REDOX Notes PDF

Oxidation Numbers (States) pg 9  an oxidation number is the charge an atom appears to have when found in a neutral molecule or charged polyatomic ion  can be used when you have a molecular compound where there are no ion charges to determine if oxidation or reduction is occurring  how d...

Oxidation Numbers (States) pg 9  an oxidation number is the charge an atom appears to have when found in a neutral molecule or charged polyatomic ion  can be used when you have a molecular compound where there are no ion charges to determine if oxidation or reduction is occurring  how do you use a change in the number? 1. if the number decreases then reduction has occurred 2. if the number increases then oxidation has occurred Rules for Assigning Oxidation Numbers: 1. In a pure element, the oxidation number is zero. Mg(s)= O, P4(s)=O, Cl2=O, Ba(s)=), S8(s)=O 2. In simple ions, the oxidation number is equal to the ion charge. Mg2+(aq)= 2+, P3-(aq)=3-, Cl- (aq)=-1, Ba(s)=), S2- (aq)=2- 3. In most compounds containing hydrogen, the oxidation number for hydrogen is +1. (Exception is the metal hydrides eg) LiH where the oxidation number of hydrogen is –1 ). 4. In most compounds containing oxygen, the oxidation number for oxygen is –2. (Exception is the peroxides eg) H2O2, Na2O2 where the oxidation number of oxygen is –1) 5. The sum of oxidation numbers of all atoms in a substance must equal the net charge of the substance. ( Zero for compounds and the charge of the polyatomic ion) eg) sum of MgO = 0 sum of PO43- = –3 6. In molecular compounds that do NOT contain hydrogen or oxygen, the more electronegative element is assigned an oxidation number equal to the negative charge it is usually has when it is in ionic compounds e.g. The oxidation number in PCl3 is -1 and in CS2, the sulfur is -2 Example What is the oxidation number (state) for the element identified in each of the following substances: a) N in N2O +1 –2 individual oxidation numbers N2 O sum of oxidation +2 –2 = 0 numbers b) N in +5 –2 NO3- N O3– +5 –6 = –1 c) C in C2H5OH –2 +1 –2+1 C2 H 5 O H –4 +5 –2+1 = 0 d) C in C6H12O6 0 +1 –2 C6 H12 O6 0 +12 –12 = 0 e) P in P2O5 +5 –2 P 2 O5 +10 -10 =0  figuring out oxidation numbers can help to identify whether a reaction is a redox reaction or not  for it to be a redox reaction, there has to be both an increase in oxidation number and a decrease in oxidation number seen in the reaction 0 +1 -1 0 +1 -1 eg) Ag(s) + NaNO3(aq) Na(s) + AgNO3(aq) Ag increases oxidized redox!!! Na decreases  reduced… +2 -2 +1 -1 +2 -1 +1 -2 PbSO4(aq) + 2 KI(aq) PbI2(s) + K2SO4(aq) nothing changes  NOT a redox reaction! electron transfer occurs in living systems eg) photosynthesis, cellular respiration also occurs in non-living systems eg) combustion, bleaching, metallurgy  F. Disproportionation  disproportionation occurs when one element is both oxidized and reduced in a reaction eg) -1 -2 0 2 H2O2(aq) 2 H2O(l) + O2(g) 0 +1 -1 Cl2(g) + 2 OH-(aq) ClO-(aq) + Cl-(aq) + H2O(l) G. Balancing Redox Reactions  sometimes most reactants and products are known but the complete reaction is not given…called a skeleton reaction 1. Half Reaction Method 1. Assign oxidation numbers (ON). 2. Balance the element that changes in oxidation number. 3. Add e– to balance the change in total oxidation number.(ON  subscript  coefficient) 4. Balance O using H2O(l). 5. Balance H using H+(aq). 6. Check that the half-reaction is balanced with respect to net charge.  Example 1: Balance the following half reaction : (+6) (+3) +6 2 +3 2 4 H+(aq) + 3 e– + CrO42-(aq) CrO2-(aq) + 2 H2O(l) +6 8 = 2 +3 4 = 1 net charge = –1 net charge = –1 (Cr is already balanced)  Example 2: Balance the following half reaction: (+6) (0) +1 +3 2 0 6 H+(aq) + 6 e– + 2 HClO2(aq) Cl2(g) + 4 H2O(l) +1 +3 4 = 0 net charge = 0 net charge = 0 Steps 1. Assign oxidation numbers. 