Chapter 19 Redox Equilibria

19.1 Redox Reactions (reduction-oxidation) is an equilibrium of the competition for e - between the 2 species reducing agent - undergo oxidation oxidizing agent - undergo reduction

19.2 Oxidation States

Ex a) KMnO 4 0 =(+1) + x + 4(-2) x=+7 oxidation number of Mn is +7 b) POCl 3 0 = x + (-2) + 3(-1) x=+5 oxidation number of P is +5 c) K 2 Cr 2 O 7 0 = 2(+1) + 2x + 7(-2) x=+6 oxidation number of Cr is +6 d) CuSO 4 0 = +2 + x + 4(-2) x=+6 oxidation number of S is +6

Ex e)CaH 2 0 = x x=-1 oxidation number of H is -1 f) Na 2 S 2 O 3 0 = 2(+1)+2x +3(-2) x=+2 oxidation number of S is +2 g) Na 2 S 2 O 8 0 = 2(+1)+2x +8(-2) x=+7 oxidation number of S is +7 h) Na 2 S 4 O 6 0 = 2(+1)+4x +6(-2 ) x=+2.5 oxidation number of S is +2.5 ??

Thiosulphate ion S 2 O Average oxidation number of S is +2

Peroxodisulphate ion S 2 O oxidation number of S is +6

Tetrathionate ion S 4 O 6 2- Average oxidation number of S is +4

Ex a)Mg(s) + H 2 O(l) → MgO(s) + H 2 (g) reducing agent oxidizing agent b)Cr 2 O 7 2- (aq)+2OH - (aq) → 2CrO 4 2- (aq) +H 2 O (l) Not a redox reaction - only conversion under diff. pH c)2CuCl (aq) → Cu (s) + CuCl 2(aq) CuCl acts as the reducing agent and the oxidizing agent at the same time.

2CuCl (aq) → Cu (s) + CuCl 2(aq) Cu + (aq) + e - → Cu (s) reduction Cu + (aq) → Cu 2+ (aq) + e - oxidation ___________________________________ Cu + (aq) + Cu + (aq) → Cu (s) + Cu 2+ (aq) Disproportionation is a chemical change in which a particular chemical species is simultaneously oxidized and reduced.

19.3 Balancing Redox Equations a)SO H 2 O → SO H + +2e - b)VO H 2 O → VO H + (not redox) c)MnO H + + 3e - → MnO 2 + 2H 2 O MnO H 2 O + 3e - → MnO 2 +2H 2 O+4OH - MnO H 2 O + 3e - → MnO 2 + 4OH -

d)MnO H + + 5e - → Mn H 2 O e)BrO H + + 6e - → Br - +3H 2 O f)N 2 H 4 → N 2 + 4H + + 4e - g)Cl 2 + 6H 2 O → 2ClO H + + 4e - h)NO H + +3e - → NO + 2H 2 O

Ex a)Br 2 +2I - → 2Br - + I 2 b)3S 2 O Cr 3+ +7H 2 O → 6SO Cr 2 O H + c)IO Fe H + → I - + 6Fe 3+ +3H 2 O

19.4Electrochemical Cells When a metal is dipped into a solution containing ions of the same metal, a metal/metal ion system is set up. M n+ (aq) + ne - → M (s) M (s) → M n+ (aq) + ne - ne - M n+ MM ne - M n+

M n+ (aq) + ne - M(s) The equilibrium position depends on - nature of the M / M n+ system, -concentration of ions -temperature Equilibrium lies on the right  oxidation ? Reduction? predominates Charge on electrode?  positive ? Negative?

a separation of charge  a potential difference between the electrode and the ions in solution metal/metal ion system is called half cell M ne - M n+ ne - M n+ M ne - M n+ M ne - M n+ M

X V connect 2 different half cells externally  e - s can flow from one electrode to another Current flow through the external conducting wire after the charge is built up between the two system, the current stop ???? Solution??? e-e- X+X+ e-e- Y+Y+ Y Metal X Metal Y X + solution Y + solution

Solutions: Using porous partition or salt bridge Functions: -complete the circuit by allowing ions flow -without extensive mixing of solutions of the 2 half-cell

combination of the two half cell systems is called electrochemical cell Also called galvanic cell is a device which converts chemical energy into electrical energy Each half-cell has a tendency to accept electrons from the other the half-cell with a stronger ability to gain electrons will win the competition

Each half-cell system has its own electrode potential that cannot be measured. Only potential difference between 2 half- cells can be measured The maximum potential difference which the cell can produce, called the electromotive force (e.m.f.)

