Electrochemistry

Oxidation – Reduction Reactions Consider the reaction of Copper wire and AgNO3(aq) AgNO3(aq) Cu(s) Ag(s)

Oxidation – Reduction Reactions If you leave the reaction a long time the solution goes blue! The blue is due to Cu2+(aq)

Oxidation-Reduction Reactions So when we mix Ag+(aq) with Cu(s) we get Ag(s) and Cu2+(aq) Ag+(aq) + 1e-  Ag(s) Cu(s)  Cu2+(aq) + 2e- The electrons gained by Ag+ must come from the Cu Can’t have reduction without oxidation (redox) Each Cu can reduce 2 Ag+ 2Ag+(aq) + 2e-  2Ag(s) 2Ag+(aq) + 2e- + Cu(s) 2Ag(s)+ Cu2+(aq) + 2e- gain electrons = reduction lose electrons = oxidation

Redox Cu/Ag E Cu electron flow Ag+ Ag+

Redox Cu/Ag E Cu2+ ΔE = e.Ecell Ag Ag e = charge on an electron Ecell = Voltage in a electrochemical cell If we could separate the two reactions we could use the energy gained by the e to do work Redox Cu/Ag E Cu2+ ΔE = e.Ecell Ag Ag

The maximum amount of energy available to do work is a definition of the free energy ΔG Redox Cu/Ag cell voltage Free energy for redox rxn E Cu2+ # electrons in redox rxn F = Faraday’s constant = 96485 C/mol Ag Ag

ΔEcell: the cell potential In a redox reaction electrons are transferred to a more stable state Most of the free energy of the reaction is due to these electrons Can we access this energy? Yes, by conducting the two half reactions in separate cells 2Ag+(aq) + 2e-  2Ag(s) Cu(s)  Cu2+(aq) + 2e- This is called the Voltaic electrochemical cell The cell potential can be measured in such a cell with a voltmeter

Where oxidation happens Voltaic Cell Cu/Ag Ecell= 0.460 V DGcell= -2F x 0.460 V cations to the cathode electrons to the cathode Anions to the anode Where oxidation happens Where reduction happens Cu(s)  Cu2+(aq) + 2e- 2Ag+(aq) + 2e-  2Ag(s)

Electrochemistry and Cells electrochemistry is the study of redox reactions that produce or require an electric current the conversion between chemical energy and electrical energy is carried out in an electrochemical cell spontaneous redox reactions take place in a voltaic cell (galvanic cell) nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

Where oxidation happens Voltaic Cell as a Battery: Electric Current Flowing Indirectly Between Atoms Electrons do work here electrons to the cathode Anions to the anode cations to the cathode Where oxidation happens Where reduction happens

Electrodes Anode (donates electrons to the cathode) electrode where oxidation occurs In a Galvanic cell it is the –ve terminal and in an electrolytic cell it is the +ve terminal anions attracted to it connected to positive end of battery in electrolytic cell loses weight in electrolytic cell Cathode (attracts electrons from the anode) electrode where reduction occurs In a Galvanic cell it is the +ve terminal and in an electrolytic cell it is the -ve terminal cations attracted to it connected to negative end of battery in electrolytic cell gains weight in electrolytic cell electrode where plating takes place in electroplating

Current and Voltage the number of electrons that flow through the system per second is the current unit = Ampere (A) 1 A of current = 1 Coulomb (C) of charge flowing by each second 1 A = 6.242 x 1018 electrons/second Electrode surface area dictates the number of electrons that can flow the difference in potential energy between the reactants and products per Coulomb of charge is called the Voltage unit = Volt (V) 1 V of force = 1 J of energy/Coulomb of charge the voltage is what drives electrons through the external circuit amount of force pushing the electrons through the wire is called the electromotive force, emf, (also units of V) the cell potential under standard conditions is called the standard emf, E°cell 25°C, 1 atm for gases, 1 M concentration of solution sum of the cell potentials for the half-reactions