2. Separate the partial net equation into two half reactions (omit any H2O(l) or H+(aq)). 3. Balance each half-reaction. 4. Multiply one or both of the equations so e lost = e gained. 5. Add the equations to produce a balanced net ionic equation 6. Simplify. Check to see if all elements and charges are balanced.  Example 1: Balance the following using oxidation numbers, assuming acidic conditions: +6 2 +4 2 +3 2 +6 2 CrO42-(aq) + SO32-(aq) CrO2-(aq) + SO42-(aq) +6 8 = 2 +4 6 = 2 +3 4 = 1 +6 8 = 2 (+6) (+3) Red 2 [ 4 H+(aq)+ 3 e–+ CrO42-(aq) CrO2-(aq)+ 2 H2O(l)] (+4) (+6) Ox 3 [ H2O(l)+ SO32-(aq) SO42-(aq)+ 2 e– + 2 H+(aq) ] 8 H+(aq) + 2 CrO42-(aq) + 3 H2O(l) + 3 SO32-(aq) 2 CrO2-(aq) + 4 H2O(l) + 3 SO42-(aq) + 6 H+(aq) Net 2 H+(aq) + 2 CrO42-(aq) + 3 SO32-(aq) 2 CrO2-(aq) + H2O(l) + 3 SO42-(aq) 2. Oxidation Number Method 1. Assign oxidation numbers. 2. Balance the substances that change in oxidation number. 3. Use a line to join the reducing agent with its corresponding product (ignore the H+(aq) and H2O(l)) and a line to join the oxidizing agent with its corresponding product. 4. On each line, write the change in oxidation number  # of atoms. 5. Multiply the RA and/or OA to balance the change in oxidation number. 6. Balance the H2O(l) and the H+(aq).  Example 1: Balance the following reaction using the oxidation number method. +7 +4 +4 +6 2 H+(aq) + 2 __ 3 SO32-(aq) __ __MnO4-(aq) + __ 2 MnO2(aq) + __ 3 SO42-(aq) + __ 1 H2O(l)  = 3  1 atom = 3  2 =6  = +2  1 atom = +2  3 =+6  Example 2: Balance the following reaction using the oxidation number method. +4 -1 +3 +5 6 3 __ H2O(l) + 3 1 Br(aq) __N2O4(g) + _ 2_ NO2 (aq) + _1 BrO3(aq) + __ 6 H+(aq)  = 1  2 atoms = 2  3 =6  = +6  1 atom = +6  Electrochemistry A. Redox Reactions electrochemistry is the branch of chemistry that studies electron transfer in chemical reactions  oxidation is a loss of electrons“LEO” eg) Mg(s) Mg2+(aq) + 2e 2Cl(aq) Cl2(g) + 2e  reduction is a gain of electrons “GER” eg) Fe3+(aq) + 3e Fe(s) Br2(l) + 2e 2Br(aq)  oxidation and reduction reactions occur together, hence the term redox  the reduction and oxidation reactions are called the half reactions  “adding” the half reactions together will give you the net ionic equation that takes place during the redox reaction  the e lost in the oxidation half reaction must equal the e gained in the reduction half reaction  you may have to multiply one or both of the half reactions to balance the e  spectator ions (ions not changing) are NOT included!  the substance that is reduced is called the oxidizing agent ( OA) (it causes the oxidation by taking e-)  the substance that is oxidizedis called the reducing agent (RA) (it causes the reduction by giving up e-) Example 1 Given the following reaction, write the half reactions and the net ionic equation. 0 1+ 1– 0 1+ 1– Na(s) + LiCl(aq) Li(s) + NaCl(aq) ox red Cl- is spectator Ox: Na(s) Na+(aq) + 1e- Red: Li+(aq) + 1e- Li(s) Net: Li+(aq) + Na(s) Li(s) + Na+(aq) Example 2 Given the following reaction, write the half reactions and the net ionic equation. 0 3+ 1– 0 2+ 1– 3 Zn(s) + 2 Au(NO3)3(aq) 2 Au(s) + 3 Zn(NO3)2(aq) ox red NO3- is spectator Ox: 3 [ Zn(s) Zn2+(aq) + 2e-] Red: 2 [ Au3+(aq) + 3e- Au(s)] Net: 2 Au3+(aq) + 3 Zn(s) 2 Au(s) + 3 Zn2+(aq) B. Spontaneous Redox Reactions  chemical reactions which occur on their own, without the input of additional energy , are called spontaneous not all reactions are spontaneous  in the table of redox half reactions (see pg 7 in Data Booklet), the strongest oxidizing agent (SOA) is at the top left and the strongest reducing agent (SRA) is at the bottom right  the redox spontaneity rule states that a