Daniel cell: Zn has a higher tendency to lose e - than Cu e - s are then pumped out from the Zn electrode through the external circuit to the cell at Cu electrode e-e- e-e-

Daniel cell: Oxidation (Anode / negative terminal) : Zn(s)  Zn 2+ (aq) + 2e - Reduction (Cathode / positive terminal): Cu 2+ (aq) +2e -  Cu(s) Overall Redox equation: Zn(s) + Cu 2+ (aq)  Zn 2+ (aq) + Cu(s)

Ex 19.5 a.An electrochemical cell consists of Ag electrode with AgNO 3 as electrolyte and Cu electrode with CuSO 4 as electrolyte. The salt bridge is made of agar gel with dissolved NaCl. However, no current flow through the external wire. Draw a diagram to show this electrochemical cell and explain why there is no current? Ag + ion from the silver nitrate solution may react with Cl - ion in the salt bridge to form silver chloride which is insoluble in water. Thus, ions cannot flow between the two system and thus the current stops.

b.Explain why salt bridge of NaCl or Na 2 SO 3 should not be used when the electrolyte is acidified KMnO 4. It is because chloride ion and sulphite ion (sulphate(IV)) ion may be oxidized by acidified KMnO 4.

Cell Diagrams A representation of an electrochemical cell using the IUPAC conventions For example, cell diagram of Daniel cell: Zn(s) ∣ Zn 2+ (aq) Cu 2+ (aq) ∣ Cu(s)E =+1.1V solid vertical line ( ∣ )= phase boundary a pair of vertical broken lines ( )= salt bridge single vertical broken line ( ) = porous partition If the salt bridge is made of KCl, Zn(s) ∣ Zn 2+ (aq) KCl Cu 2+ (aq) ∣ Cu(s) E=+1.1V E represents the e.m.f. of the cell in volts

L.H.S. :Anode (negative electrode): Zn(s)  Zn 2+ (aq) + 2e - R.H.S.:Cathode (positive electrode): Cu 2+ (aq) +2e -  Cu(s) By convention, the electrode on LHS is considered as the anode while the one on RHS is considered to be the cathode sign of the e.m.f. indicates the polarity of the right-hand electrode For a +ve e.m.f., the reaction proceeds from left to right Zn(s) ∣ Zn 2+ (aq) Cu 2+ (aq) ∣ Cu(s)E=+1.1V Meaning???

Ex Cu(s) ∣ Cu 2+ (aq) Zn 2+ (aq) ∣ Zn(s)E = ? E = -1.1V that means the reaction proceeds from RHS to LHS or Cu + Zn 2+  Cu 2+ + Zn is not spontaneous. But Cu 2+ + Zn  Cu + Zn 2+ is spontaneous.

Ex Sn(s) ∣ Sn 2+ (aq) Cu 2+ (aq) ∣ Cu(s) E=+0.50V Oxidation (anode): Sn(s)  Sn 2+ (aq) + 2e - Reduction (cathode) Cu 2+ (aq) + 2e -  Cu(s) Overall equation: Sn(s) + Cu 2+ (aq)  Sn 2+ (aq) + Cu(s) (a)

Cu(s) ∣ Cu 2+ (aq) Mg 2+ (aq) ∣ Mg (s) E=-2.70V Oxidation (anode): Mg(s)  Mg 2+ (aq) + 2e - Reduction (cathode) Cu 2+ (aq) + 2e -  Cu(s) Overall equation: Mg(s) + Cu 2+ (aq)  Mg 2+ (aq) + Cu(s) (b)

Ni(s) ∣ Ni 2+ (aq) Zn 2+ (aq) ∣ Zn (s) E=-0.70V Oxidation (anode): Zn(s)  Zn 2+ (aq) + 2e - Reduction (cathode) Ni 2+ (aq) + 2e -  Ni(s) Overall equation: Zn(s) + Ni 2+ (aq)  Zn 2+ (aq) + Ni(s) (c)

Ni(s) ∣ Ni 2+ (aq) Ag + (aq) ∣ Ag(s) E=+0.95V Oxidation (anode): Ni (s)  Ni 2+ (aq) + 2e - Reduction (cathode) Ag + (aq) + e -  Ag (s) Overall equation: Ni(s) + 2Ag + (aq)  Ni 2+ (aq) + 2Ag(s) (d)

