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8/10/2019 Electrochemistry Slides Upload

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Electrochemistry has a vital role in electro refining, battery technology, corrosioncontrol & metal finishing.

An electrochemical cell consists of two electrodes or metallic conductors, in contactwith an electrolyte, an ionic conductor.The two electrodes may share the same compartment. If the electrolytes are different,the two compartments may be joined by a salt bridge.


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Electrochemical cell consists of two electrodes & an electrolyte in a compartment.The two electrodes may share the same compartment. If the electrolytes are different,the two compartments may be joined by a salt bridge, which is a tube containing aconcentrated electrolyte soln. in agar jelly that completes the electrical circuit toenable the cell to function. An electrochemical cell can act as Galvanic cell orelectrolytic cell.

In a galvanic cell, anode is the site for oxidation & cathode is the site for thereduction. The -vely charged electrons flow from anode to cathode and e -s areattracted to the +ve cathode from the ve anode thru the external circuit.


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A galvanic cell produces electricity as a result of the spontaneous redox reaction thatoccurs.Galvanic cell generally consists of two electrodes dipped in two electrolyte solns.which are separated by a porous diaphragm or connected through a salt bridge. Itconverts stored up CE in the form of electroactive materials directly into EE. In agalvanic cell, anode is -ve; and cathode is +ve.Galvanic cells are used as portable source of electrical energy.

Galvanic cell is represented by as follows:Anode is written on the LHS & cathode is written on the RHS of cell representationThe anode is represented by writing the metal (or solid phase) first & then theelectrolyte along with the concentration in the bracket. The two are separated by avertical line.Anode representation:Metal/ Metal ion (concentration) i.e. M/M n+ (c)

ZnZn2+ or Zn; Zn 2+ or Zn ZnSO4 (1M) or Zn; ZnSO 4 (1M)The cathode is represented by metal ion first & then the metal (or solid phase)thereafter. The two are separated by a vertical line.Cathode representation:

Mn+ (c) / M i.e. Metal ion (concentration) / Metale.g. Cu 2+Cu or Cu2+; Cu or Cu 2+ (1M); Cu or CuSO 4 (1M)/Cu

4. A salt bridge is indicated by two vertical lines, separating the two half-cells.


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e.g. Zn ZnSO4 (1M) CuSO4(1M)/CuCu(s) / Cu 2+ (aq) Ag+(aq) / Ag (s)


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Daniel cell is a typical example for a galvanic cell & it may be represented as:Zn ZnSO4 (1M) CuSO4(1M)/Cu

At the anode: Zn Zn2+ + 2e -At the cathode: Cu 2+ + 2e - Cu

Net reaction:Zn(s)+ Cu 2+(aq) Zn2+(aq)+ Cu(s)


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An electrolytic cell is an electrochemical cell in which a non-spontaneous redox rxn isdriven by an external source of electrical energy (or a battery ) i.e. EC-a singlearrangement of two electrodes in one or two electrolytes where non-spontaneousredox rxn is forcibly carried out by sending EE into the cellElectrolytic cell involves conversion of EE into CE.In an electrolytic cell, anode is +ve (electrode which attracts the anions) & cathode is ve (electrode which attracts the cations).This type of cells have wide ranging applications in purification of metals, inelectrodeposition of a metal on to the surface.


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The emf cannot be measured accurately using a voltmeter :

As a part of the cell current is drawn thereby causing a change in the emf.

As a part of the emf is used to overcome the internal resistance of the cell.

The E Cell depends on :the nature of the electrodes.temperature.concentration of the electrolyte solutions.


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Potential at the junction of the two electrolytes, arising from unequal migration ofcations in one direction & anions in the other.In a Daniel cell, Cu 2+ ions diffuse into ZnSO 4 (aq) & Zn 2+ ions diffuse into CuSO 4 (aq) because Cu 2+ ions are slightly more mobile than Zn 2+ ions. This produces a smallexcess +ve charge on ZnSO 4(aq) side of the junction & equal - ve charge on theCuSO 4 (aq) side.LJPs are of the order of a few millivolts. It can be minimized by SB : KCl, KNO 3,

NH 4 NO 3 whose ions have same migration velocities. As ions move almost with equalspeed to anode & cathode compartments Junction potential almost reduced to zero.