Cell Notation shorthand description of Voltaic cell anode | electrolyte || electrolyte | cathode oxidation half-cell on left (anode), reduction half-cell on the right (cathode) single | = phase barrier if multiple electrolytes in same phase, a comma is used rather than | often use an inert electrode double line || = salt bridge

Write out the cell notation for this cell anode | electrolyte || electrolyte | cathode Fe(s) | Fe2+(aq) || MnO4-(aq), Mn2+(aq), H+(aq) | Pt(s)

Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) Practice - Sketch and Label the Voltaic Cell Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) Write the Half-Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions. Fe(s) | Fe2+(aq) || Pb2+(aq) | Pb(s) anode cathode Pb2+(aq) + 2 e−  Pb(s) Ered = −0.13 V Fe2+(aq) + 2 e−  Fe(s) Ered = -0.45 V The anode is where oxidation happens Reverse the reaction for Fe2+(aq) and flip the sign to get Eox Ecell = Eox + Ered = 0.45V + -0.13V = 0.32V Fe(s)  Fe2+(aq) + 2 e− Eox = +0.45 V Pb2+(aq) + 2 e−  Pb(s) Ered = −0.13 V

ox: Fe(s)  Fe2+(aq) + 2 e− Eox = +0.45 V red: Pb2+(aq) + 2 e−  Pb(s) Ered = −0.13 V tot: Pb2+(aq) + Fe(s)  Fe2+(aq) + Pb(s) Ecell = +0.32 V

Measuring the Tendency to Reduce: Standard Reduction Potential when two half-cells are connected, one side will oxidize and one side will reduce, but which will do what? clearly the electrons will flow so that the half-reaction with the stronger tendency to reduce will reduce, making the other half-cell reaction oxidize we cannot measure the absolute tendency of a half-reaction to reduce, we can only measure it relative to another half-reaction To measure the tendency to reduce we measure the E (the voltage) in a voltaic cell where one of the electrodes is a standard hydrogen electrode (see right) We assign a potential difference = 0.0 V to the following reaction 2H+(aq) +2e-  H2(g) Eored = 0.0V E°cell = Eoox + Eored = Eoox + 0.0 V = Eoox In so doing scientists can obtain standard reduction potentials as follows

The measured voltage can then be placed in a table like this one, where we write the voltage for the reduction reaction

Half-Cell Potentials ΔE°cell = E°oxidation + E°reduction SHE reduction potential is defined to be exactly 0 V half-reactions with a stronger tendency toward reduction than the SHE have a + value for E°red half-reactions with a stronger tendency toward oxidation than the SHE have a - value for E°red ΔE°cell = E°oxidation + E°reduction E°oxidation = -E°reduction when adding E° values for the half-cells, do not multiply the half-cell E° values (voltage is the energy per unit charge), even if you need to multiply the half-reactions to balance the equation ΔGocell=-nFΔE°cell Since ΔGocell < 0 to be spontaneous ΔE°cell > 0 for a redox reaction to be spontaneous if ΔE°cell < 0 the reaction will not be spontaneous

red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l) Calculate E°cell for the reaction at 25°C Al(s) + NO3−(aq) + 4 H+(aq)  Al3+(aq) + NO(g) + 2 H2O(l) Separate the reaction into the oxidation and reduction half-reactions ox: Al(s)  Al3+(aq) + 3 e− red: NO3−(aq) + 4 H+(aq) + 3 e−  NO(g) + 2 H2O(l) find the Eo for each half-reaction from the standard reduction table. If you need the oxidation potential of one Eoox=-Eored and sum to get Eocell Eoox = −Eored = +1.66 V Eored = +0.96 V Eocell = (+1.66 V) + (+0.96 V) = +2.62 V The ½ rxn with the largest reduction potential is the reduction reaction in the cell The ½ rxn with the smallest reduction potential is reversed and sign flipped reverse

red: Mg2+(aq) + 2 e−  Mg(s) Predict if the following reaction is spontaneous under standard conditions Fe(s) + Mg2+(aq)  Fe2+(aq) + Mg(s) Separate the reaction into the oxidation and reduction half-reactions ox: Fe(s)  Fe2+(aq) + 2 e− red: Mg2+(aq) + 2 e−  Mg(s) look up the relative positions of the reduction half-reactions red: Mg2+(aq) + 2 e−  Mg(s) Ered = -2.37V ox: Fe(S)  Fe2+(aq) + 2 e− Eox =+0.45V Ecell = Eox + Ered = 0.45V – 2.37V = -1.92V Since Ecell < 0 the cell reaction is not spontaneous as written reverse

