge goes up.
      4. red 1/2 rxn:  Cr3+  +  3e  g  Cr.  This is reduction since it gains electrons and the charge goes down.
      5. We need the same number of electrons on both sides so they will cancel out when we add the 1/2 reactions together to get the overall balanced reaction.  So we need to multiply the Be reaction by 3 (give 6 electrons) and the Cr reaction by 2 (gives 6 electrons).
      6. The balanced reaction is  2 Cr3+(aq) + 3 Be(s)  g 2 Cr(s) + 3 Be2+(aq).  The 6 electrons cancelled.  Note the total charge on each side of the arrow (+6) is the same.  We did not create or destroy charge. 
  2. Galvanic Cells section 18.1 (chemical energy to electrical energy)
    1. If solid Be is put in a Cr(NO3)3 solution (which of course contains Cr3+ and NO3- ions) this reaction will occur:  2 Cr3+(aq) + 3 Be(s)  g 2 Cr(s) + 3 Be2+(aq).  If we could harness the electron transfer we could use the electron flow (current) as electricity.  We can do this by separating the reactants from each other and connecting by a wire so the electrons will flow along the wire, which is electricity.  Then appliances can be connected to this wire so that the electrons will flow through their wires and operate them.  Figure 18.1 shows another great example
    3. Above is a drawing of a galvanic cell for the reaction we balanced.  
      1. Solid Be at the anode is "jumping off" into solution becoming Be2+ ions.  The two released electrons CAN NOT SWIM so they travel through the conducting wire to the solid Cr electrode.  Oxidation occurs at the anode.  The anode mass is decreasing as it dissolves into ions.  
      2. Electrons are pushed into the solid Cr cathode and their negative charge attracts Cr3+ ions from the solution.  The Cr3+ ions join with 3 electrons to make solid Cr.  Reduction occurs at the cathode.  The cathode mass is increasing as more solid Cr is formed. 
      3. Very quickly the anode solution has a positive charge build up from the new Be2+ions being formed - there is an excess of positive charge with no negative to balance it.  The cathode solution however is losing Cr3+ ions as they become neutral solid Cr atoms so the negative nitrate ions are still there without positive charge to balance it out - a build up of negative charge.  So the battery ceases to operate due to charge build up.
      4. Hello Salt Bridge   The salt bridge contains a soluble salt solution so there are + and - ions present.  The anions are pulled / attracted to the positive anode solution so they travel through the bridge to the anode side.  The cations are pulled / attracted to the negative cathode solution so they travel to the cathode side.  Now the galvanic cell has voltage again as electrons flow via the wire from the anode to the cathode since there is no longer a charge buildup.
      5. The flow of electrons is measured in volts and we get a voltage reading from the voltmeter.  This voltage is called the cell potential (E) or the electromotive force (EMF) of the cell. 
      6. How does a battery die?   1) the anode can all dissovle and be gone 2) the cathode solution can all be gone 3) the salt bridge can run out of ions
      7. If you make a bigger battery will it have more volts?  No, it will just last longer!
    4. Slow galvanic cell demo
    5. Animations of several galvanic cells
    6. Figure 18.2 shows another galvanic cell all worked out.  Try example 18.1 and problem 18.1
  3. Short Hand Notation section 18.2. 
    1. The cell above can be represented by short hand notation which is  anode | anode ion || cathode ion | cathode.  This is very helpful as the anode g anode ion is the oxidation 1/2 rxn and the cathode ion  g cathode is the reduction 1/2 rxn.  So they are listed in order of oxidation reactant, product, reduction reactant, product.  For our cell above, the notation is Be(s) | Be2+ || Cr3+ | Cr(s).  All we really need is this notation to draw a complete cell.  (Hint hint)
    2. Not all half reactions involve a solid metal turning into cations. Become familiar with half reactions that use an inert electrode like C or Pt as discussed in class.  These will help - Figure 18.4, example 18.2, problems 18.2-4
  4. Standard Cell Potentials Eo section 18.4 (standard is gases at 1 atm, solutions at 1 mol/L, temp at 298 K)
    1. Reduction 1/2 rxn potentials (Eored) are in tables.  If you need the potential for the ox 1/2 rxn, just write the reduction rxn backwards and change the sign of Eored.  So if you have the 1/2 rxns for your galvanic cell, you can find the cell potential Eocell by adding the 1/2 rxn potentials together.  Note a volt = Joule / Coulomb (V=J/C or J=CV or C=J/V)
    2. For our reaction above, Be  g  Be2+ + 2e- has a Eo  = 1.85 V.  Cr3+  +  3e  g  Cr has a Eo  = -0.74 V.  So the overall reaction has an Eocell = 1.11 Volts.  Notice we do NOT multiply the 1/2 rxn cell potentials by the numbers used to balance the overall reaction.  This is because a larger cell (battery) has the same voltage as a smaller cell - it just runs longer.  
