Chapter 18 Electrochemistry Chemistry II

Redox Reaction one or more elements change oxidation number all single displacement, and combustion, some synthesis and decomposition

Redox Reaction always have both oxidation and reduction split reaction into oxidation half-reaction and a reduction half-reaction aka e - transfer reactions half-reactions include e - oxidizing agent is reactant molecule that causes oxidation contains element reduced reducing agent is reactant molecule that causes reduction contains the element oxidized

Oxidation & Reduction oxidation: ox number of an element increases element loses e - compound adds O compound loses H half-reaction has e - as products reduction: ox number of an element decreases element gains e - compound loses O compound gains H half-reactions have e - as reactants

Rules for Assigning Oxidation States rules are in order of priority 1. free elements have an oxidation state = 0 Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 2. monatomic ions have an oxidation state equal to their charge Na = +1 and Cl = -1 in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0

Rules for Assigning Oxidation States 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO 3 –, (+5) + 3(-2) = (a) Group I metals have an oxidation state of +1 in all their compounds Na = +1 in NaCl (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl 2

Rules for Assigning Oxidation States 5. in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority NonmetalOxidation StateExample FCF 4 H+1CH 4 O-2CO 2 Group 7ACCl 4 Group 6A-2CS 2 Group 5A-3NH 3

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 CH O 2 → CO H 2 O – oxidation reduction

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

Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H + → 3 S + 2 NO + 4 H 2 O MnO HBr → MnBr 2 + Br H 2 O

Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO HBr  MnBr 2 + Br H 2 O ox agred ag oxidation reduction oxidation reduction red agox ag

Common Oxidizing Agents

Common Reducing Agents

Balancing Redox Reactions 1. assign oxidation numbers a)determine element oxidized and element reduced 2. write ox. & red. half-reactions, including e - a)ox. electrons on right, red. electrons on left of arrow 3. balance half-reactions by mass a)first balance elements other than H and O b)add H 2 O where need O c)add H + where need H d)neutralize H + with OH - in base 4. balance half-reactions by charge a)balance charge by adjusting e - 5. balance e - between half-reactions 6. add half-reactions 7. check

Ex 18.3 – Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Assign Oxidation States I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) Separate into half-reactions ox: red: Assign Oxidation States Separate into half-reactions ox: I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s)

Ex 18.3 – Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Balance half- reactions by mass ox: I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) Balance half- reactions by mass ox: 2 I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) Balance half- reactions by mass then O by adding H 2 O ox: 2 I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) Balance half- reactions by mass then H by adding H + ox: 2 I  (aq)  I 2(aq) red: 4 H + (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) Balance half- reactions by mass in base, neutralize the H + with OH - ox: 2 I  (aq)  I 2(aq) red: 4 H + (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) 4 H + (aq) + 4 OH  (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) + 4 OH  (aq) 4 H 2 O (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) + 4 OH  (aq) MnO 4  (aq) + 2 H 2 O (l)  MnO 2(s) + 4 OH  (aq)

Ex 18.3 – Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Balance Half- reactions by charge ox: 2 I  (aq)  I 2(aq) + 2 e  red: MnO 4  (aq) + 2 H 2 O (l) + 3 e   MnO 2(s) + 4 OH  (aq) Balance electrons between half- reactions ox: 2 I  (aq)  I 2(aq) + 2 e  } x 3 red: MnO 4  (aq) + 2 H 2 O (l) + 3 e   MnO 2(s) + 4 OH  (aq) } x 2 ox: 6 I  (aq)  3 I 2(aq) + 6 e  red: 2 MnO 4  (aq) + 4 H 2 O (l) + 6 e   2 MnO 2(s) + 8 OH  (aq)

Ex 18.3 – Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Add the Half- reactions ox: 6 I  (aq)  3 I 2(aq) + 6 e  red: 2 MnO 4  (aq) + 4 H 2 O (l) + 6 e   2 MnO 2(s) + 8 OH  (aq) tot: 6 I  (aq) + 2 MnO 4  (aq) + 4 H 2 O (l)  3 I 2(aq) + 2 MnO 2(s) + 8 OH  (aq) Check Reactant CountElement Product Count 6I6 2Mn2 12O 8H8 88 charge 88