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1. Prevents any net charge accumulation in the two electrolytes2. Physically separates the two electrolytes but completes the electrical circuit by

allowing ions carrying charge to move from one half-cell to the other3. Reduces LJP to a minimum by nearly equal diffusion rates of positive & negative

ions [ Diffusion, Migration, Convection ]


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Construction: It consists of an H-shaped glass vessel with the lower ends of botharms closed, having platinum wire scaled into the bottom of the each arm. The +veelectrode contains mercury over which a paste of mercurous sulfate (Hg 2SO 4) &mercury is placed. The -ve electrode consists of an amalgam of Cd & Hg, containingabout 12-15 % Cd by weight. Over both electrodes are sprinkled some crystals ofsolid CdSO 4. 8/3 H 2O are added to keep the electrolyte solution satd at alltemperatures. The remaining part of the cell is filled with a saturated solution ofcadmium sulphate that acts as an electrolyte. The cupper ends of the tube are closedwith corks & sealing wax.

Cell representation:Cd-Hg/Cd 2+// Hg 2SO 4/Hg

Working: Anode: Cd (s) Cd 2+(aq) + 2e - Cathode: Hg2SO 4(s) + 2e - 2Hg (l)+ SO 42-(aq)CR: Cd(s) + Hg 2SO 4(s) CdSO 4(s) + 2Hg(l)

Explanation:-----------Weston Cadmium Cell :

*The emf of the cell is 1.01807 V at 298 K *(E/T) = 4x10 -5 volt/degree*Emf remains const. for many years *Reproducible value of emf


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Application : as a std.cell in accurate emf measurements of other cells


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Single Electrode potential (E): The potential of single electrode is the difference of potential between a metal and the solution of its salt. i.e the tendency of an electrodeto lose or gain electrons when in contact with its own ions in the soln.In a cell, the electrode with lower EP has less tendency to undergo reduction & acts asanode. The electrode with higher electrode potential has higher tendency to undergoreduction & acts as cathode.Example: In case of Zn in ZnSO 4 soln. acquires a -ve chargeM(s) Mn+ + ne -(aq) (1) Di ssoluti on r eactionMetal shows the tendency to go into the solution as metal ion by losing electrons.Mn+ (aq) + ne -(aq) M(s) (2) Deposition reactionOxidation potential vs Reduction potentialBy convention, E.P. Reduction potential

Standard E.P. (E o) std.state c = 1 M, P = 1atm.T = 298 K

If the dissolution rxn is faster than deposition If the deposition rxn is faster than dissolution

Formation of H elmh oltz double layer /electrical double layer (EDL) at the junctionof the electrode & the soln.The P.D. across the EDL is the cause for the EP.


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Absolute EPs cannot be measured i.e. it is impossible to determine absolute half-cell potential Why?All voltage measuring devices differences in potentialVoltmeter-2 knobs ( EPs cannot be measured with one wire !)

one contact test electrode--other contact with the soln in the electrode compartment via another conductorSecond contact solid/soln interface acts as second half cellRedox rxn must occur if charge is to flow & potential is to be measuredPotential is associated with this second rxnAbsolute half-cell potential is not obtainedDifference between the half-cell potential of test electrode & a half cell made up ofthe second contact & the soln.We can only determine the relative value of electrode potential, if we can fixarbitrarily the potential of any one electrode.

The numerical values of electrode potentials of different electrodes are relative valuesw.r.t SHE by arbitrarily fixing its potential as zero.Therefore, single electrode potentials of electrodes are referred to as potentials on thehydrogen scale.


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Sign of EP :A +ve sign on the EP value indicates that the half cell acts as cathode & aceptselectrons from the hydrogen electrode.A ve sign on the electrode potential value indicates that the half cell acts as anode byreleaseing electrons to the hydrogen electrode.

According to latest convention, all single electrode potential values reported representreduction potentials.

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The relation between E & G:The maximum electrical work as per thermodynamics, chemical reaction can do isgiven by the value G for the reaction.

We, max = G----------------- (1)The net electrical energy & the emf of the cell produced in a cell is the product of thequantity of electricity that passes through the cell & the emf of the cell.

