Electrochemistry Applications of Redox

Cell Potential Oxidizing agent pull the electron. Reducing agent push the electron. The push or pull (“driving force”) is called the cell potential Ecell Also called the electromotive force (emf) Unit is the volt(V) = 1 joule of work/coulomb of charge Measured with a voltmeter Measures the ability for a ½ cell to attract electrons ie: to be reduced. If ½ cell undergoes reduction Ecell > 0 If ½ cell undergoes oxidation Ecell < 0

Standard Hydrogen Electrode This is the reference all other oxidations are compared to Eº = 0 º indicates standard states of 25ºC, 1 atm, 1 M solutions. H2 in H+ Cl- 1 M HCl

0.76 H2 in Cathode Anode H+ Cl- Zn+2 SO4-2 1 M ZnSO4 1 M HCl

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 = EºZn® 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) 1. Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 V 2. Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V Since the iron electrode has a higher Eº value, the forward reaction is observed. Reaction 2 must be written in reverse and the sign changed. Cu(s) ® Cu+2(aq)+2e- Eº = -0.34 V Balance the charges for each ½ cell. 2Fe+3(aq) + 2e-® 2Fe+2(aq) Cu(s) ® Cu+2(aq)+2e- Add the ½ cells and the E º 2Fe+3(aq) + Cu(s) ® 2Fe+2(aq) + Cu+2(aq)+ Eº = -0.34 V + 0.77 = 0.43V

Line Notation solid½Aqueous½½Aqueous½solid Anode on the left½½Cathode 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)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)

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

Practice Completely describe the galvanic cell based on the following half-reactions under standard conditions. MnO4- + 8 H+ +5e- ® Mn+2 + 4H2O Eº=1.51 Fe+3 +3e- ® Fe(s) Eº=0.036V p. 708 #10-15 p.709 #1-3, 7, 8

Batteries are Galvanic Cells Car batteries are lead storage batteries. Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O Dry Cell Zn + NH4+ +MnO2 ® Zn+2 + NH3 + H2O Alkaline Zn +MnO2 ® ZnO+ Mn2O3 (in base) NiCad NiO2 + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2

Corrosion Rusting - spontaneous oxidation. Most structural metals have reduction potentials that are less positive than O2 . Fe ® Fe+2 +2e- Eº= 0.44 V O2 + 2H2O + 4e- ® 4OH- Eº= 0.40 V Fe+2 + O2 + H2O ® Fe2 O3 + H+ Reaction happens in two places.

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

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.

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

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

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

Other uses Electroysis of water. Separating 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.

Potential, Work and DG emf = potential (V) = work (J) / Charge(C) AP Potential, Work and DG 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 = DG

Potential, Work and DG DGº = -nFE º AP Potential, Work and DG DGº = -nFE º if E º < 0, then DGº > 0 spontaneous if E º > 0, then DGº < 0 nonspontaneous In fact, reverse is spontaneous. Calculate DGº for the following reaction: Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq) Fe+2(aq) + e-® Fe(s) Eº = 0.44 V Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V

Cell Potential and Concentration AP 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 Ecell will be greater or less than Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M if [Al+3] = 1.0 M and [Mn+2] = 1.5M if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

The Nernst Equation E = Eº - RTln(Q) nF DG = DGº +RTln(Q) AP The Nernst Equation DG = DGº +RTln(Q) -nFE = -nFEº + RTln(Q) E = Eº - RTln(Q) nF 2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Eº = 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 0 = Eº - RTln(K) nF Eº = RTln(K) nF AP The Nernst Equation As reactions proceed concentrations of products increase and reactants decrease. Reach equilibrium where Q = K and Ecell = 0 0 = Eº - RTln(K) nF Eº = RTln(K) nF nFEº = ln(K) RT

Calculating plating Have to count charge. AP 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+