Practice - Balance the Equation H 2 O 2 + KI + H 2 SO 4 ® K 2 SO 4 + I 2 + H 2 O

oxidation reduction ox:2 I -  I 2 + 2e - red:H 2 O 2 + 2e H +  2 H 2 O tot2 I - + H 2 O H +  I H 2 O H 2 O KI + H 2 SO 4  K 2 SO 4 + I H 2 O

Practice - Balance the Equation ClO Cl -  Cl 2 (in acid)

Practice - Balance the Equation ClO Cl - → Cl 2 (in acid) oxidation reduction ox:2 Cl - → Cl e - } x5 red:2 ClO e H + → Cl H 2 O} x1 tot10 Cl ClO H + → 6 Cl H 2 O ClO Cl H + → 3 Cl H 2 O

Electrical Current current of a liquid in a stream, = amount of water that passes by in a given period of time electric current = amount of electric charge that passes a point in a given period of time whether as e - flowing through a wire or ions flowing through a solution

Redox Reactions & Current redox reactions involve the transfer of e - from one substance to another therefore, redox reactions have the potential to generate an electric current in order to use that current, we need to separate the place where oxidation is occurring from the place that reduction is occurring

Electric Current Flowing Directly Between Atoms

Electric Current Flowing Indirectly Between Atoms

Electrochemical 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 aka galvanic cells nonspontaneous redox reactions can be made to occur in an electrolytic cell by the addition of electrical energy

Electrochemical Cells redox reactions kept separate half-cells e - flow in a wire along and ion flow in solution constitutes an electric circuit requires a conductive metal or graphite electrode to allow the transfer of e - through external circuit ion exchange between the two halves of the system electrolyte

Electrodes Anode electrode where oxidation occurs anions attracted to it connected to positive end of battery in electrolytic cell loses weight in electrolytic cell Cathode electrode where reduction occurs 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

Voltaic Cell the salt bridge is required to complete the circuit and maintain charge balance

Current and Voltage # e - that flow through the system per second is the current unit = Ampere 1 A of current = 1 Coulomb of charge per second 1 A = x e - /sec. Electrode surface area dictates the number of e - that can flow

Current and Voltage the difference in potential energy between the reactants and products is the potential difference unit = Volt 1 V of force = 1 J of energy/Coulomb of charge the voltage needed to drive electrons through the external circuit amount of force pushing the electrons through the wire is called the electromotive force, emf

Cell Potential the difference in potential energy between the anode the cathode in a voltaic cell is called the cell potential cell potential depends on the relative ease with which the oxidizing agent is reduced at the cathode and the reducing agent is oxidized at the anode the cell potential under standard conditions is called the standard emf, E° cell 25°C, 1 atm for gases, 1 M concentration of solution E cell = E OX + E RED

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

Fe(s) | Fe 2+ (aq) || MnO 4  (aq), Mn 2+ (aq), H + (aq) | Pt(s)

Standard Reduction Potential a half-reaction with a strong tendency to occur has a large + half-cell potential two half-cells are connected, e - will flow so that the half-reaction with the stronger tendency will occur we cannot measure the absolute tendency of a half-reaction, we can only measure it relative to another half-reaction select as a standard half-reaction the reduction of H + to H 2 under standard conditions, which we assign a potential difference = 0 v standard hydrogen electrode, SHE

Half-Cell Potentials 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 E° 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, even if you need to multiply the half-reactions to balance the equation When E° cell > 0 reaction may be spontaneous

Ex 18.4 – Calculate E  cell for the reaction at 25  C Al (s) + NO 3 − (aq) + 4 H + (aq)  Al 3+ (aq) + NO (g) + 2 H 2 O (l) Separate the reaction into the oxidation and reduction half- reactions ox:Al (s)  Al 3+ (aq) + 3 e − red:NO 3 − (aq) + 4 H + (aq) + 3 e −  NO (g) + 2 H 2 O (l) find the E  for each half-reaction and sum to get E  cell E  ox = −E  red = v E  red = v E  cell = E  ox + E  red E  cell = (+1.66 v) + (+0.96 v) = v

