electrochemistry
TRANSCRIPT
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References:1. Engg.Chemistry by Jain and Jain
2. Engg.Chemistry by Dr. R.V.Gadag and Dr. A.Nithyananda Shetty
3. Principles of Physical Chemistry by Puri and Sharma
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• Electrochemistry is a branch of chemistry which
deals with the properties and behavior of
electrolytes in solution and inter-conversion of
chemical and electrical energies.
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• An electrochemical cell can be defined as a single arrangement of two electrodes in one or two electrolytes which converts chemical energy into electrical energy or electrical energy into chemical energy.
• It can be classified into two types: Galvanic Cells. Electrolytic Cells.
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Galvanic Cells:
A galvanic cell is an electrochemical cell that
produces electricity as a result of the
spontaneous reaction occurring inside it.
Galvanic cell generally consists of two
electrodes dipped in two electrolyte solutions
which are separated by a porous diaphragm or
connected through a salt bridge. To illustrate a
typical galvanic cell, we can take the example
of Daniel cell.
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Daniel Cell.
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At the anode: Zn → Zn 2+ + 2e-
At the cathode: Cu 2+ + 2e- → Cu
Net reaction: Zn(s)+Cu 2+ (aq)→ Zn 2+ (aq)+ Cu(s)
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• ELECTROLYTIC CELL
An electrolytic cell is an electro –chemical cell in
which a non- spontaneous reaction is driven by
an external source of current although the
cathode is still the site of reduction, it is now the
negative electrode whereas the anode, the site
of oxidation is positive.
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Representation of galvanic cell.
• Anode Representation:
Zn│Zn2+ or Zn ; Zn2+
Zn │ ZnSO4 (1M) or Zn ; ZnSO4 (1M)
• Cathode Representation:
Cu2+/Cu or Cu2+ ;Cu
Cu2+ (1M) ; Cu or CuSO4(1M)/Cu
• Cell Representation: Zn │ ZnSO4 (1M)║ CuSO4(1M)/Cu
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Liquid Junction Potential.
• Difference between the electric potentials developed in the two solutions across their interface .
Ej = Ø soln, R - Ø soln,L
Eg: *Contact between two different electrolytes (ZnSO4/ CuSO4). *Contact between same electrolyte
of different concentrations(0.1M HCl / 1.0 M HCl).
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Salt Bridge.
• The liquid junction potential can be reduced (to about 1 to 2 mV) by joining the electrolyte compartments through a salt bridge.
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Function Of Salt Bridge.
It provides electrolytic contact between the two electrolyte solutions of a cell.
It avoids or at least reduces junction potential in galvanic cells containing two electrolyte solutions in contact.
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Emf of a cell.
• The difference of potential, which causes a current to flow from the electrode of higher potential to one of lower potential.
Ecell = Ecathode- Eanode
• The E Cell depends on:the nature of the electrodes. temperature.concentration of the electrolyte
solutions.
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• Standard emf of a cell(Eo cell) is defined as the
emf of a cell when the reactants & products of
the cell reaction are at unit concentration or unit
activity, at 298 K and at 1 atmospheric pressure.
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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.
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• The emf of the cell Ex is proportional to the length AD.
Ex α AD
• The emf of the standard cell Es is proportional to the length AD1.
Es α AD1
Ex ═ ADEs AD1
Ex = AD x Es
AD1
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Standard Cell.
It is one which is capable of giving constant and reproducible emf.
It has a negligible temperature coefficient of the emf.
The cell reaction should be reversible.
It should have no liquid junction potential.
Eg: Weston Cadmium Cell. The emf of the cell is 1.0183 V at 293 K and 1.0181 V at 298 K.
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• Weston Cadmium Cell
Sealed wax
CorkSoturated solution of
CdSO4.8/3H2OCdSO4.8/3H2O
crystals
Cd-Hg
12-14% Cd
Paste of Hg2SO4
Mercury, Hg
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• Cell representation:
Cd-Hg/Cd2+// Hg2SO4/Hg
At the anode:
Cd (s) → Cd2+ + 2e-
At the cathode:
Hg2SO4(s) + 2e- → 2 Hg (l)+ SO42-(aq)
Cell reaction:
Cd + Hg22+ → Cd2+ + 2Hg
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Origin of single electrode potential.
• Consider Zn(s)/ ZnSO4
Anodic process: Zn(s) → Zn2+(aq)
Cathodic process: Zn2+(aq) → Zn(s)
At equilibrium: Zn(s) ↔ Zn2+(aq)
Metal has net negative charge and solution has equal positive charge leading to the formation of an Helmholtz electrical layer.
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Single electrode potential.
Electric layer on the metal has a potential Ø (M).
Electric layer on the solution has a potential Ø (aq)
• Electric potential difference between the electric double layer existing across the electrode /electrolyte interface of a single electrode or half cell.
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De-electronation Electronation
Helmholtz double layer
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MEASUREMENT OF ELECTRODE POTENTIAL.
