Electrochemistry Applications of Redox

Review  Oxidation reduction reactions involve a transfer of electrons.  OIL- RIG  Oxidation Involves Loss  Reduction Involves Gain  LEO-GER  Lose Electrons Oxidation  Gain Electrons Reduction

Applications  Moving electrons is electric current.  8H + +MnO Fe +2 +5e -  Mn Fe +3 +4H 2 O  Helps to break the reactions into half reactions.  8H + +MnO e -  Mn +2 +4H 2 O  5(Fe +2  Fe +3 + e - )  In the same mixture it happens without doing useful work, but if separate

H + MnO 4 - Fe +2  Connected this way the reaction starts  Stops immediately because charge builds up.

H + MnO 4 - Fe +2 Galvanic Cell Salt Bridge allows current to flow

H + MnO 4 - Fe +2 e-e-  Electricity travels in a complete circuit  Instead of a salt bridge

H + MnO 4 - Fe +2 Porous Disk

Reducing Agent Oxidizing Agent e-e- e-e- e-e- e-e- e-e- e-e- AnodeCathode

Cell Potential  Oxidizing agent pushes the electron.  Reducing agent pulls the electron.  The push or pull (“driving force”) is called the cell potential E cell  Also called the electromotive force (emf)  Unit is the volt(V)  = 1 joule of work/coulomb of charge  Measured with a voltmeter

Zn +2 SO M HCl Anode M ZnSO 4 H + Cl - H 2 in Cathode

1 M HCl H + Cl - H 2 in Standard Hydrogen Electrode  This is the reference all other oxidations are compared to  E º = 0  º indicates standard states of 25ºC, 1 atm, 1 M solutions.

Cell Potential  Zn(s) + Cu +2 (aq)  Zn +2 (aq) + Cu(s)  The total cell potential is the sum of the potential at each electrode.  Eº cell = Eº Zn  Zn +2 + Eº Cu +2  Cu  We can look up reduction potentials in a table.  One of the reactions must be reversed, so change it sign.

Cell Potential  Determine the cell potential for a galvanic cell based on the redox reaction.  Cu(s) + Fe +3 (aq)  Cu +2 (aq) + Fe +2 (aq)  Fe +3 (aq) + e -  Fe +2 (aq) Eº = 0.77 V  Cu +2 (aq)+2e -  Cu(s) Eº = 0.34 V  Cu(s)  Cu +2 (aq)+2e - Eº = V  2Fe +3 (aq) + 2e -  2Fe +2 (aq) Eº = 0.77 V  0.77 V V = 0.43V

Homework  and 17.28

Line Notation  solid  Aqueous  Aqueous  solid  Anode on the left  Cathode on the right  Single line different phases.  Double line porous disk or salt bridge.  If all the substances on one side are aqueous, a platinum electrode is indicated.  For the last reaction  Cu(s)  Cu +2 (aq)  Fe +2 (aq),Fe +3 (aq)  Pt(s)

Galvanic Cell  The reaction always runs spontaneously in the direction that produced a positive cell potential.  Four things for a complete description. 1)Cell Potential 2)Direction of flow 3)Designation of anode and cathode 4)Nature of all the components- electrodes and ions

Practice  Completely describe the galvanic cell based on the following half-reactions under standard conditions.  MnO H + +5e -  Mn H 2 O Eº=1.51  Fe +2 +2e -  Fe(s) Eº=0.44V  Reverse the smaller valued reaction  Fe(s)  Fe +2 +2e -  Eº= V

Solution  Balance the cell reaction:  Balance the electrons to balance the equation: The First reaction requires 5 e-, the second one supplies 2 e-, so 10 must be exchanged.  2(MnO H + +5e -  Mn H 2 O)  5(Fe(s)  Fe +2 +2e - )  Add the reactions together

Solution Continued  2MnO 4 - (aq) + 16 H + (aq)+5Fe (s)  5 Fe +2 (aq) + 2 Mn +2 (aq) + 8H 2 O (l)  Determine the Cell Potential: 2(1.51) + 5(-0.44) = 3.02 – 2.2 = 0.8  Fe(s) l Fe +2 (aq) ll MnO 4 - (aq), Mn +2 (aq) l Pt(s)

More Homework  and 17.30

Potential, Work and  G  emf = potential (V) = work (J) / Charge(C)  E = work done by system / charge  E = -w/q  Charge is measured in coulombs.  -w = qE  Faraday = 96,485 C/mol e -  q = nF = moles of e - x charge/mole e -  w = -qE = -nFE =  G

Potential, Work and  G   Gº = -nFE º  if E º 0 nonspontaneous  if E º > 0, then  Gº < 0 spontaneous  In fact, reverse is spontaneous.  Calculate  Gº for the following reaction:  Cu +2 (aq)+ Fe(s)  Cu(s)+ Fe +2 (aq)  Fe +2 (aq) + 2e -  Fe(s) E º = V  Cu +2 (aq)+2e -  Cu(s) E º = 0.34 V