spontaneous reaction occurs if the oxidizing agent is above the reducing agent in the table of redox half reactions Try These: For each of the following combinations of substances, state whether the reaction would be spontaneous or non-spontaneous: Cr3+(aq) with Ag(s) non-spontaneous I2(s) with K(s) spontaneous H2O2(l) with Au3+(aq) spontaneous Sn2+(aq) with Cu(s) non-spontaneous Fe2+(aq) with H2O (l) non-spontaneous (both ways) C. Predicting Redox Reactions  we will be predicting the strongest or most dominating reaction that occurs when substances are mixed (other reactions do take place because of atomic collisions!) Steps: 1. List all species present as reactants  dissociate soluble ionic compounds and acids  do not dissociate molecular compounds  include H+(aq) ions if it is acidic  always include H2O(l) 2. Identify each as OA or RA (***some can be both so memorize them… Fe2+, Cr2+ , Sn2+ , H2O) 3. Identify the SOA and SRA using the table. 4. Write out the half reactions for the SOA and SRA. 5. Determine the net ionic reaction 6. Determine spontaneity (SOA must be higher than SRA to be spontaneous) Example 1 Predict the most likely redox reaction when chromium is placed into aqueous zinc sulphate. Cr(s) Zn2+(aq) SO42-(aq) H2O(l) S RA S OA OA with H2O(l) OA/RA SOA (Red): Zn2+(aq) + 2e- Zn(s) SRA (Ox): Cr(s) Cr2+(aq) + 2e- spont Net: Zn2+(aq) + Cr(s) Zn(s) + Cr2+(aq) Example 2 Predict the most likely redox reaction when silver is placed into aqueous cadmium nitrate. Ag(s) Cd2+(aq) NO3-(aq) H2O(l) S RA S OA OA with H+ (aq) OA/RA SOA (Red): Cd2+(aq) + 2e- Cd(s) SRA (Ox): 2 [ Ag(s) Ag+(aq) + e- ] nonspont Net: Cd2+(aq) + 2 Ag(s) Cd(s) + 2 Ag+(aq)  Example 3 Predict the most likely redox reaction when potassium permanganate is slowly poured into an acidic iron (II) sulphate solution. K+(aq) MnO4-(aq) H+(aq) Fe2+(aq) H2O(l) SO42-(aq) OA S OA with H+ (aq) OA OA/S RA OA/RA OA with H+ (aq), H2O(l) SOA (Red): MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) SRA (Ox): 5 [ Fe2+(aq) Fe3+(aq) + e- ] spont Net: MnO4-(aq) +8H+(aq) + 5Fe2+(aq) Mn2+(aq) + 4H2O(l) + 5 Fe3+(aq) D. Generating Redox Tables  you can be given data for certain ions and elements then be asked to generate a redox table like the one on pg 7 of you Data Booklet (a smaller version!)  you may have to generate a table from real or fictional elements and ions  the tables that we use are all written as reduction half reactions Example 1 Generate a redox table given the following data:  indicates no reaction  indicates a reaction Cu2+(aq) Zn2+(aq) Pb2+(aq) Ag+(aq) Cu(s)     Zn(s)     Pb(s)     Ag(s)      Redox Table SOA Ag+(aq) + e- Ag(s) Cu2+(aq) + 2e- Cu(s) Pb2+(aq) + 2e- Pb(s) Zn2+(aq) + 2e- Zn(s) SRA Put the oxidizing agents in order from strongest to weakest. Ag+(aq), Cu2+(aq), Pb2+(aq), Zn2+(aq) Put the reducing agents in order from strongest to weakest. Zn(s), Pb(s), Cu(s), Ag(s) Example 2: Generate a redox table given the following data: Cu(s) + Ag+(aq) Cu2+(aq) + Ag(s) Zn2+(aq) + Ag(s) no reaction Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) Hg(l) + Ag+(aq) no reaction Label each as OA or Ra Redox Table SOA Hg2+(aq) + 2e- Hg(l) Ag+(aq) + e- Ag(s) Cu2+(aq) + 2e- Cu(s) Zn2+(aq) + 2e- Zn(s) SRA  Example 2 (continued): Put the oxidizing agents in order from weakest to strongest. Zn2+(aq), Cu2+(aq), Ag+(aq), Hg2+(aq) Put the reducing agents in order from weakest to strongest. Hg(l), Ag(s), Cu(s), Zn(s) Example 3: Generate a redox table given the following data: 2X-(aq) + Y2(g) spontaneous reaction 2Z-(aq) + Y2 (g) no reaction 2Z-(aq) + W2 (g) spontaneous reaction Label each as OA or RA Redox Table SOA W2 (g) + 2e- 2W-(aq) Z2 (g) + 2e- 2Z-(aq) Y2 (g) + 2e- 2Y-(aq) X2 (g) + 2e- 2X-(aq) SRA  Example 3 (continued): Put the