Types of Half-cell 1. Metal in contact with an aqueous solution of its metal ion Zn 2+ (aq) +2e - Zn (s) Cell diagram : Zn(s) ∣ Zn 2+ (aq) Zn 2+ (aq) ∣ Zn(s)

2. Metal in contact with its insoluble salt and an aqueous solution of the anion AgCl (s) + e - Ag (s) +Cl - (aq) Cell diagram : Pt(s) ∣ [Ag(s) + Cl - (aq)], AgCl(s) AgCl(s), [Ag(s) + Cl - (aq)] ∣ Pt(s) PbSO 4(s) + 2e - Pb (s) +SO 4 2- (aq) Cell diagram : Pt(s) ∣ [Pb(s) + SO 4 2- (aq,1M)], PbSO 4 (s) PbSO 4 (s), [Pb(s) + SO 4 2- (aq,1M)] ∣ Pt(s)

3. An inert electrode (e.g. Pt) and a gas in contact with an aqueous solution from which the gas can be generated 2H + (aq) + 2e - H 2(g) Cell diagram : Pt(s) ∣ H 2 (g) ∣ 2H + (aq) or Pt(s) [H 2 (g)] ∣ 2H + (aq) 2H + (aq) ∣ H 2 (g) ∣ Pt(s) or 2H + (aq) ∣ [H 2 (g)] Pt(s)

O 2(g) +4e - +2H 2 O (l) 4OH - (aq) Cell diagram : Pt(s) ∣ 4OH - (aq), [O 2 (g) + 2H 2 O(l)] [O 2 (g) + 2H 2 O(l)], 4OH - (aq) ∣ Pt(s)

4. An inert electrode and a solution containing both the oxidized and reduced forms of the species Fe 3+ (aq) + e - ===== Fe 2+ (aq) Cell diagram : Pt(s) ∣ Fe 2+ (aq), Fe 3+ (aq) Fe 3+ (aq), Fe 2+ (aq) ∣ Pt(s)

I 2(aq) + 2e - 2I - (aq) Cell diagram : Pt(s) ∣ 2I - (aq), I 2 (aq) I 2 (aq), 2I - (aq) ∣ Pt(s) MnO 4 - (aq) +8H + (aq) +5e - Mn 2+ (aq) +4H 2 O (l) Cell diagram : Pt(s) ∣ [ Mn 2+ (aq) +4H 2 O (l) ],[MnO 4 - (aq) +8H + (aq) ] [MnO 4 - (aq) +8H + (aq) ], [ Mn 2+ (aq) +4H 2 O (l) ] ∣ Pt(s) The most reduced form is written next to the inert electrode (i.e.Pt)

19.5 Standard Electrode and Relative Electrode Potentials from E.m.f Measurements Standard Electrode Electrode potential of a half cell cannot be measured. standard hydrogen electrode (s.h.e.) -selected as the reference electrode -its electrode potential is assigned as 0V at 25 o C, 1atm and 1.0M H +. With the reference, the relative scale of electrode potential is established for different systems in order to compare their tendency to release e - s

Standard hydrogen electrode: A. Inlet for 1atm H 2 at 25 o C B. Platinum coated with platinum black C. 1M H + D. outlet for H 2

has an electrode made of a piece of platinum coated with finely divided platinum black function of this special electrode: –catalyses the half-cell reaction : H 2 (g) 2H + (aq) + 2e - –provides a surface on which the hydrogen can be adsorbed –provides an electrical connection to the voltmeter with the concentration of HCl(aq) of 1 mol dm -3 a slow stream of pure hydrogen gas at 1atm is bubbled over the platinized surface

cell diagram Pt(s) ∣ H 2 (g) ∣ 2H + (aq) or Pt(s) [H 2 (g)] ∣ 2H + (aq) Half -cell reaction : Reduction: 2H + (aq) + 2e - ===> H 2 (g) Oxidation: H 2 (g) ===> 2H + (aq) + 2e -

As the potential of an electrode system depends on, and. The electrode potential of the reference electrode can only be assigned as 0V under a specified conditions: H 2 gas at 1atm [H + ] = 1 mol dm -3 under temperature of 298K When the hydrogen electrode is used under the above conditions, the hydrogen electrode is known as a standard hydrogen electrode (s.h.e.)

s.h.e. is always assumed to be the anode of the system. (the left hand side in the cell diagram) Apart from hydrogen electrode, a calomel electrode can also be used: Cell diagram : Pt(s) ∣ [2Hg(l) + 2Cl - (aq,1M)], Hg 2 Cl 2 (s) Equation : Hg 2 Cl 2 (s) + 2e -  2Hg(l) + 2Cl - (aq)