We = QE Joules. ------- (2)If n moles of electrons are transferred in the cell reactions, the quantity of electricity =n FTherefore, electrical energy= nFE, ---------- (3)Electrical energy produced = decrease in free energy for cell reaction i.e. - G =nFE

or G =-nFE Joules------- (4)A cell in which the overall reaction has not reached chemical equilibrium can do

electrical work as the reaction drives electrons through an external circuit.The cell does net work at the expense of decrease in free energy. If we know emf E,G can be calculated. The larger the value of CP, the further the reaction is fromeqmThe sign of CP tells us the direction in which rxn. must shift to reach eq m. Relation between E & H :From Gibbs Helmholtz equation: - G =-H- T [(G)/ T]P

G = H + T [(G)/ T]P ------- (5)-nEF = H + T {(-nEF)/ T} p


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nFE = - H +TnF( E/T)P { as n & F are constants}H =TnF( E/T)P nFE

Enthalpy Change H = nF[T( E / T ) P E] -------- (6)If we know, emf & temp.coeff. of emf, H can be calculated The heats of reaction calculated from emf measurements is almost equal to valuesderived from thermal measurements.Case- 1, If ( E/ T)P =0, nFE = - H .The electrical energy is equal to the heat ofreactions.Case- 2, If ( E/ T)P >0, nFE > - H. The additional energy will come either fromsurrounding or the temperature of the cell would fall.Case- 3, If ( E/ T)P


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Solution:1- G: G =- n FEn = 2 for the cell reaction; F = 96,500 CE= 1.0181 V at 298 KG =-2 x 96,500 x 1.0181 J = -196.5 KJ

H: H =nF [ T ( E /T)P E](E/T)p = 1.0181 1.0183 / 298-293

= -0.0002 / 5 = -0.00004VK -1

T = 298 KH = 2 x 96,500 { 298 x (-0.00004) 1.0181)

= -198. 8 KJS: S =nF (E / T) P

= 2 x 96,500 x (-0.00004) = -7.72JK -1

Soln. 2. Here n = 2, E = 0.6753 V at 298 K and 0.6915 at 273 K. (E/T)p = (0.6753-0.6915) V/ (298-273) K = - 0.00065 V/K Now at 298 K, H =nF [ T ( E /T)P E]

= 2 mol x 96,500 C mol -1{ [298 K x (- 0.00065 V/K)] 0.6753 V }

= -167. 7 KJG at 298 K ; G =- n FE = -2 x 96,500 x 0.6753 = -130.33 KJS =nF(E / T) P

= 2 x 96,500 x (0-00065) = -125.45 J.K -1


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Derivation:Consider a general redox reaction:

Mn+(aq) + ne - M(s)---- (1)We know that, G =Go +RT ln K ---- (2)and G =-nFE ----- (3)G =Go +RT ln[M]/[M n+]----- (4)-nFE= -nFE o + RT ln [M]/[M n+]---- (5)

E= E o RT/nF ln 1/[M n+]------ (6)E=E o- 2.303 RT/nF log 1/[M n+]--- (7)

At 298K,E= E o-0.0592/n log 1/[M n+]------- (8)

E= Electrode potential at some moment in timeE0= Standard electrode potentialR= Universal gas constant ( joules per mol K)T= Temperature in Kelvin scaleF= Faraday constant (96,500 Coulumbs)n = Number of electrons transferred in the half reaction/ cell reactions[M n+]= Concentration of metal ion at that moment in time (moles per litre).

The Nernst equation for the emf of the cell isEcell = E ocell 2.303RT/nF log [C] c [D] d /[A] a [B] b At 298 K,


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Application of the Nernst equation:It can be used to calculate the potential of a cell that operates under non-standardconditions.It can be used to measure the equilibrium constant for a reaction. At eqm. the overallcell potential for the reaction is zero. i.e. E=0

0 = E o - RT/nF lnKcKc = e nFE0 / RT

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Solution :1:

Cell reaction: Fe + Cu 2+ Fe2+ + CuECell = E 0Cell -0.0592/2 log [Fe 2+ ]/[Cu 2+]

ECell = 0.78 + 0.0296 log 10/1=0.8096V

Soln. 2:Cell representation: Zn/ Zn 2+((1M)//Ag +(10M) /Ag

Cell reaction: Zn + 2Ag + Zn2+ + 2Ag

ECell = E 0Cell -0.0592/2 log [Zn 2+ ]/[Ag +]2

ECell = 1.56 + 0.0592 log 10/1.0=1.6192 V


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Construction: Pure mercury at the bottom of a tube. Paste of mercurous chloride(calomel) filled with satd soln of KCl. Pt-wire sealed at its end fixed into main tube( Pt-wire remains dipped into the mercury )The electrode can be represented as ; Hg