oxidizing agents in order from strongest to weakest. W2(g), Z2(g), Y2(g), X2(g) Put the reducing agents in order from strongest to weakest. X-(aq), Y-(aq), Z-(aq), W-(aq) H. Redox Stoichiometry 1. Calculations  stoichiometry can be used to predict or analyze a quantity of a chemical involved in a chemical reaction  in the past we have used balanced chemical equations to do stoich calculations  we can now apply these same calculations to balanced redox equations Example 1 What is the mass of zinc is produced when 100 g of chromium is placed into aqueous zinc sulphate. Cr(s) Zn2+(aq) SO42-(aq) H2O(l) RA OA OA with H2O(l) OA/RA SRA SOA SOA (Red): Zn2+(aq) + 2e- Zn(s) SRA (Ox): Cr(s) Cr2+(aq) + 2e- Net: Cr(s) + Zn2+(aq) Zn(s + Cr2+(aq) ) Cr(s) + Zn2+(aq) Zn(s) + Cr2+(aq) m = 100 g m=? M = 52.00 g/mol M = 65.39 g/mol n=m n = 1.923… mol x 1/1 M = 1.923… mol = 100 g 52.00 g/mol m = nM = 1.923… mol = (1.923…mol)(65.39 g/mol) = 125.75 g = 126 g  Example 2 What volume of 1.50 mol/L potassium permanganate is needed to react with 500 mL of 2.25 mol/L acidic iron (II) sulphate solution? K+(aq) MnO4-(aq) H+(aq) H2O(l) Fe2+(aq) SO42-(aq) OA OA with H+(aq)OA OA/RA OA/RA OA with H+(aq) SOA SRA OA with H2O(l) SOA (Red): MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l) SRA (Ox): 5 [ Fe2+(aq) Fe3+(aq) + e- ] Net: MnO4-(aq) +8H+(aq) + 5Fe2+(aq) Mn2+(aq)+ 4H2O(l) + 5Fe3+(aq) MnO4-(aq)+8H+(aq) + 5Fe2+(aq) Mn2+(aq)+ 4H2O(l) + 5Fe3+(aq) v=? v = 0.500 L c = 1.50 mol/L c = 2.25 mol/L n = 1.125 mol x 1/5 n = cv = 0.225 mol = (2.25 v= n mol/L)(.500L) C = 1.125 mol v = 0.225 mol 1.50 mol/L = 0.150 L 2. Titrations  a titration is a lab process used to determine the volume of a substance needed to react completely with another substance  this volume can then be used to calculate an unknown concentration using stoichiometry  one reagent ( titrant- OA ) is slowly added to another (sample - RA ) until an abrupt change ( endpoint ) occurs, usually in colour  in redox titrations, two common oxidizing agents are used because of their colour and strength: 1. permanganate ions (MnO4-(aq)) – purple 2. dichromate ions (Cr2O72-(aq)) – orange  as long as the sample (RA) in the flask is reacting with the permanganate ions (dichromate ions) the sample will be colourless (green)  when the reaction is complete, any unreacted permanganate ions will turn the sample purple (pink) (with dichromate, sample goes from orange to green)  the volume of titrant (OA) needed to reach the endpoint is called the equivalence point  the concentration of the titrant must be accurately known  the concentration of the permanganate solution must be calculated against a primary standard (a solution of known concentration) before it can be used in a titration itself  this is done just prior to the titration Example Find the concentration of (standardize) the KMnO4(aq) solution by titrating 10.00 mL of 0.500 mol/L acidified tin (II) chloride with the KMnO4(aq). Trial 1 2 3 4 Final Volume 18.40 35.30 17.30 34.10 (mL) Initial Volume 1.00 18.40 0.60 17.30 (mL) Volume of KMnO4(aq).(mL) 17.40 16.90 16.70 16.80 Endpoint pink light light light Colour pink pink pink  endpoint average is calculated by using 3 volumes within 0.20 mL Endpoint average = 16.90 mL + 16.70 mL + 16.80 mL 3 = 16.80 mL  Analysis:  determine net ionic redox equation K+(aq) MnO4-(aq) H+(aq) H2O(l) Sn2+(aq) Cl-(aq) OA OA with H+(aq)OA OA/RA OA/RA RA SOA SRA SOA (Red): 2 [ MnO4-(aq) + 8H+(aq) + 5e- Mn2+(aq) + 4H2O(l)] SRA (Ox): 5 [ Sn2+(aq) Sn4+(aq) + 2 e- ] Net: 2MnO4-(aq)+ 16H+(aq)+ 5Sn2+(aq) 2Mn2+(aq)+ 8H2O(l) + 5Sn4+(aq)   use net redox equation to calculate KMnO4(aq) concentration 2MnO4-(aq) +16H+(aq) +5Sn2+(aq) 2Mn2+(aq)+ 8H2O(l) +5Sn4+(aq) v = 0.01680 L v = 0. 