Ex 18.4a – Predict if the following reaction is spontaneous under standard conditions Fe (s) + Mg 2+ (aq)  Fe 2+ (aq) + Mg (s) Separate the reaction into the oxidation and reduction half- reactions ox:Fe (s)  Fe 2+ (aq) + 2 e − red: Mg 2+ (aq) + 2 e −  Mg (s) look up the E  half- reactions E  cell = E  ox + E  red = = since E  cell = -ve the reaction is NOT spontaneous as written [Mg 2+ reduction is below Fe 2+ reduction, the reaction is NOT spontaneous as written]

the reaction is spontaneous in the reverse direction Mg (s) + Fe 2+ (aq)  Mg 2+ (aq) + Fe (s) ox:Mg (s)  Mg 2+ (aq) + 2 e − red:Fe 2+ (aq) + 2 e −  Fe (s) sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode

Practice - Sketch and Label the Voltaic Cell Fe(s) | Fe 2+ (aq) || Pb 2+ (aq) | Pb(s), Write the Half- Reactions and Overall Reaction, and Determine the Cell Potential under Standard Conditions.

ox: Fe(s)  Fe 2+ (aq) + 2 e − E  = V red: Pb 2+ (aq) + 2 e −  Pb(s) E  = −0.13 V tot: Pb 2+ (aq) + Fe(s)  Fe 2+ (aq) + Pb(s) E  = V Spontaneous

Predicting Whether a Metal Will Dissolve in an Acid acids dissolve in metals if the reduction of the metal ion is easier than the reduction of H + (aq) metals whose ion reduction reaction lies below H + reduction on the table will dissolve in acid All have +ve E ox E cell = E ox + 0 = +ve

E° cell, ΔG° and K for a spontaneous reaction one that proceeds in the forward direction with the chemicals in their standard states ΔG° < 1 (negative) E° > 1 (positive) K > 1 ΔG° = −RTlnK = −nFE° cell n is the number of electrons F = Faraday’s Constant = 96,485 C/mol e −

Example Calculate ΔG° for the reaction I 2(s) + 2 Br − (aq) → Br 2(l) + 2 I − (aq) since  G° is +, the reaction is not spontaneous in the forward direction under standard conditions Answer: Solve: Concept Plan: Relationships: I 2(s) + 2 Br − (aq) → Br 2(l) + 2 I − (aq)  G , (J) Given: Find: E° ox, E° red E° cell  G° ox: 2 Br − (aq) → Br 2(l) + 2 e − E° = −1.09 v red: I 2(l) + 2 e − → 2 I − (aq) E° = v tot: I 2(l) + 2Br − (aq) → 2I − (aq) + Br 2(l) E° = −0.55 v

E° cell, ΔG° and K ΔG° = −RTlnK = −nFE° cell E° cell = RT x ln K nF R = J/mol.K lnK = 2.303log K F = 96,485 C/mol e - E° cell = logK n

Example Calculate K at 25°C for the reaction Cu (s) + 2 H + (aq) → H 2(g) + Cu 2+ (aq) since  < 1, the position of equilibrium lies far to the left under standard conditions Answer: Solve: Concept Plan: Relationships: Cu (s) + 2 H + (aq) → H 2(g) + Cu 2+ (aq)  Given: Find: E° ox, E° red E° cell  ox: Cu (s) → Cu 2+ (aq) + 2 e − E° = −0.34 v red: 2 H + (aq) + 2 e − → H 2(aq) E° = v tot: Cu (s) + 2H + (aq) → Cu 2+ (aq) + H 2(g) E° = −0.34 v