Relative Electrode Potential from e.m.f. Measurements e.m.f. for the cell is a measure of the relative tendencies of half cells to release or gain e - s e.m.f. of a cell depends on temperature, concentration and also pressure all the measurement should be measured under the standard conditions: -temp = 298K (25 o C) -concentration of all solution = 1.0M -pressure of all gases = 1.0 atm

standard electrode potential (or standard reduction potential / E o ) is the electrode potential (e.m.f.) of a half cell measured when connected to a standard hydrogen electrode under the standard conditions

Cell diagram : Pt(s) ∣ H 2 (g, 1atm) ∣ 2H + (aq,1M) M n+ (aq,1M) ∣ M(s) E o =? Assumed Anode? Assumed Cathode? If s.h.e. undergoes oxidation, sign of E o ?

For the standard reduction potential of the half equation: Cu 2+ (aq) + 2e -  Cu(s)E o = +0.34V Meaning?? Pt(s) ∣ H 2 (g, 1atm) ∣ 2H + (aq,1M) Cu 2+ (aq,1M) ∣ Cu(s) E o = +0.34V Anode: H 2 (g)  2H + (aq) + 2e - Cathode: Cu 2+ (aq) + 2e -  Cu(s) Overall reaction: H 2 (g) + Cu 2+ (aq)  2H + (aq) + Cu(s)

for the following half equation: Ni 2+ (aq) + 2e -  Ni(s) E o = -0.25V cell diagram: Pt(s) ∣ H 2 (g, 1atm) ∣ 2H + (aq,1M) Ni 2+ (aq,1M) ∣ Ni (s) E o = -0.25V The value of E o is, the polarity of the nickel electrode is. That means nickel is the and undergoes. Anode: Ni(s)  Ni 2+ (aq) + 2e - Cathode 2H + (aq) + 2e -  H 2 (g) Overall Reaction: Ni(s) + 2H + (aq)  Ni 2+ (aq) + H 2 (g)

e.m.f. of this cell, E o cell = E R.H.S. - E L.H.S. The electrode system with the greatest negative electrode potential is the strongest reducing agent. By comparing the value of the standard reduction potentials of different electodes, the electrochemical series can be obtained.

19.6 Uses of Standard Reduction Potentials A cell will always run spontaneously in the direction that produces a positive e.m.f. For an electrochemical cell: Cu(s) ∣ Cu 2+ (aq,1M) Ag + (aq, 1M) ∣ Ag(s) E o cell =+0.462V 2Ag + (aq) + Cu(s) → Cu 2+ (aq) +2Ag(s)E o cell = V From data book, Cu 2+ (aq) + 2e - → Cu (s) E o =+0.337V Ag + (aq) + e - → Ag(s)E o = V Anode: Cu(s) → Cu 2+ (aq) + 2e - E o =-0.337V Cathode: 2Ag + (aq) + 2e - → 2Ag(s) E o = V Ag + (aq)+Cu(s) → Cu 2+ (aq)+2Ag(s) E o cell = V

The value of electrode potential is not changed when a half-reaction is multiplied by an integer because a standard reduction potential is an intensive property The e.m.f. are usually shown for the reduction half reaction. For an oxidation half reaction, the sign of the e.m.f. must be reversed P.10

Ex (a) Ni(s) ∣ Ni 2+ (aq) [NO 3 - (aq)+3H + (aq)],[HNO 2 (aq)+H 2 O(l)] ∣ Pt(s) Ni 2+ (aq) → Ni(s) E o = -0.25V NO 3 - (aq) → HNO 2 (aq)E o = +0.94V E o cell = E o cathode - E o anode E o cell = (+0.94V) - (-0.25V) = +1.19V Ni (s) +NO 3 - (aq) +3H + (aq) → Ni 2+ (aq) +HNO 2(aq) +H 2 O (l)

(b) Pt(s) ∣ 2H 2 SO 3 (aq),[4H + (aq) + S 2 O 6 2- (aq)] Cr 3+ (aq) ∣ Cr(s) Given: S 2 O 6 2- (aq) → H 2 SO 3 (aq) E o = +0.57V Cr 3+ (aq) → Cr(s) E o = -0.74V E o cell = E o cathode - E o anode E o cell = (-0.74V) - (+0.57V) = -1.31V S 2 O 6 2- (aq) +4H + (aq) + 2e - → 2H 2 SO 3(aq) x 3 Cr (s) → Cr 3+ (aq) + 3e - x S 2 O 6 2- (aq) +12H + (aq) + 2Cr (s) → 6H 2 SO 3(aq) + 2Cr 3+ (aq)