    3. All potentials are compared to SHE - the standard hydrogen electrode.  We let the Eored for H2  g 2H+ + 2e be equal to zero volts.  All other reduction 1/2 reaction potentials are compared to SHE.
    4. Note a positive Eocell means the products are favored - the reaction will go forward to reach equilibrium.  So a negative Eocell  means the reactants are favored - the reaction doesn't really go forwards.  In order for a battery to work, we need a positive cell potential. If you have a negative cell potential it means you switched the anode and cathode and the reaction really goes in reverse - what you thought was oxidized is really reduced and what you thought was reduced is really oxidized.  See Table 18.1, Appendix D and problem 18.6
  5. Using the Eored tables. section 18.5
    1. As Eored gets more and more positive, it means the reaction really likes to go in that direction, reduction is natural - reduction get easier and easier.  So looking at the table, F2(g) really wants to be reduced into 2F- ions, it is a great oxidizing agent.  Eored = 2.87 V.  This makes sense as fluorine is the most electronegative element and really really wants to gain one electron to be like the noble gas Neon.  The higher the Eored the easier to reduce and the better the reactant can be an oxidizing agent. (Note this means the product is really hard to oxidize, the rxn does not want to reverse and be an oxidation rxn.  Thus F- ions are terrible reducing agents)
    2. As Eored gets more and more negative, it means the reaction really wants to go in the reverse direction, reduction will be unnatural - but oxidation get easier and easier.  So looking at the table, Li(s) really wants to be oxidized into Li+ ions.  Eored = -3.05 V so Eoox = +3.05 V.  This makes sense as lithium is a very unstable element and really really wants to lose one electron to be like the noble gas Helium. The lower the Eored the harder to reduce, so the reaction would rather reverse and be oxidation, and the better the product (reactant once written in reverse) can be an reducing agent.
    3. Consider these reduction 1/2 reactions from the Eored Table:
      1. Au3+  +  3e  g  Au(s)   1.50V
      2. Br2(l)  + 2e  g  2Br -     1.07V
      3. Pb2+  +  2e  g  Pb(s)    -0.13V
      4. Ni2+ +  2e  g  Ni(s)      -0.25V
        1. First notice they are all reductions, the charge went down, written this direction.  The reactants are all reduced:  oxidizing agents.  The products are all oxidized:  reducing agents.
        2. What is the best reducing agent?  Ni(s)  (This means the easiest to oxidize so we must look at the reverse rxn, the oxidation)
        3. What is the worst reducing agent?  Au(s) (Again look at the reverse reaction and what doesn't want to go reverse)
        4. What is the best oxidizing agent?  Au3+    (This means easiest to reduce so what reactant has the largest number)
        5. What is the worst oxidizing agent?  Ni2+
        6. Will Ni(s) react with Au3+?  That would give a Eocell = +0.25 + 1.50 = 1.75V so yes.
        7. Will Br - react with Pb2+?  That would give a Eocell = -1.07 -0.13 = -1.20V so no.
        8. What combination would give the highest voltage?  Au3+ + Ni(s)  Eocell = 1.50 + 0.25 = 1.75V
    4. Try examples 18.4-5 and problems 18.7-9
  6. How Eocell , K and DGo relate sections 18.3 and 18.8
    1. DGo = -nFEocell where n is the number of electrons transferred in the balanced overall cell reaction and F is Faraday's constant = 96,500 J/Vmol e or 96,500 C/mol e
    2. Eocell = (RT/nF) (ln K)    where ln K is in the numerator and R = 8.314 J/molK  See Figure 18.7
    3. If any one of these (Eocell , DGo , K) is known, the other two can be found.  
    4. Calculate DGo for the galvanic cell depicted above.  We already found that Eocell = 1.11 V.  So DGo = (-6 mol e)(96,500 J/Vmol e)(1.11 V)(kJ / 1000J) = -643 kJ for the balanced reaction. 