01000 L C=? c = 0.500 mol/L n = 0.00500 mol x n = cv 2/5 = (0.500 mol/L)(0.01000 L) = 0.00200 mol = 0.00500 mol C= n v C = 0.00200 mol 0.01680 L = 0.119 mol/L I. Electrochemical Cells 1. Voltaic Cells  electric cells are devices that convert chemical energy into electrical energy  in redox reactions, e- are transferred from the oxidized substance to the reduced substance  the transfer of e- can occur through a conducting wire separating the two substances in containers called half cells  a voltaic cell is an arrangement where two half cells are joined so that the e- and ions can move between them  electrodes are made of good conducting materials so e- can flow…can be the metal of the solution or inert such as carbon  the electrolyte is a solution that contains ions and will transmit ions (charged particles)  the electrode where oxidation occurs is called the anode  if the anode is a metal, it loses mass as the cell operates  the anode is labelled as negative since it is the electrode where the electrons originate  the anions move to the anode since this electrode loses electrons (leaving a net positive charge in the electrode)  the electrode where reduction occurs is called the cathode  if the cathode is a metal, it gains mass as the cell operates  the cathode is labelled as positive since the anode is labelled negative  the cations move to the cathode since this electrode accepts electrons (leaving a net negative charge in the electrode)  electrons flow from the anode (LEOA) to the cathode (GERC) through a connecting wire  ions must be able to move to their attracting electrode (either through the porous cup or a salt bridge ) otherwise a buildup of charge will occur opposing the movement of e-  the flow of ions through the solution and e- through the wire maintains overall electrical neutrality 2. Standard Reduction Potentials  reduction potentials are the ability of a half cell to attract e-  these potentials are measured using a voltmeter  each half reaction listed in the Data Booklet has an E value measured in volts assigned to it  all values in the table are arbitrarily assigned based on a standard  the hydrogen cell half reaction has been set as the standard and has an E value of 0.00 V 3. Predicting Voltage of a Voltaic Cell  the standard cell potential (Enet) is determined by adding the E values for the two half reactions  the sign on the E value for the oxidation half reaction must be reversed  if you multiply an equation to balance e-, you DO NOT multiply the E value (voltage is independent of number of e- transferred)  a positive E net is a spontaneous reaction  a negative E net is a nonspontaneous reaction  Example Calculate the E net for the reaction of Zn(s) with CuSO4(aq). Zn(s) Cu2+(aq) SO42-(aq) H2O(l) S RA S OA OA with H2O(l) OA/RA SOA (Red): Cu2+(aq) + 2e- E = +0.34 V Cu(s) SRA (Ox): Zn(s) Zn2+(aq) + 2e- E = +0.76 V Net: Zn(s) + Cu2+(aq) Cu(s + Zn2+(aq) Enet = +1.10 V 4. Shorthand Notation Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s) OR Zn(s) / Zn2+(aq) // Cr2O72-(aq) , H+(aq) , Cr3+(aq)/ C(s) ***anode // cathode  line (/) separates phases  double line(//) represents the porous cup or salt bridge and separates the two half reactions  comma (,) separates chemical species in the same phase 5. Drawing Cells  when drawing a cell from the shorthand notation, you have to be able to label the cathode, anode, positive terminal, negative terminal, electrolytes, direction of e flow, and directions of cation and anion flow  you also have to show and label the reduction half reaction, oxidation half reaction and net reaction including E values, E net and spontaneity Example Draw and fully label the following electrochemical cell: Al(s)/ Al3+(aq) // Ni2+(aq) / Ni(s) e- V Al(s) Ni(s) anode cathode negative