Keeping track of electrons Redox reactions involve the transfer of electrons from the donor to the acceptor but sometimes it is hard to tell eg (2H2O22H2O + O2) The donor loses electrons and is oxidized The acceptor acquires electrons and is reduced To know if a reaction is a redox reaction we need a way to keep track of how many valence electrons each element has We define the oxidation state of an element to be +n (n integer) if it has n less electrons than it does as the free atom, and –p if it has p more electrons than it does in the free atom

Assigning Oxidation Numbers The sum of the oxidation numbers Q of all the atoms in a compound/ion up to the charge on the compound/ion. This means …. free elements have an oxidation state = 0 QNa = 0 and QCl = 0 in 2 Na(s) + Cl2(g) monatomic ions have an oxidation state equal to their charge QNa = +1 and QCl = -1 in NaCl The sum of the oxidation numbers of all the atoms in a neutral compound is 0 the sum of the oxidation numbers of all the atoms in a polyatomic ion equals the charge on the ion (a) Group I metals have an oxidation state of +1 in all their compounds QNa = +1 in NaCl (b) Group II metals have an oxidation state of +2 in all their compounds QMg = +2 in MgCl2 F has an oxidation number QF = -1 H has an oxidation number QH = +1 O has an oxidation number QO = -2

Determine the oxidation states of the following S in SO42-(aq) Na in NaF(aq) Fe in FeCl3(aq) S in S2O32-(aq) Cr in CrO42-(aq)

Oxidation and Reduction oxidation occurs when an atom’s oxidation state increases during a reaction reduction occurs when an atom’s oxidation state decreases during a reaction rule 4 rule 1 rule 5 rule 4 CH4 + 2 O2 → CO2 + 2 H2O -4 +1 0 +4 –2 +1 -2 oxidation reduction

Oxidation–Reduction 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) oxidation and reduction must occur simultaneously if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl2(g) → 2 Na+Cl–(s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl2 is the oxidizing agent

Identify the Oxidizing and Reducing Agents in Each of the Following 3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4 H2O MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O

Identify the Oxidizing and Reducing Agents in Each of the Following red ag ox ag 3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4 H2O MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O +1 -2 +5 -2 +1 0 +2 -2 +1 -2 reduction oxidation ox ag red ag +4 -2 +1 -1 +2 -1 0 +1 -2 oxidation reduction

Common Oxidizing Agents

Common Reducing Agents

Balancing Redox Reactions: Aqueous Solution assign oxidation numbers determine element oxidized and element reduced write ox. & red. half-reactions, including electrons ox. electrons on right, red. electrons on left of arrow balance half-reactions by mass first balance elements other than H and O add H2O where need O add H+1 where need H neutralize H+ with OH- in base balance half-reactions by charge balance charge by adjusting electrons balance electrons between half-reactions add half-reactions check

Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution Assign Oxidation States Separate into half-reactions ox: I-(aq)  I2(aq) red: MnO4-(aq)  MnO2(s) Assign Oxidation States I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) Separate into half-reactions ox: red:

Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution Balance half-reactions by mass in base, neutralize the H+ with OH- ox: 2 I-(aq)  I2(aq) red: 4 H+(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) 4 H+(aq) + 4 OH-(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) + 4 OH-(aq) 4 H2O(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) + 4 OH-(aq) MnO4-(aq) + 2 H2O(l)  MnO2(s) + 4 OH-(aq) ` Balance half-reactions by mass then H by adding H+ ox: 2 I-(aq)  I2(aq) red: 4 H+(aq) + MnO4-(aq)  MnO2(s) + 2 H2O(l) Balance half-reactions by mass ox: 2 I-(aq)  I2(aq) red: MnO4-(aq)  MnO2(s) Balance half-reactions by mass ox: I-(aq)  I2(aq) red: MnO4-(aq)  MnO2(s) Balance half-reactions by mass then O by adding H2O ox: 2 I-(aq)  I2(aq) red: MnO4-(aq)  MnO2(s) + 2 H2O(l)

Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution Balance Half-reactions by charge ox: 2 I-(aq)  I2(aq) + 2 e- red: MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq) Balance electrons between half-reactions ox: { 2 I-(aq)  I2(aq) + 2 e- } x 3 red: {MnO4-(aq) + 2 H2O(l) + 3 e-  MnO2(s) + 4 OH-(aq) } x 2 ox: 6 I-(aq)  3 I2(aq) + 6 e- red: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq)

Balance the equation: I-(aq) + MnO4-(aq)  I2(aq) + MnO2(s) in basic solution Add the Half-reactions ox: 6 I-(aq)  3 I2(aq) + 6 e- red: 2 MnO4-(aq) + 4 H2O(l) + 6 e-  2 MnO2(s) + 8 OH-(aq) tot: 6 I-(aq)+ 2 MnO4-(aq) + 4 H2O(l)  3 I2(aq)+ 2 MnO2(s) + 8 OH-(aq) Check Reactant Count Element Product 6 I 2 Mn 12 O 8 H 8- charge

Your turn Your Turn

MnO4-(aq) + HSO3-(aq)  Mn2+(aq) + SO42-(aq) Balance the following reaction in acid   MnO4-(aq) + HSO3-(aq)  Mn2+(aq) + SO42-(aq) In the above reaction what is the oxidation number of Mn in MnO4-(aq)? What is the oxidation number of Mn in Mn2+(aq) What is the oxidation number of S in HSO3-(aq)? What is the oxidation number of S in SO42-(aq)? In the above reaction is MnO4-(aq) the oxidizing or reducing agent? 20. When permanganate is reacted with hydrogen sulfite in base instead of acid we get MnO4-(aq) + HSO3-(aq)  MnO2(s)+ SO42-(aq) Balance the above equation  

The Relationship between ΔGo and Eocell Remember ΔGo is the maximum work the system can do Eocell is the cell potential energy (standard emf) per unit of charge Since the potential energy is the maximum amount of work that can be done on the surrounding we can write The total charge q = nF, n = number moles of electrons in the balanced equation and F is Faraday’s constant. Then we can write Δ 𝐺 𝑐𝑒𝑙𝑙 =Δ 𝐺 𝑐𝑒𝑙𝑙 𝑜 +𝑅𝑇𝑙𝑛𝑄 −𝑛𝐹 𝐸 𝑐𝑒𝑙𝑙 =−𝑛𝐹 𝐸 𝑐𝑒𝑙𝑙 𝑜 +𝑅𝑇𝑙𝑛𝑄 𝐸 𝑐𝑒𝑙𝑙 = 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 𝑅𝑇 𝑛𝐹 𝑙𝑛𝑄= 𝐸 𝑐𝑒𝑙𝑙 𝑜 − 0.0592𝑉 𝑛 𝑙𝑜𝑔𝑄 𝐸 𝑐𝑒𝑙𝑙 𝑜 = 0.0592𝑉 𝑛 𝑙𝑜𝑔𝐾 Divide b.s. by -nF At equilibrium E = 0

E° at Nonstandard Conditions

Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq) You are given a Zn/Cu Voltaic cell. Zn(s)|1M ZnSO4(aq)||IM CuSO4|Cu(s)   The net reaction in this cell is Cu2+(aq) + Zn(s)  Cu(s) + Zn2+(aq) 1. Which metal is the anode? a. Cu b. Zn 2. Which metal is the cathode? 3. The measured DEocell = 1.10V at 298K 4. What is DGo for this reaction? 5. Is the reaction spontaneous? 6. What is K for this reaction when equilibrium is reached? 7. Does equilibrium favor reactants or products?  8. What is the DEcell for the following cell Zn(s)|2M ZnSO4(aq)||0.1M CuSO4|Cu(s) ?  9. What is the free energy DG for this reaction at 298K?