Nonstandard Conditions - the Nernst Equation Relationship between E cell (nonstandard) and E° cell (standard) ΔG = ΔG° + RT ln Q Subs. ΔG° = -nFE° cell into above eqn. -nFE cell = -nFE° cell + -RTlnQ(divide by -nF) E cell = E° cell - (RT/nF) log Q R = J/mol.K, (RT/nF)lnQ = (0.0592/n)logQ E cell = E° - (0.0592/n) logQ Called the Nernst equation

Nonstandard Conditions - the Nernst Equation E cell = E° cell - (0.0592/n) log Q at 25°C 1. when Q = 1 (std. conditions) E cell = E° cell 2. At equilibrium,Q = K, E cell = E° cell - (0.0592/n) log Kand (0.0592/n) log K = E° cell E cell = 0 Potential reaches zero as concentrations approach equilibrium Used to calculate E when concentrations not 1 M

E  at Nonstandard Conditions Reactant conc. > standard conditions Product conc. < standard conditions … reaction shifts right

Example Calculate E cell at 25°C for the reaction 3 Cu (s) + 2 MnO 4 − (aq) + 8 H + (aq) → 2 MnO 2(s) + 3 Cu 2+ (aq) + 4 H 2 O (l) units are correct, E cell > E° cell as expected because [MnO 4 − ] > 1 M and [Cu 2+ ] < 1 M Check: Solve: Concept Plan: Relationships: 3 Cu (s) + 2 MnO 4 − (aq) + 8 H + (aq) → 2 MnO 2(s) + Cu 2+ (aq) + 4 H 2 O (l) [Cu 2+ ] = M, [MnO 4 − ] = 2.0 M, [H + ] = 1.0 M E cell Given: Find: E° ox, E° red E° cell E cell ox: Cu (s) → Cu 2+ (aq) + 2 e − } x 3E° = −0.34 v red: MnO 4 − (aq) + 4 H + (aq) + 3 e − → MnO 2(s) + 2 H 2 O (l) } x 2 E° = v tot: 3 Cu (s) + 2 MnO 4 − (aq) + 8 H + (aq) → 2 MnO 2(s) + Cu 2+ (aq) + 4 H 2 O (l)) E° = v

Concentration Cells it is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different the difference in energy is due to the entropic difference in the solutions the more concentrated solution has lower entropy than the less concentrated electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anode reduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode

when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow Concentration Cell Cu 2+ (aq) + 2e - → Cu(s) 0.34V Cu(s) → Cu 2+ (aq) + 2e V Cu 2+ (aq) + Cu(s) → Cu(s) + Cu 2+ (aq) E° cell = E° red + E° ox = 0V Cu(s)  Cu 2+ (aq) (1 M)  Cu 2+ (aq) (1 M)  Cu(s)

Concentration Cell when the cell concentrations are different, e - flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode) Cu(s)  Cu 2+ (aq) (0.010 M)  Cu 2+ (aq) (2.0 M)  Cu(s) Cell potential E cell calculated using Nernst eqn. E cell = E° - (0.0592/n) logQ E cell = E° - (0.0592/n) log([OX]/[RED]) = 0.068V

LeClanche’ Acidic Dry CellDry Cell electrolyte in paste form ZnCl 2 + NH 4 Cl  or MgBr 2 anode = Zn (or Mg) Zn(s) → Zn 2+ (aq) + 2 e - cathode = graphite rod MnO 2 is reduced 2 MnO 2 (s) + 2 NH 4 + (aq) + 2 H 2 O(l) + 2 e - → 2 NH 4 OH(aq) + 2 Mn(O)OH(s) cell voltage = 1.5 v expensive, nonrechargeable, heavy, easily corroded

Alkaline Dry Cell same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste anode = Zn (or Mg) Zn(s) → Zn 2+ (aq) + 2 e - cathode = brass rod MnO 2 is reduced: 2 MnO 2 (s) + 2 NH 4 + (aq) + 2 H 2 O(l) + 2 e - → 2 NH 4 OH(aq) + 2 Mn(O)OH(s) cell voltage = 1.54 v longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc

Lead Storage Battery 6 cells in series electrolyte = 30% H 2 SO 4 anode = Pb Pb(s) + SO 4 2- (aq) → PbSO 4 (s) + 2 e - cathode = Pb coated with PbO 2 PbO 2 is reduced: PbO 2 (s) + 4 H + (aq) + SO 4 2- (aq) + 2 e - → PbSO 4 (s) + 2 H 2 O(l) cell voltage = 2.09 v rechargeable, heavy

NiCad Battery electrolyte is concentrated KOH solution anode = Cd Cd(s) + 2 OH - (aq) → Cd(OH) 2 (s) + 2 e - E 0 = 0.81 v cathode = Ni coated with NiO 2 NiO 2 is reduced: NiO 2 (s) + 2 H 2 O(l) + 2 e - → Ni(OH) 2 (s) + 2OH - E 0 = 0.49 v cell voltage = 1.30 v rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown

Ni-MH Battery electrolyte is concentrated KOH solution anode = metal alloy with dissolved hydrogen oxidation of H from H 0 to H + M∙H(s) + OH - (aq) → M(s) + H 2 O(l) + e - E° = 0.89 v cathode = Ni coated with NiO 2 NiO 2 is reduced: NiO 2 (s) + 2 H 2 O(l) + 2 e - → Ni(OH) 2 (s) + 2OH - E 0 = 0.49 v cell voltage = 1.30 v rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad

Lithium Ion Battery electrolyte is concentrated KOH solution anode = graphite impregnated with Li ions cathode = Li - transition metal oxide reduction of transition metal work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode rechargeable, long life, very light, more environmentally friendly, greater energy density

Fuel Cells like batteries in which reactants are constantly being added so it never runs down! Anode and Cathode both Pt coated metal Electrolyte is OH – solution Anode Reaction: 2 H OH – → 4 H 2 O(l) + 4 e - Cathode Reaction: O H 2 O + 4 e - → 4 OH –

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 = anode the - terminal of the battery = cathode cations attracted to the cathode, anions to the anode cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized some electrolysis reactions require more voltage than E tot, called the overvoltage

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

Electrochemical Cells in all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode in voltaic cells, anode is the source of electrons and has a (−) charge cathode draws electrons and has a (+) charge in electrolytic cells electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery electrons are forced toward the anode, so it must have a source of electrons, the − terminal of the battery

Electrolysis electrolysis is the process of using electricity to break a compound apart electrolysis electrolysis is done in an electrolytic cell electrolytic cells can be used to separate elements from their compounds generate H 2 from water for fuel cells recover metals from their ores

Electrolysis of Water

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 H 2  2 H 2 O + 2 e - → H OH - E° = stand. cond. E° = pH 7 possible anode reactions oxidation of anion to element oxidation of H 2 O to O 2  2 H 2 O → O 2 + 4e - + 4H + E° = stand. cond. E° = pH 7 oxidation of electrode  particularly Cu  graphite doesn’t oxidize half-reactions that lead to least negative E tot will occur unless overvoltage changes the conditions

Electrolysis of NaI (aq) with Inert Electrodes possible oxidations 2 I - → I e - E° = −0.54 v 2 H 2 O → O 2 + 4e - + 4H + E° = −0.82 v possible reductions Na + + e - → Na 0 E° = −2.71 v 2 H 2 O + 2 e - → H OH - E° = −0.41 v possible oxidations 2 I - → I e - E° = −0.54 v 2 H 2 O → O 2 + 4e - + 4H + E° = −0.82 v possible reductions Na + + e - → Na 0 E° = −2.71 v 2 H 2 O + 2 e - → H OH - E° = −0.41 v overall reaction 2 I − (aq) + 2 H 2 O (l) → I 2(aq) + H 2(g) + 2 OH - (aq)

Faraday’s Law the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs charge that flows through the cell = current x time

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

Corrosion corrosion is the spontaneous oxidation of a metal by chemicals in the environment since many materials we use are active metals, 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 promote rusting enhances current flow acids promote rusting lower pH = lower E° red

Preventing Corrosion 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 nails

Sacrificial Anode