positive terminal terminal Ni2+ (electrolyte) anions Al3+ (electrolyte) cations  Al(s) Al3+(aq) Ni(s) Ni2+(aq) H2O(l) S RA OA RA S OA OA/RA SOA (Red): 3 [ Ni2+(aq) + 2e- Ni(s)] E = –0.26 V SRA (Ox): 2 [ Al(s) Al3+(aq) + 3e- ] E = +1.66 V Net: 2 Al(s) + 3 Ni2+(aq) 3 Ni(s + 2 Al3+(aq) )Enet = +1.40 V spontaneous:yes J. Commercial Cells  batteries are made by connecting two or more voltaic cells in series (one after the other)  the voltage of the battery is the sum of the individual cells  there are many types of batteries:  a) Dry Cell  common 1.5 V and 9 V batteries of clocks, remote controls, noisy kids toys etc. Cathode (Red): 2 MnO2(s) + H2O(l) + 2e- Mn2O3(aq) + 2 OH-(aq) E= +0.79 V Anode (Oxid): Zn(s) Zn2+(aq) + 2e- E = +0.76 V Net: 2 MnO2(s)+ H2O(l) + Zn(s) Mn2O3(aq) + 2 OH-(aq) + Zn2+(aq) Enet = +1.55 V  the OH- produced causes irreversible side reactions to occur making recharging impossible  b) Nickel-Cadmium  one type of rechargeable battery Cat (Red): 2 NiO(OH)(s)+ 2 H2O(l) + 2e- 2 Ni(OH)2(s) +2 OH-(aq) E= +0.49 V An (Oxid): Cd(s) + 2 OH-(aq) 2 Cd(OH)2(s) + 2e- E = +0.76 V Net: 2 NiO(OH)(s)+ 2 H2O(l) + Cd(s) 2 Ni(OH)2(s)+ 2 Cd(OH)2(s) Enet = +1.25 V c) Lead Storage Battery  typical car battery where lead serves as the anode, and lead coated with lead dioxide serves as the cathode  both electrodes dip into an electrolyte solution of sulfuric acid  six cells are connected in series Cat (Red): PbO2(s)+ HSO4-(aq) + 3 H+(aq) + 2e- PbSO4(s)+ 2 H2O(l) E= +1.68 V An (Oxid): Pb(s) + HSO4-(aq) PbSO4(s) + H+(aq) + 2e- E = +0.36 V Net: Pb(s)+ PbO2(s) + 2 H+(aq) + 2 HSO4-(aq) 2 PbSO4(s)+ 2 H2O(l) Enet = +2.04 V d) Fuel Cells  cells where reactants are continuously supplied  the energy from this reaction can be used to run machines  one type is the hydrogen-oxygen fuel cell  hydrogen gas is pumped in at the anode while oxygen gas is pumped in at the cathode (which both have a lot of surface area)  pressure is used to push the H2 through a platinum catalyst which splits the H2 into 2H+ and 2e-   the 2e- move through an external circuit towards the cathode generating electrical energy  the O2 is also pushed through the platinum catalyst forming two oxygen atoms  the H+ ions and oxygen atoms combine to form water Cathode (Red): O2(g) + 4 H+(aq) + 4e- 2 H2O(l) E=+1.23 V Anode (Oxid): 2 H2(g) 4 H+(aq) + 4e- E = 0.00 V Net: O2(g) + 2 H2(g) 2 H2O(l) Enet = +1.23 V Hydrogen-oxygen Fuel Cell  need a source of hydrogen…reformers are used to convert CH4 or CH3OH into H2 and CO2  unfortunately, CO2 is a greenhouse gas  about 24-32% efficient where gas-powered car is about 20% efficient K. Electrolytic Cells 1. The Basics  in an electrolytic cell, electricalenergy is used to force a nonspontaneous chemical reaction to occur (opposite of a voltaic cell)  these reactions have a negative Enet  commonly used to electroplate metals (eg. gold, silver, bronze, chromium etc), recharge batteries, and split compounds into useful gases (eg. H2, O2, Cl2 etc)  the electrolytic cell is hooked up to a battery or power supply (instead of load or external circuit) so the flow of e- is “pushed” by an outside force  the cathode of the electrolytic cell is connected to the anode of the battery and therefore is negative  the anode of the electrolytic cell is connected to the cathode of the battery and therefore is positive  Voltaic Cells Electrolytic Cells  chemical to electrical energy  electrical to chemical energy  cathode + anode –  cathode – anode +  usually contains porous  does not (usually) contain a cup or salt bridge porous cup or salt bridge  Enet is positive (spont)  Enet is negative (nonspont)  has a voltmeter or external  has a power supply load  e– flow from anode to cathode  oxidation at anode  reduction at cathode  