It is not possible to determine experimentally the potential of a single electrode.
It is only the difference of potentials between two electrodes that we can measure by combining them to give a complete cell.
By arbitrarily fixing the potential of reversible hydrogen electrode as zero it is possible to assign numerical values to potentials of the various other electrodes.
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Sign Of Electrode Potential.
The electrode potential of an electrode:
Is positive: If the electrode reaction is reduction when coupled with the standard hydrogen electrode
Is negative: If the electrode reaction is oxidation when coupled with standard hydrogen electrode. According to latest accepted conventions, all single electrode potential values represent reduction tendency of electrodes.
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• when copper electrode is combined with SHE, copper
electrode acts as cathode and undergoes reduction
hydrogen electrode acts as anode.
H2(g) → 2H+ +2e- (oxidation)
Cu2+ +2e- → Cu (reduction)
Hence electrode potential of copper is assigned a
positive sign. Its standard electrode potential is 0.34 V.
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• When zinc is coupled with S.H.E. zinc electrode
acts as anode and hydrogen electrode acts as
cathode.
Zn → Zn2+ +2e-
2H+ + 2e-→ H2.
Hence, electrode potential of zinc is negative.
The standard electrode potential of zinc
electrode is -0.74 V.
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Nernst Equation.
• It is a quantitative relationship between electrode potential and concentration of the electrolyte species.
• Consider a general redox reaction:
Mn+(aq) + ne- → M(s) ----(1)
We know that, ΔG =-nFE ----- (2)
ΔGo=-nFEo-----(3) ΔG =ΔGo +RT ln K
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ΔG =ΔGo +RT ln K
ΔG =ΔGo +RT ln[M]/[Mn+]-----(4)
-nFE= -nFEo + RT ln [M]/[Mn+]----(5)
E= Eo – RT/nF ln 1/[Mn+]------(6)
E=Eo- 2.303 RT/nF log 1/[Mn+]---(7)
At 298K,
E= Eo-0.0592/n log 1/[Mn+]-------(8)
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1. A galvanic cell consists of copper plate
immersed in 10 M solution of CuSO4 and iron
plate immersed in 1M FeSO4 at 298K. If
E0cell=0.78 V, write the cell reaction and calculate
E.M.F. of the cell.
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Solution:
Cell reaction:
Fe + Cu2+ ↔ Fe2+ + Cu
ECell= E0Cell-0.0592/2 log [Fe2+ ]/[Cu2+]
ECell= 0.78 + 0.0296 log 10/1
=0.8096V
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Calculate E.M.F. of the zinc – silver cell at 25˚C
when [Zn2+ ] = 1.0 M and [Ag+] = 10 M
(E0cell=1.56V at 25˚C). Write the cell
representation and cell reaction
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• Solution:• Cell representation• Zn/ Zn2+((1M)//Ag+(10M) /Ag• Cell reaction:
Zn + 2Ag+ ↔ Zn2+ + 2Ag
ECell= E0Cell-0.0592/2 log [Zn2+ ]/[Ag+]2
ECell= 1.56 + 0.0592 log 10/1.0
=1.6192 V
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The emf of the cell
Mg│ Mg 2+ (0.01M)║ Cu 2+ /Cu is measured to
be 2.78 V at 298K. The standard eletrode
potential of magnesium electrode is -2.37
V. Calculate the electrode potential of
copper electrode
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Cell reaction:
Mg + Cu2+ ↔ Mg2+ + Cu
E= Eo-0.0592/n log 1/[Mn+]
EMg= EoMg-0.0592/2 log 1/[Mg2+]
=-2.4291V
Ecell=ECu-EMg
2.78 = ECu-[-2.429]
ECu =2.78-2.429
=0.3509 V
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The emf of the cell
Cu│ Cu 2+ (0.02M)║ Ag+ /Ag is measured to
be 0.46 V at 298K. The standard eletrode
potential of copper electrode is 0.34 V.
Calculate the electrode potential of silver.
electrode
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Energetics of Cell Reactions.
• Net electrical work performed by the cell reaction of a galvanic cell:
W= QE ------(1)
Charge on 1mol electrons is F(96,500)Coulombs.
When n electrons are involved in the cell reaction,
the charge on n mole of electrons = nF
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Q = nF
Substituting for Q in eqn (1)
W = nFE ----------(2)
The cell does net work at the expense of
- ΔG accompanying. ΔG = -nFE
- ΔG = nFE
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From Gibbs – Helmholtz equation.
ΔG = ΔH + T [δ(ΔG)/ δT]P ------- (2)
-nFE = ΔH – nFT (δ E/ δT)P
ΔH = nFT (δ E/ δ T)P – nFE
ΔH = nF[T(δ E/ δT)P –E]
• We know that, [δ (∆G)/ δT]P = - ΔS
ΔS = nF (δE/ δT)P
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Problem: Emf of Weston Cadmium cell is 1.0183 V at 293 K and 1.0l81 V at 298 K.