Prediction of the Feasibility of Redox Reactions more -ve E o ====> more reducing the system more +ve E o ====> more oxidizing the system By comparing E o values, an electrochemical series is formed. sign on the e.m.f. of cell ===> to predict the cell reaction Simiarly, use to predicting whether the reaction is feasible or not. overall redox reaction with E o cell = E o cathode - E o anode > 0 is thermodynamically (energetically) feasible. The redox reaction will run spontaneously

Example 19.1 Pb(s) + Mn 2+ (aq) → Pb 2+ (aq) + Mn(s) feasible? Cell diagram:Pb(s) ∣ Pb 2+ (aq) Mn 2+ (aq) ∣ Mn(s) Half-equations: Pb 2+ (aq) +2e -  Pb(s) E o = -0.13V Mn 2+ (aq) + 2e -  Mn(s) E o = -1.18V Δ E o = E o R - E o L = (-1.18V) - (-0.13V) = -1.05V The negative sign shows that the given reaction will not proceed spontaneously or say the reaction is not feasible.

Ex a)Br 2 (aq) + 2I - (aq) → 2Br - (aq) + I 2 (aq) E o cell =(+1.07V) - (+0.54V) = +0.53V As the emf is positive, the reaction will proceed spontaneously. (the reaction is feasible) b) 2MnO 4 - (aq) + 16H + (aq) + 10Br - (aq) → 8H 2 O(l) + 2Mn 2+ (aq) + 5Br 2 (aq) E o cell =(+1.51V) - (+1.07V) = +0.44V As the emf is positive, the reaction will proceed spontaneously. (the reaction is feasible)

Limitations of Predictions Made Using E o Values Is it true that all the redox reactions with E o > 0 will take place??? A cell e.m.f. only gives information about the feasibility of a redox reaction from an energetic point of view. It cannot tell how fast a feasible reaction is likely to proceed. Therefore, some feasible redox reactions may not appear to take place just because they are too slow. The cell e.m.f. may be not taken under standard conditions.

A a rule of thumb, if the cell e.m.f. > 0.4V, it is quite safe to say that the prediction is valid. The electrochemical series deals only with aqueous solutions of ions, and reducing tendencies may be quite different in other solvents or for reactions involving gases and solids

Testing Predictions about the Feasibility of Redox Reactions By simple tests or simple observation For example, to test whether the reaction between Br 2 and KI is feasible 3ml aqueous solutions (about 0.1M) of both Br 2 and KI can be mixed in test tubes 1,1,1-trichloroethane is added and the lower layer became purple Deductions : I 2 is present The reaction has taken place orange purple

19.7Secondary Cell and Fuel Cell a device that stores chemical energy for later releases as electricity. (chemical energy ==> electrical energy) Basically, there are 3 types of cells and batteries: 1. Primary Cells 2. Secondary Cells 3. Fuel Cells

1.Primary Cells - galvanic cells - cannot be recharged after reaching the eqm 2.Secondary Cells - rechargeable - a non-equilibrium mixture of reactants is produced by an external source of electricity in the charging process 3.Fuel Cells - a type of primary cell - the reactants are continuously applied from outside while the cell is in use

Lead Acid Accumulator most common secondary cell used as motor car storage battery Anode : Lead Cathode : Lead(IV) coated with lead dioxide Electrolyte : sulphuric acid a number of cathode alternate with several anode plates ==> increase current

Anode : Pb(s) + SO 4 2- (aq) → PbSO 4 (s) + 2e - Cathode:PbO 2 (s)+4H + (aq)+SO 4 2- (aq)+2e - → PbSO 4 (s)+2H 2 O(l) Cell reaction :Pb(s)+PbO 2 (s)+2H 2 SO 4 (aq) → 2PbSO 4 (s)+2H 2 O(l) V Pb PbO 2 e-e- H 2 SO 4 (aq)