(l) /Hg

2Cl

2 (satd)/KCl -(satd/xM)

The calomel electrode can act as anode or cathode depending on the nature of otherelectrode of the cell.When it acts anode, the electrode reaction is2Hg(l) + 2Cl - (aq) Hg2Cl2 (s)+ 2e - When it acts as cathode, the electrode reaction is,Hg2Cl2(s) + 2e - 2Hg (l) + 2 Cl -(aq)The net reversible electrode reaction is,Hg2Cl2(s) + 2e - 2 Hg (l) + 2 Cl -(aq)

Electrode potential is given byE = E o - 0.0591 log [Cl -] at 298 KIts electrode potential depends on the concentration of KCl solution.Uses:Since the electrode potential is a constant it can be used as a secondary referenceelectrode to replace the inconvenient SHE for potential measurements.The test electrode, Zn(s) /Zn 2+(aq) is coupled with a satd calomel electrode.Zn(s)/Zn 2+ (aq )Cl- (satd soln.) /Hg 2Cl2(s) / Hg (l)The emf of the so formed cell is determined experimentally by potentiometric method.

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Then E cell = E cathode E anode = 0.2444 Ezn

Ezn = 0.2444 E cell

To determine the pH of a solution cell:

H+ /pt,H 2 (g)//Hg(l) / Hg 2Cl2 (s)

E cell = E cathode E anode = 0.2422+ 0.0592 pH

pH = (E cell- 0.2422) /0.0592

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It is less prone to contamination because the mercury/ mercurous chloride interfaceis protected inside a tube not in direct contact with the electrolyte.If we use the Calomel electrodes above 50 o C , the mercurous chloride breaks down,yielding unstable readings. This is important if substantial temperature changes occurduring a measurement.

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Construction: The electrode consists of a thin glass membrane, typically about 0.03to 0.1 mm thick, sealed onto one end of a heavy walled glass tube. A special varietyof glass (corning 0l5 glass with approximate composition 20 % Na 2O, 6 % CaO & 72% SiO

2) is used which has low melting point & high electrical resistance. The glass

bulb is filled with a solution of constant pH (0.1 M HCl). A small volume of saturatedsilver chloride is contained in the tube. A silver wire in this solution forms asilver/silver chloride reference electrode which is connected to one of the terminals ofa potential measuring device. The internal reference electrode is a part of the glasselectrode & it is not the pH sensing element. Only the potential that occurs betweenthe outer surface of the glass bulb & the test solution responds to pH changes.Working: The glass is a partially hydrated aluminosilicate containing sodium orcalcium ions. The hydration of a pH sensitive glass membrane involves an ion-exchange rxn. between singly charged cations in the interstices of the glass lattice &

protons from the solution. The process involves univalent cations exclusively becausedivalent cations are too strongly held within the silicate structure to exchange withions in the solution & hence immobile. The ion-exchange reaction can be written as

H + + Na + Na + + H + Solution glass solution glass

The Na + ions on the glass membrane are exchanged for H + ions in the solution. The

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potential of the electrode is controlled by the difference between the hydrogen ionconcentration inside & outside the thin glass membrane. Since the H + ionconcentration inside the electrode is constant, the electrodes potential varies onwith the concentration of H + in the solution outside. Thus the potential arises from thedifference in positions of ion-exchange equilibrium on each of the two surfaces. Thesurface exposed to the solution having the higher H + concentration becomes positivewith respect to the other surface. This charge difference or potential serves as theanalytical parameter when the pH of the solution on one side of the membrane is heldconstant. Evidently the selectivity of glass electrodes is related both to the ability ofthe various monovalent cations to penetrate into the glass membrane and to the degreeof attraction of the cations to the negative sites within the glass.

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The overall potential of the glass electrode has three components.1) The boundary potential E b, which varies with the pH of the analyte soln. It is madeup of two potentials, E 1 & E 2 which develop at the two surface of the glass membranei.e. the potential developed at the inner glass surface & the potential developed at theouter glass surface.E b = E 1-E2 _______________________ (1)Where E b is the boundary potential

E1 = potential developed at the interface between the exterior of theglass & the analyte soln.