Practice  Cu +2 (aq)+ Fe(s)  Cu(s)+ Fe +2 (aq)  Fe(s)  Fe +2 (aq) + 2e - E º = 0.44 V  Cu +2 (aq)+2e -  Cu(s) E º = 0.34 V  0.44 V V = 0.78V   Gº = -nFE º   Gº = -(2 mol e-)96,485 C/mol e-(0.78J/C)   Gº = -1.5 x10 5 J

Practice Predicting Spontaneity  Will 1M HNO 3 dissolve gold metal to form a 1M Au 3+ solution?  HNO 3 half reaction:  NO H + + 3e -  NO + 2H 2 O E=0.96V  Au 3+ Half Reaction:  Au  Au e - - E = -1.50V  Au(s) + NO 3 - (aq) + 4H + (aq)  Au 3+ (s) + NO(g) + 2H 2 O(l)  E cell = 0.96V – 1.50V = -0.54V  Since E cell is negative, it will NOT occur under standard conditions.

Homework  17.37, 17.39, 17.49

Cell Potential and Concentration  Qualitatively - Can predict direction of change in E from LeChâtelier.  2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s)  Predict if E cell will be greater or less than Eº cell if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.0 MLESS  if [Al +3 ] = 1.0 M and [Mn +2 ] = 1.5M GREATER  if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.5 M SLIGHTLY GREATER

Homework  17.55

The Nernst Equation   G =  Gº +RTln(Q)  -nFE = -nFEº + RTln(Q)  E = Eº - RT ln(Q) nF  2Al(s) + 3Mn +2 (aq)  2Al +3 (aq) + 3Mn(s) E º = 0.48 V  Always have to figure out n by balancing.  If concentration can gives voltage, then from voltage we can tell concentration.

The Nernst Equation  As reactions proceed concentrations of products increase and reactants decrease.  Reach equilibrium where Q = K and E cell = 0  0 = Eº - RT ln(K) nF  Eº = RT ln(K) nF  nFEº = ln(K) RT

Example Using Nernst  Describe the cell based on the following:  VO H + + e -  VO H 2 O E o = 1.00V  Zn e -  Zn E o = -0.76V  T = 25 C  [VO 2 + ] = 2.0 M[H + ] = 0.50 M  [VO 2 2+ ] = 0.01 M[Zn 2+ ] = 0.1 M

Starting Solution  In Order to balance the electrons we must double the first equation, and reverse the second equation, so we end up with:  2VO H + + 2e -  2VO H 2 O  Zn  Zn e -  2VO 2 + (aq) + 4H + (aq) + Zn (s)  2VO 2 2+ (aq) + 2H 2 O (l) + Zn 2+ (aq)  E cell = 1.76 V  Note: Because we are using the Nernst Equation we use base values for E

Final Solution  At 25 C R =  E = E cell –R/n log (Q)  E = 1.76 –(0.0521/2) log ([ Zn 2 +][ VO 2 2+ ] 2 ) [ VO 2 + ] 2 [ H + ] 4  E = 1.76 –(0.0521/2) log ([0.10][0.010] 2 ) [2.0] 2 [0.50] 4  E = 1.76 –(0.0521/2) log 4.0 x  E = = 1.89V

Homework  and 17.63

Batteries are Galvanic Cells  Car batteries are lead storage batteries.  Pb +PbO 2 +H 2 SO 4  PbSO 4 (s) +H 2 O  Dry Cell Zn + NH 4 + +MnO 2  Zn +2 + NH 3 + H 2 O  Alkaline Zn +MnO 2  ZnO+ Mn 2 O 3 (in base)  NiCad  NiO 2 + Cd + 2H 2 O  Cd(OH) 2 +Ni(OH) 2

Corrosion  Rusting - spontaneous oxidation.  Most structural metals have reduction potentials that are less positive than O 2.  Fe  Fe +2 +2e - Eº= 0.44 V  O 2 + 2H 2 O + 4e -  4OH - Eº= 0.40 V  Fe +2 + O 2 + H 2 O  Fe 2 O 3 + H +  Reaction happens in two places.

Water Rust Iron Dissolves- Fe  Fe +2 e-e- Salt speeds up process by increasing conductivity

Preventing Corrosion  Coating to keep out air and water.  Galvanizing - Putting on a zinc coat  Has a lower reduction potential, so it is more. easily oxidized.  Alloying with metals that form oxide coats.  Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

 Running a galvanic cell backwards.  Put a voltage bigger than the potential and reverse the direction of the redox reaction.  Used for electroplating. Electrolysis

1.0 M Zn +2 e-e- e-e- Anode Cathode 1.10 Zn Cu 1.0 M Cu +2

1.0 M Zn +2 e-e- e-e- Anode Cathode A battery >1.10V Zn Cu 1.0 M Cu +2

Calculating plating  Have to count charge.  Measure current I (in amperes)  1 amp = 1 coulomb of charge per second  q = I x t  q/nF = moles of metal  Mass of plated metal  How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag +

Other uses  Electroysis of water.  Seperating mixtures of ions.  More positive reduction potential means the reaction proceeds forward.  We want the reverse.  Most negative reduction potential is easiest to plate out of solution.