Alternatives, Anode : Pb(s) + HSO 4 - (aq) → PbSO 4 (s) +H + (aq) + 2e - Cat:PbO 2 (s)+3H + (aq)+HSO 4 - (aq)+2e - → PbSO 4 (s)+2H 2 O(l) Cell rxn :Pb (s) +PbO 2(s) +2HSO 4 - (aq) +2H + (aq) → 2PbSO 4(s) +2H 2 O (l) Cell diagram: Pb(s) ∣ [Pb(s)+SO 4 2- (aq)],PbSO 4 (s) ∣ H 2 SO 4 (aq) ∣ [ PbO 2 (s)+SO 4 2- (aq)+4H + (aq)],[PbSO 4 (s)+2H 2 O(l)] ∣ Pb(s) Alternatives, Pb(s) ∣ PbSO 4 (s) ∣ H 2 SO 4 (aq) ∣ PbO 2 (s) ∣ PbSO 4 (s) ∣ Pb(s)

No current Reasons: - both electrode coated with PbSO 4 - density of liquid drops (conc. H 2 SO 4 drops) After discharging for some time, V PbSO 4

When charging: Cathode :PbSO 4 (s) + 2e - → Pb(s) + SO 4 2- (aq) Anode :PbSO 4 (s) + 2H 2 O(l) → PbO 2 (s) + 4H + (aq) + |SO 4 2- (aq) + 2e Overall :2PbSO 4 (s) + 2H 2 O(l) → Pb(s)+PbO 2 (s)+2H 2 SO 4 (aq) Pb(s) + PbO 2 (s) + 2H 2 SO 4 (aq) 2PbSO 4 (s) + 2H 2 O(l) discharge charge

Hydrogen-Oxygen Fuel Cell is a primary cell converts the chemical energy of continuous supply of reactants (fuels) into electrical energy. KOH(aq) as electrolyte e-e- 1. Stream of H 2 (fuel) 2. H 2 diffuses through the porous anode (e.g. Ni). When is comes into contact with the electrolyte, KOH (aq), adsorbed H 2 is oxidized: 2OH - (aq) + H 2(g) → 2H 2 O (l) + 2e - 4. O 2 diffuses through the porous cathode (e.g. Ni). Adsorbed O 2 is reduced: O 2(g) + 2H 2 O (l) + 4e - → 4OH - (aq) 3. Stream of O 2 (oxidizing agent) 5. e - s flow through an external circuit from anode to cathode. AnodeCathode Overall:2H 2 (g) + O 2 (g) → 2H 2 O(l)

Function of porous nickel electrodes Acts as electrical conductors connecting to external circuit acts as catalyst for the reaction allows adsorption of reactants(fuels) and oxidizing agent for reaction

Fuel cells differ from other electrochemical cells do not store reactants and electrolytes electrodes do not undergo permanent change reactants are gases product and operating procedures are non-polluting (only water is produced) high efficiency of energy production (  continuous supply of fuel) can supply energy at a constant rate Disadvantages: High cost (limited in spacecrafts)

19.8 Corrosion of Iron and its Prevention Rusting - corrosion of iron an electrochemical process the slow deterioration of iron to hydrated iron(III) oxide this oxide is permeable to both air and water  can’t protect iron from further corrosion (like Al 2 O 3 )

Anode:Fe(s) → Fe 2+ (aq) + 2e - Cathode:O 2 (g) + 2H 2 O(l) + 4e - → 4OH - (aq) Fe 2+ (aq) + 2OH - (aq) → Fe(OH) 2 (s) 2Fe(OH) 2 (s) + 1/2O 2 (g) + (x-2)H 2 O(l) → Fe 2 O 3 xH 2 O(s) Overall : 4Fe 2+ (aq) + O 2 (g) + (4+2n)H 2 O(l) → 2Fe 2 O 3 nH 2 O(s) + 8H + (aq)

Factors that speed up rusting a low pH - presence of acid, or even CO 2, NO 2 and SO 2 in rain water Presence of electrolytes (i.e. NaCl) - increases the conductivity High temperature - increase the rate of chemical reaction in contact with less reactive metal such as Cu and Sn Sharply pointed regions - serves as anode

Prevention of Rusting 1.Protective Coating A protective layer is coated on the surface of iron to exclude oxygen and water a)Painting b) Oiling and greasing

c) Galvanizing (zinc-plating) d) Tin-plating e) Electroplating

2. Sacrificial protection 3. Alloying 4. Cathodic protection iron is treated as the cathode and connected to the negative terminal of a battery

Ex a)i) (+1.23)-(-0.45) = +1.68V ii) (+0.40)-(-0.45) = +0.85V b)As E o cell in (i) is more positive, iron lose electrons more readily to form Fe(II) ions. Thus, Iron would rust more rapidly under acidic conditions. c)Iron would rust more rapidly in sea water than in pure water because the process of corrosion can be speeded up by electrolyte like NaCl, which increases the conductivity of the solution.

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