Voltaic Cells as Batteries Use chemical reactions to provide a source of energized electrons that can power electronics Batteries in everything, phones, laptops, tablets, watches, cars Actively seeking batteries that are light, powerful, and retain their effectiveness after many recharge cycles Molten metal batteries developed for storing electricity in power stations (infinitely re-chargable) Huge area of scientific and commercial interest Batteries are being made ever smaller for energy efficient processor applications Recent breakthroughs using gold nanowires that are gel coated make batteries that will last longer than the devices they are in Nanotechnology, even biotechnology using viruses! A benign bacteria virus serves as a template in water to grow manganese oxide nanowires that, when combined with palladium, increases the energy potential of a lithium battery

Voltaic Cells as Batteries Lead Storage Battery Anode: Pb Pb  Pb2+ + 2e- Pb(s) + SO42-(aq)  PbSO4(s) + 2 e- Cathode: Pb coated with PbO2 Pb4+ +2e- Pb2+ PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e- PbSO4(s) + 2 H2O(l) Electrolyte: 30% H2SO4 cell voltage = 2.09 V 6 in series makes 12V rechargeable, heavy

Voltaic Cells as Batteries Lithium Ion Battery anode = graphite impregnated with Li ions cathode = Li - transition metal oxide Electrolyte is an organic solvent like dimethylcarbonate, with a Li salt like LiPF6 rechargeable, long life, very light, more environmentally friendly, greater energy density High surface area electrodes Tiny ions that move fast between electrodes

Voltaic Cells as Batteries Why do we use Li-ion batteries in our computers and phones and not say Pb Acid Batteries? We want to make batteries as portable (ie small and light) as possible

H2 Fuel Cell like batteries in which reactants are constantly being added so it never runs down! Anode and Cathode both Pt coated metal (may change with cheaper technology) Electrolyte is OH– solution Anode Reaction: 2 H2 + 4 OH– → 4 H2O(l) + 4 e- Cathode Reaction: O2 + 4 H2O + 4 e- → 4 OH– Toyota Mirai available in CA $499/month 312 miles per tank

Electrolysis electrolysis is the process of using electricity to make non-spontaneous redox reactions happen electrolysis is done in an electrolytic cell electrolytic cells can be used to separate elements from their compounds generate H2 from water for fuel cells recover metals from their ores Make halogen gasses from halide salts

Electrolytic Cell uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction must be DC source the + terminal of the battery connects to the anode the - terminal of the battery connects to the cathode cations attracted to the cathode, anions to the anode Oxidation at the anode, reduction at the cathode some electrolysis reactions require more voltage than Etot, called the overvoltage

Applications of electrolysis: electroplating In electroplating, the work piece is the cathode. Cations are reduced at cathode and plate to the surface of the work piece. The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution Gold plated terminals in electronics (Au is the most conductive of all metals)

Electrolysis of Pure Compounds must be in molten (liquid) state electrodes normally graphite cations are reduced at the cathode to metal element anions oxidized at anode to nonmetal element

Electrolysis of NaCl(l)

Mixtures of Ions when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode least negative or most positive E°red when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode least negative or most positive E°ox

Electrolysis of Aqueous Solutions Complicated by more than one possible oxidation and reduction possible cathode reactions reduction of cation to metal reduction of water to H2 2 H2O + 2 e-1  H2 + 2 OH-1 E° = -0.83 V @ stand. cond. E° = -0.41 V @ pH 7 possible anode reactions oxidation of anion to element oxidation of H2O to O2 2 H2O  O2 + 4e-1 + 4H+1 E° = -1.23 V @ stand. cond. E° = -0.82 V @ pH 7 oxidation of electrode particularly Cu graphite doesn’t oxidize half-reactions that lead to least negative Etot will occur unless overvoltage changes the conditions