    5. Calculate K for this cell :  Ni(s) | Ni2+ || Br2(l) | Br -(aq) | Pt(s)
      1. First what does this mean?  It means that at the anode we have solid nickel becoming nickel 2+ ions. At the cathode we have bromine becoming bromide ions and since neither of these are solids, we need the inert Pt metal electrode to carry the electrons to the liquid bromine so the bromine can gain the electrons to make the bromide ions. 
      2. You balance the 1/2 reactions and make sure you get 2 electrons being transferred and the total reaction Ni(s) + Br2(l) g Ni2+ + 2 Br -(aq)
      3. The Eocell = 1.32 V. 
      4. Now solve for K:  1.32 V = [(8.314 J/molK)(298K) / (2 mol e)(96500 J/Vmol e)] ln K
      5. Solving we get ln K = 102.83  and taking e of both sides K = 4.54 x 1044
      6. Just for those who are curious:  Eventually the exponent will be greater than 99 and our calculators don't show more than 2 digits in a calculator - so must solve like this: pretend you get  ln K = 300, If you take e of both sides you get ERROR
        So ln K = 150 + 150 (divide the large number of 300 by 2 so you have smaller numbers)
        take each number to e
        K = (1.39 x 1065)(1.39 x 1065) and solve by doing 1.39 squared then adding the exponents
        K = 1.93 x 10130    COOL!!!
    6. When a reaction favors products, Eocell is positive, DGo is negative, K is > 1
    7. When a reaction favors reactants, Eocell is negative, DGo is positive, K is < 1
    8. Try examples 18.3 and 18.9, and problems 18.5 and 18.13-14
  7. Concentration and Eocell  section 18.6
    1. What if all the solutions are not at 1M?  Then we have a non standard Ecell 
    2. Hello Nernst equation E = Eocell - (RT/nF) ln Q     Think of this as the non standard potential is related to the standard potential minus a correction factor that takes into account temperature, pressure and concentrations.
    3. At equilibrium E becomes zero and Q becomes K.  At equilibrium for a redox reaction no more electrons will flow.  So we want a cell potential with a large K (very positive potential) so that the reaction must go forwards for a long time before reaching equilibrium.  We want products heavily favored.  
    4. Try example 18.6-7 and problem 18.10
    5. Calculate the cell potential for the above pictured cell if [Be(NO3)2] = 0.88M and [Cr(NO3)3] = 0.14M.  We need the balanced reaction:  2 Cr3+(aq) + 3 Be(s)  g 2 Cr(s) + 3 Be2+(aq)  Now plug into the Nernst equation:  E = 1.11V - (8.314 J/molK)(298K) / (6 mol e)(96500 J/Vmol e) ln [(0.88)3 / (0.14)2] = 1.09 V.  
  8. Electrolysis sections 18.11 and 18.13
    1. Table 18.2
    2. Electrical energy from an external source (outlet or other battery) is used to force a redox reaction to go in the non spontaneous direction.  This is called an electrolytic cell.
    3. Example:  melted (molten) salt.  2NaCl(l) g  2Na(s) + Cl2(g).  Ecell = - 4 V.  We must have MORE than 4 volts to drive this reaction forwards, naturally it goes backwards so that the voltage is + 4 volts.  This is how we make elemental sodium.  (Normally salt does NOT decompose into its elements! Ever smell chlorine gas in your salt shaker???)
    4. Example:  electrolysis of water.  2H2O(l)  g 2H2(g) + O2(g).  Ecell = -1.23 V.  Again, water does not normally decompose into hydrogen and oxygen elemental gases! Thank goodness or we'd blow up.
    5. Example:  aqueous salt.  2NaCl + 2H2O  g  Cl2 + H2 + 2NaOH.  Ecell = -2.19 V  Again, salt water does not normally make chlorine gas, hydrogen gas and sodium hydroxide!  But by hooking salt and water up to more than 2.19 volts, we can drive this reaction forwards. This is actually how we make very pure NaOH. 
    6. Units:  current is measured in amps (A).  A = Coulomb per second (C/s)  Remember that 
    7. How many grams of copper can be collected in 1.00 hour by a current of 1.62 A from a CuSO4 solution?  Solution:  Change hours to seconds since A is per seconds.  1.00 hr (60 min/hr)(60 sec/min) = 3600 sec.  Now multiply by A.  3600 sec(1.62 C/sec) = 5832 C.  Now use Faraday's constant to convert to mole e.  5832 C(mol e / 96,500C) = 0.06044 mol e.  Now how many moles e are in one mol of copper?  0.06044 mol e (mol Cu / 2 mol e) = 0.03022 mol Cu.  Now change to grams.  0.03022 mol Cu(63.5 g/mol) = 1.92 g Cu.  Whew! In a nutshell:  get seconds (1.62 C/sec)(mol e / 96,500C)(mol Cu / 2 mol e)(63.5 g/mol) = 1.92 g Cu.  