cations migrate to cathode  anions migrate to anode  some processes are used in industry to produce gases, for example: 1. the Hall-Heroult cell for producing Al …aluminum oxide is electrolyzed using carbon electrodes …liquid aluminum is collected http://images.google.ca/imgres?imgurl=http://wps.prenhall.com/wps/media/objects/602/616516/Media_Assets/Chapter18/Text_Im ages/FG18_18.JPG&imgrefurl=http://wps.prenhall.com/wps/media/objects/602/616516/Chapter_18.html&h=756&w=1600&sz=183& tbnid=4cFJrFlK4noQXM:&tbnh=70&tbnw=150&hl=en&start=11&prev=/images%3Fq%3Dhall- heroult%2Bcell%26svnum%3D10%26hl%3Den%26lr%3D 2. a chlor-alkali plant for producing chlorine gas …a saturated sodium chloride solution is electrolyzed …chlorine gas is formed and collected at the anodes http://www.cheresources.com/chloralk.shtml  Example 1 An electric current is passed through a solution of nickel (II) nitrate using inert electrodes. Predict the anode and cathode reactions, overall reaction, and minimum voltage required. Ni2+(aq) NO3-(aq) H2O(l) OA OA with H+(aq) OA/RA SOA SRA Cathode SOA(Red): 2 [Ni2+(aq) + 2e- Ni(s) ] E = -0.26 V Anode SRA(Ox): 2 H2O(l) O2(g) + 4 H+(aq) + 4e- E = -1.23 V Net: 2 Ni2+(aq) + 2 H2O(l) 2 Ni(s)+ O2(g) + 4 H+(aq) Enet = -1.49 V  Example 2 An electric current is passed through a solution of potassium iodide using inert electrodes. Predict the anode and cathode reactions, overall reaction, and minimum voltage required. K+(aq) I-(aq) H2O(l) OA RA OA/RA SRA SOA Cathode SOA(Red):2 H2O(l) +2 e- H2(g) + 2 OH- (aq)E = -0.83 V Anode SRA(Ox): 2 I-(aq) I2(s) + 2e- E = -0.54 V Net: 2 H2O(l) + 2 I-(aq) H2(g) + 2 OH-(aq) + I2(s) Enet = -1.37 V  Example 3 An electric current is passed through a solution of copper(II) sulphate using a carbon electrode and a metal electrode. Predict the anode and cathode reactions, overall reaction, and minimum voltage required. Cu2+(aq) SO42-(aq) H2O(l) OA OA/RA OA/RA SOA SRA Cathode SOA(Red): 2 [ Cu2+(aq) + 2 e- ] E = +0.34 V Cu(s) Anode SRA(Ox): 2 H O O + 4 H + + 4e- E = -1.23 V 2 (l) 2(g) (aq) Net: 2 H2O(l) + 2 Cu2+(aq) 2 Cu(s) + O2(g) + 4 H+(aq) Enet = -0.89 V *** Note: chlorine (ions) is an exception to our rules… when water and chlorine are competing as reducing agents, water is the stronger RA but chloride ions is chosen because the transfer of e- from H2O to O2 is more difficult …called overvoltage  Example 4 An electric current is passed through a solution of sodium chloride. Predict the anode and cathode reactions, overall reaction, and minimum voltage required. Na+(aq) Cl-(aq) H2O(l) OA RA OA/RA SRA SOA Cathode SOA(Red):2 H2O(l) +2 e- H2(g) + 2 OH- (aq)E = -0.83 V Anode SRA(Ox): 2 Cl-(aq) Cl2(g) + 2e- E = -1.36 V Net: 2 H2O(l) + 2 Cl-(aq) H2(g) + Cl2(g) + 2 OH- (aq) Enet = -2.19 V 2. Quantitative Study of Electrolysis  quantitative analysis (stoich) provides information on necessary quantities, current and/or time for electrolytic reactions  the unit for charge (q) is the Coulomb (C)  one e- carries 1.60 x 10-19 C of charge  this means that one mole of e- carry 9.65 x 104 C of charge  9.65 x 104 C/mol is called the Faraday constant (F) (see Data Booklet pg 3)  ne- = q q = It F ne- = number of moles of electrons (mol) where: q = charge in Coulombs (C) F = Faraday constant = 9.65 x 104 C/mol I = current in C/s or Amperes (A) t = time in seconds (s)  the above equations can be combined into one equation: ne- = It F  we can use these equations in stoichiometric calculations for current, time, mass, moles of a substance and moles e- Electrolysis is useful for: 1) Recharging batteries 2) Electroplating metals (gold, silver, bronze etc. ) 3) Production of useful gases (oxygen, chlorine, hydrogen etc) Example 1 An electrochemical cell caused a 0.0720 mol of e- to flow through