Electrolysis of NaI(aq): Inert Electrodes Write down ions/molecules/metals present and don’t forget water Here Na+, I-, H2O are what is present (the electrodes here are inert) Write down the reduction reactions involving these species I2(aq)+2e-  2I-(aq) Ered = 0.54V O2(g) + 4e-1 + 4H+1(aq) 2 H2O(l) Ered = 0.82V 2 H2O + 2 e-1  H2 + 2 OH-1 Ered = 0.41V Na+(aq) + e-  Na(s) Ered = -2.71V Rearrange the equations so the reactants are on the left (reverse the sign of E if you flip the equation – these will be the oxidation reactions possible oxidations (anode –ve terminal) 2 I-1  I2 + 2 e-1 E° = −0.54 v 2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v possible oxidations 2 I-1  I2 + 2 e-1 E° = −0.54 v 2 H2O  O2 + 4e-1 + 4H+1 E° = −0.82 v possible reductions (cathode +ve terminal) Na+1 + 1e-1  Na0 E° = −2.71 v 2 H2O + 2 e-1  H2(g) + 2 OH-1E° = -0.41V possible reductions Na+1 + 1e-1  Na0 E° = −2.71 v 2 H2O + 2 e-1  H2 + 2 OH-1 E° = −0.41 v brown liquid overall reaction 2 I−(aq) + 2 H2O(l)  I2(aq) + H2(g) + 2 OH-1(aq) bubbles Ecell = -0.54V – 0.41V = -0.95V The battery needs to at least be 1V to make this happen

Electrolysis in aqueous solutions: Overpotentials There are (potentially) competing processes in the electrolysis of an aqueous solution consider now NaCl(aq) with C electrodes Cathode Anode We choose those reactions that are thermodynamically favored But…chlorine is evolved at the anode!

Overpotentials Thermodynamics true at equilibrium but when the process has a current flowing kinetics plays a role Overpotential represents the additional voltage that must be applied to drive the process In the NaCl(aq) solution the overpotential for evolution of oxygen is greater than that for chlorine, and so chlorine is evolved preferentially The limiting process in electrolysis is usually diffusion of the ions in the electrolyte Overpotential will depend on the electrolyte and electrode Driving the cell at the least current will give rise to the smallest overpotential

Faraday’s Law the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current (A), and the length of time (s) the cell runs charge = current x time Q = I x t

Calculate the mass of Au that can be plated in 25 min using 5 Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au3+(aq) + 3 e− → Au(s) 3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g Given: Find: t(s), amp charge (C) mol e− mol Au g Au Concept Plan: Relationships: Solve: Check: units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−

Corrosion corrosion is the spontaneous oxidation of a metal by chemicals in the environment since many materials we use are active metals (metals that like to make ions), corrosion can be a very big problem

Rusting rust is hydrated iron(III) oxide moisture must be present water is a reactant required for flow between cathode and anode electrolytes (sea-water) promote rusting enhances current flow acids (acid rain) promote rusting lower pH = lower E°red

O2(g) + 2H2O(l) + 4e-  4OH-(aq) Rusting Anode (inside the droplet): Fe(s)  Fe2+(aq) + 2e- Cathode (at the surface of the droplet: O2(g) + 2H2O(l) + 4e-  4OH-(aq) Within the droplet Fe2+(aq) + 2OH-(aq)  Fe(OH)2(s) Rust is then quickly produced by the oxidation of the precipitate at the edge of the droplet. 4Fe(OH)2(s) + O2(g)  2Fe2O3 •H2O(s) + 2H2O(l)

Preventing Corrosion Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V one way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment paint some metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding another method to protect one metal is to attach it to a more reactive metal that is cheap sacrificial electrode galvanized steel (Zinc coating) Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V

Sacrificial Anode Zn  Zn2+ + 2e- 0.76V Fe  Fe2+ + 2e- 0.45V Galvanic anode on the hull of a ship