Presentation on theme: "Electrochemistry Applications of Redox. Review l Oxidation reduction reactions involve a transfer of electrons. l OIL- RIG l Oxidation Involves Gain l."— Presentation transcript:
Electrochemistry Applications of Redox
Review l Oxidation reduction reactions involve a transfer of electrons. l OIL- RIG l Oxidation Involves Gain l Reduction Involves Loss l LEO-GER l Lose Electrons Oxidation l Gain Electrons Reduction
Applications l Moving electrons is electric current. 8H + +MnO Fe +2 +5e - Mn Fe +3 +4H 2 O l Helps to break the reactions into half reactions. 8H + +MnO e - Mn +2 +4H 2 O 5(Fe +2 Fe +3 + e - ) l In the same mixture it happens without doing useful work, but if separate
H + MnO 4 - Fe +2 l Connected this way the reaction starts l 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- l Electricity travels in a complete circuit l Instead of a salt bridge
Cell Potential l Oxidizing agent pushes the electron. l Reducing agent pulls the electron. The push or pull (driving force) is called the cell potential E cell l Also called the electromotive force (emf) l Unit is the volt(V) l = 1 joule of work/coulomb of charge l 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 l This is the reference all other oxidations are compared to E º = 0 l º indicates standard states of 25ºC, 1 atm, 1 M solutions.
Cell Potential Zn(s) + Cu +2 (aq) Zn +2 (aq) + Cu(s) l The total cell potential is the sum of the potential at each electrode. E º cell = E º Zn Zn +2 + E º Cu +2 Cu l We can look up reduction potentials in a table. l One of the reactions must be reversed, so change it sign.
Cell Potential l 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) E º = 0.77 V Cu +2 (aq)+2e - Cu(s) E º = 0.34 V Cu(s) Cu +2 (aq)+2e - E º = V 2Fe +3 (aq) + 2e - 2Fe +2 (aq) E º = 0.77 V
Line Notation solid Aqueous Aqueous solid Anode on the left Cathode on the right l Single line different phases. l Double line porous disk or salt bridge. l If all the substances on one side are aqueous, a platinum electrode is indicated. l For the last reaction Cu(s) Cu +2 (aq) Fe +2 (aq),Fe +3 (aq) Pt(s)
Galvanic Cell l The reaction always runs spontaneously in the direction that produced a positive cell potential. l 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 l 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 +3 +3e - Fe(s) Eº=0.036V
Potential, Work and G l emf = potential (V) = work (J) / Charge(C) E = work done by system / charge E = -w/q l Charge is measured in coulombs. -w = qE l Faraday = 96,485 C/mol e - l q = nF = moles of e - x charge/mole e - w = -qE = -nFE = G
Potential, Work and G Gº = -nFE º if E º 0 spontaneous if E º > 0, then Gº < 0 nonspontaneous l In fact, reverse is spontaneous. Calculate Gº for the following reaction: Cu +2 (aq)+ Fe(s) Cu(s)+ Fe +2 (aq) Fe +2 (aq) + e - Fe(s) E º = 0.44 V Cu +2 (aq)+2e - Cu(s) E º = 0.34 V
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 E º cell if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.0 M l if [Al +3 ] = 1.0 M and [Mn +2 ] = 1.5M l if [Al +3 ] = 1.5 M and [Mn +2 ] = 1.5 M
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 l Always have to figure out n by balancing. l If concentration can gives voltage, then from voltage we can tell concentration.
The Nernst Equation l 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
Batteries are Galvanic Cells l 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) l NiCad NiO 2 + Cd + 2H 2 O Cd(OH) 2 +Ni(OH) 2
Corrosion l Rusting - spontaneous oxidation. l 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 + l Reaction happens in two places.
Water Rust Iron Dissolves- Fe Fe +2 e-e- Salt speeds up process by increasing conductivity
Preventing Corrosion l Coating to keep out air and water. l Galvanizing - Putting on a zinc coat l Has a lower reduction potential, so it is more. easily oxidized. l Alloying with metals that form oxide coats. l Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.
l Running a galvanic cell backwards. l Put a voltage bigger than the potential and reverse the direction of the redox reaction. l 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 l Have to count charge. Measure current I (in amperes) l 1 amp = 1 coulomb of charge per second q = I x t l q/nF = moles of metal l Mass of plated metal l How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag +
Other uses l Electroysis of water. l Seperating mixtures of ions. l More positive reduction potential means the reaction proceeds forward. l We want the reverse. l Most negative reduction potential is easiest to plate out of solution.