a wire. Calculate the charge. ne- = 0.0720 mol ne- = q F = 9.65 x 104 C/mol F 0.0720 mol = q 9.65 x 104 C/mol q = 6948 C = 6.95  103 C Example 2 Determine the number of moles of electrons supplied by a dry cell supplying a current of 0.100 A to a radio for 50.0 minutes. I = 0.100 A (C/s) ne- = It t = 50.0 min  60 F s/min = (0.100 C/s)(3000 s) = 3000 s 9.65 x 104 C/mol F = 9.65 x 104 C/mol = 0.00311 mol Example 3 If a 20.0 A current flows through an electrolytic cell containing molten aluminum oxide for 1.00 hours, what mass of Al(l) will be deposited at the cathode? Al3+(l) + 3 e- Al(l) ne- = It n = 0.746…mol  1/3 F = 0.248…mol = (20.0 A)(1.00 h  3600 M = 26.98 g/mol s/h) m = nM 9.65 x 104 C/mol = = 0.746…mol (0.248…mol)(26.98g/mol) = 6.71 g L. Rust and Corrosion  corrosion can be viewed as the process of returning metals to their natural state (ore)  the metal is oxidized causing the loss of structural integrity  most metals develop a thin oxide coating which then protects their internal atoms against further oxidation  commonly, the oxide coating will scale off leaving new metal exposed an susceptible to corrosion  salt will speed up the oxidation by acting as a salt bridge  O2(g) Fe(OH)2(s) H2O rust droplet cathode Fe(s) anode Cathode SOA(Red): O2(g) + 2H2O(l) + 4e- 4 OH-(aq) Anode SRA(Ox): 2 [ Fe(s) Fe2+(aq) + 2e-] O2(g) + 2H2O(l) + 2Fe(s) 4 OH-(aq) + 2 Fe2+(aq) Net: O2(g) + 2H2O(l) + 2Fe(s) 2 Fe(OH)2(s) M. Prevention of Corrosion  applying a coating of paint to protect metal from oxidation  other metals (eg. Zn, Cr, Sn) can be plated onto metals that you don’t want to corrode (eg. steel (Fe))  this coating is of a metal that is a stronger reducing agent than the metal that is to be protected…the coating metal will react instead and is called the sacrificial anode Fe Zn coating  this method is also called cathodic protection  has been in use before the science of electrochemistry was developed  Sir Humphrey Davy first used cathodic protection on British naval ships in 1824!  can be used to protect any metal but steel (iron) is most commonly protected  we use steel (iron) for many structures such as buried fuel tanks, septic tanks, pipelines, hulls of ships, bridge supports etc  to protect these structures by cathodic protection, an active metal (eg. Mg, Zn, Al) is connected by a wire to the structure  because the attached metal is stronger reducing agent a than the iron in the steel, the more active metal supplies the e- for reduction and therefore the steel (iron) becomes the cathode and is protected Fe (tank) more active metal eg) Mg, Zn, Al  another protection method is alloying pure metals, which changes their reduction potential  stainless steel contains chromium and nickel, changing steels reduction potential to one characteristic of noble metals like gold (basically unreactive)  electroplating is the process of depositing the neutral metal on the cathode by reducing metal ions in solution  an object can be plated by making it the cathode in an electrolytic cell containing ions of the plating metal Summary of Metal Protection: 1) Paint 2) Plating with non-reactive meals such as Au, Ag or Cr 3) Galvanizing : coat iron with zinc which reactions with O2 and CO2 to produce zinc oxide and zinc carbonate (theses compounds resists corrosion). 4) Cathodic Protection: Make iron the cathode and protect it from water and air (use a SRA such as Mg for the anode). 5) Alloys: solid solutions of metals which when together makes a very weak RA (brass=copper/zinc; stainless steel (iron, carbon, Cr, etc), ; bronze=tin/copper). 6) Electroplating: a plating process that uses electrical current to reduce cations of a desired material from a solution to coat a conductive object with a thin layer of material