    8. Figure 18.20 and example 18.10

    Study Hard and Good Luck.

Practice Problem

Create a functioning galvanic cell, draw the cell, label all parts, find the cell potential, write the 1/2 rxns and overall reaction given: Al3+ (aq)   g Al(s)   Eored = -1.66V              and               Cu2+ (aq)   g Cu(s)   Eored = +0.34V  

  1. All this is Just FYI - not covered in this class. 
  2. A variation on non standard cells is the Concentration Cell where both the anode and cathode contain the same chemicals but in different concentrations.  For example Cu solid in CuSO4 solution for both the anode and cathode but one solution is 0.3M and the other is 1.2M.
    1. How can this work?  Eocell for this would be zero as the ox and red 1/2 rxns cancel out.  But it is NOT standard.  The driving force behind the reaction is that both solutions are connected by the salt bridge so they want to have the same concentration. (Similar to the force behind osmosis).  The dilute solution wants more ions to become more concentrated - so the solid on that side will oxidize and make ions = anode.  The concentrated solution wants less ions to become more diluted - so the ions reduce by gaining electrons and become part of the solid cathode. 
    2. The anode reaction is Cu(s) g  Cu2+ (dilute) + 2e  which is oxidation as electrons are lost.  The cathode reaction is Cu2+ (conc) + 2e g  Cu(s)  which is reduction as electrons are gained.  Overall the reaction is Cu2+ (conc)  g Cu2+ (dilute)  since the 2 electrons and solid Cu cancel out.  
    3. If a concentration cell were made of copper electrodes and copper sulfate solutions of 0.13M and 1.2M calculate the cell potential.  Well we already have the overall cell reaction and the 1/2 reactions worked out.  We also know that Eocell is zero (that is if both solution were 1M there would be no force driving the reaction since the solution concentrations are already equal) So we plug into the Nernst equation:  E = 0 - [(8.314 J/molK)(298K) / (2 mol e)(96500 J/Vmol e)]  ln [(0.13) / (1.2)] = 0.029V.  
  3. Batteries
    1. Dry Cell - normal AA, D batteries.  Solid Zn is the anode in a Zn2+ paste.  The cathode is complex.  Ecell = 1.5 V.
    2. Mercury battery - used in pacemakers, hearing aids, watches.  Solid Zn anode, HgO cathode.
    3. Car battery - lead storage battery.  Solid Pb anode, PbO2 cathode, both immersed in sulfuric acid solution.  We put 6 of these cells in series so the voltage adds up.  Each is 2V so the total is 12V.  Car batteries are rechargeable.
    4. Lithium batteries - all solids.  Li solid anode and TiS2 solid cathode connected by a polymer than conducts only ions, not electrons, so the electrons are forced to more along a wire like in a normal battery.  Rechargeable.
    5. Fuel Cells.  Anode is H2 gas and cathode is O2 gas.  The product is water (environmentally friendly isn't it?)  Ecell = 1.23 V. This cell needs two inert metals to conduct the electrons from the anode to the cathode.  Inert means they don't react - they are not involved in the cell reaction.  Problem - must continue to supply the reactant gases to the cell.  Currently in use in space - the astronauts drink the water produced!  Seriously. 
  4. Corrosion
    1. Rust, tarnish and other damage to metals.  Costs about $100 billion each year to fix, clean up, repair, etc...
    2. Rust is the most common.  Fe(s) becomes Fe2+ while O2 reacts to form H2O.  Ecell = 1.67 V. So yeah, rust forms spontaneously.  The iron ions react with water and oxygen gas to make Fe2O3, iron(III) oxide commonly known as rust.  This happens even easier in humid areas, where roads are salted during winter, and where acid rain falls.  
    3. How to solve corrosion?  Primary methods are to coat the metal with paint, oxide material, or another metal which can be sacrificed to take the damage instead.  A tin can is really iron with a coating of tin.  Galvanized iron is iron with a coating of zinc.