E2 = Potential developed at the interface between the internal solution& the interior of the glass. The boundary potential is related to the concentration ofhydrogen ion in each of the solution by the Nernst-like equation.E b = E 1 E2 = 0.0592 log C l / C 2 _________________ (2)For a glass pH electrode the hydrogen ion concentration of the internal solution is heldconstant. So eqn. (2) becomesE b = K + 0.0592 log C 1 __________________( 3) (Recall pH= -log [H +]Asymmetry potential include the following;

(i) Differing conditions of strain in the two glass surfaces duringmanufacture

(ii) Mechanical abrasion on the on the outer surface during use(iii) Chemical etching of the outer surface during use.

The asymmetry potential changes slowly with time.


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The glass electrode potential can be written in the equation form asEG = E b + E Ag/ AgCl + E asym __________________ (4)Substitution of eqn (3) for E b, givesEG = K + 0.0592 log C 1 + E Ag/AgCl + E asym

= K 0.0592 log pH + E Ag/AgCl + E asym ___ (5)EG = E oG 0.0592 pH __________________ (6)

where E oG = K + E Ag/AgCl + E asym. a combination of three constant terms = constantTo measure the hydrogen ion concentration of the test solution, the glass electrode(indicator electrode) must be combined with an external reference electrode, which isrequired for all kinds of ion-selective electrode determinations.

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The glass electrode has an emf that changes with hydrogen ion concentration, i.e theglass electrode is the most important indicator electrode for hydrogen ion. It is usedfor the measurement of pH under many conditions & normally calomel electrode isused as reference electrode to complete the cell.E cell = E glass E calomel where E glass = the E.P. of glass electrode.

E calomel = the E.P. of the SCEE cell = E oG 0.0592 pH 0.2444The E oG value of a glass electrode can be determined by dipping the glass electrode ina solution of known pH. Typical fields are the clinical & food analysis, environmentalmonitoring (industrial waste acidity of rain) & process control (fermentation, boilerwater, galvanization & precipitation)

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Soln: E o g = 0.0592pH +E cell + E cal. = 0.0592 x 2.8 + 0.24 + 0.2422

=0.648 V pH = E o

g-E

cell E

cal. /0.0592

= 0.648 -0.26-0.2422/0.0592= 2.46


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General rule : The ve half cell has the lower conc of ionsThe +ve half cell has the higher conc of ionsThe CR proceeds in the direction that equalizes the concs in the half-cellsChange in conc. due to current flow- the higher conc. is reduced & the lower conc. isincreased -Eq m is achieved when the two concns. are equal .

e.g. Ag(s)/Ag +(C1)//Ag +(C2)/Ag(s)Zn(s)/Zn 2+(0.1M)//Zn 2+(1.0 M)/Zn(s)


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Working:Anode: Ni (s) Ni 2+[C1] + 2e

(Lower concentration reaction)Cathode: Ni 2+[C2] + 2e Ni (s)

(Higher concentration reaction) Net reaction: Ni 2+[C2] Ni 2+[C 1]There is a change in concentration due to current flow. The higher concentration isreduced and the lower concentration is increased. The equilibrium is attained whenthe two concentration are equal.Emf of the concentration cell :Ecell = E cathode -Eanode

= {E 0 + (2.303RT)logC 2 }- {E 0+(2.303RT)logC 1}At 298 K,Ecell = 0.0592/ n log [C 2/C1]Conclusions:i) When the two solns. are at the same concentrationslog [C 2/C1] = 0, No electricity flowsii) When [C 2/C1] > 1, log [C 2/C1] is positive and E is positive. Electrode in contactwith lower electrolyte concentration acts as anode and electrode in contact with higherelectrolyte concentration acts as cathode. When the cell is in operation, the

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Solution 1:Ecell = 0.0591/n log C 2/C1

0.09 =(0.0591/2) log ( x / 0.001)x =1.097M

Solution 2: Ecell=(2.303 RT/nF) log C 2/C1 0.029 = 2.303RT/n) log (0.01/0.001)0.029 =0.057 x 1/ nn = 0.057/0.029 = 2Valency of mercurous ions is 2, Hg 2 2+

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