Calculate ∆G, ΔH and ΔS of the cell reaction at 298 K.
Solution:- ∆G: ∆G = - n FE
n = 2 for the cell reaction; F = 96,500 C E= 1.0181 V at 298 K
∆G = -2 x 96,500 x 1.0181 J = -196.5 KJ
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• ∆H: ∆H = nF [ T (δE /δT)P – E] • (δE/δT)p = 1.0181 – 1.0183 / 298-293 = -
0.0002 / 5• = -0.00004VK-1• T = 298 K• ∆H = 2 x 96,500 { 298 x (-0.00004) –
1.0181)• = -198. 8 KJ• ΔS: ΔS = nF (δE / δT) P• = 2 x 96,500 x (0.00004) = -7.72JK-1
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Classification of Electrodes.
Gas electrode ( Hydrogen electrode).Metal-metal insoluble salt (Calomel
electrode).Ion selective electrode.(Glass electrode).
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Gas electrode.
• It consists of gas bubbling over an inert metal wire or foil immersed in a solution containing ions of the gas.
• Standard hydrogen electrode is the primary reference electrode, whose electrode potential at all temperature is taken as zero arbitrarily.
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Construction.
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• Representation: Pt,H2(g)/ H+
• Electrode reaction: H+ + e- 1/2 H2(g)
The electrode reaction is reversible as it can
undergo either oxidation or reduction depending
on the other half cell.
• If the concentration of the H+ ions is 1M,
pressure of H2 is 1atm at 298K it is called as
standard hydrogen electrode (SHE).
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Applications.
• To determine electrode potential of other unknown electrodes.
• To determine the pH of a solution.
E=Eo- 2.303 RT/nF log [H2]1/2/[H+] = 0 -0.0591 log 1/[H+] = -0.0591pH.
Cell Scheme: Pt,H2,H+(x)// SHE
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• The emf of the cell is determined.
• E (cell) = E (c) – E(A)
= 0 – (- 0.0592 pH)
E (cell) = 0.0592 pH
pH = E(cell)/ 0.0592
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Limitations.
• Constuction and working is difficult.• Pt is susceptible for poisoning.• Cannot be used in the presence of
oxidising agents.
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Metal –metal salt ion electrode.
• These electrodes consist of a metal and a sparingly soluble salt of the same metal dipping in a solution of a soluble salt having the same anion.
Eg: Calomel electrode.
Ag/AgCl electrode.
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Construction.
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• Representation: Hg; Hg2Cl2 / KCl
It can act as anode or cathode depending on the
nature of the other electrode.
• As anode: 2Hg + 2Cl- → Hg2Cl2 + 2e-
• As Cathode: Hg2Cl2 + 2e- → 2Hg + 2 Cl-
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E = Eo – 2.303 RT/2F log [Cl-)2
= Eo -0.0591 log [Cl-] at 298 K
Its electrode potential depends on the concentration of
KCl.
Conc. of Cl- Electrode potential
0.1M 0.3335 V
1.0 M 0.2810 V
Saturated 0.2422 V
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Applications.
• Since the electrode potential is a constant it can be used as a secondary reference electrode.
• To determine electrode potential of other unknown electrodes.
• To determine the pH of a solution.
Pt,H2/H+(X) // KCl,Hg2Cl2,Hg
pH = E(cell) – 0.2422/ 0.0592
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Ion Selective Electrode.
• It is sensitive to a specific ion present in an electrolyte.
• The potential of this depends upon the activity of this ion in the electrolyte.
• Magnitude of potential of this electrode is an indicator of the activity of the specific ion in the electrolyte.*This type of electrode is called indicator electrode.
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• Glass Electrode:`
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Scheme of typical pH glass electrode
1. a sensing part of electrode,2. a bulb made from a specific glass sometimes electrode contain small amount
of AgCl precipitate inside the glass electrode 3 internal solution, usually 0.1M HCl for pH electrodes 4.internal electrode, usually silver chloride electrode or calomel electrode 5.body of electrode, made from non- conductive glass or plastics. 6.reference electrode, usually the same type as 4 7.junction with studied solution, usually made
from ceramics or capillary with asbestos or quartz fiber.
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• The hydration of a pH sensitive glass
membrane involves an ion-exchange reaction
between singly charged cations in the
interstices of the glass lattice and protons
from the solution.
H+ + Na+ Na+ + H+
Soln. glass soln. glass
Eg = Eog – 0.0592 pH
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Electrode Potential of glass electrode.
The overall potential of the glass electrode has three components:
The boundary potential Eb,
Internal reference electrode potential Eref.
Asymetric potential Easy.- due to the difference in response of the inner and outer surface of the glass bulb to changes in [H+].
Eg = Eb + Eref. + Easy.
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• Eb = E1 – E2
= RT/nF ln C1 – RT/nF ln C2
= L + RT/nF ln C1
Eb depends upon [H+]
Eg = Eb + EAg/AgCl + Easy.
= L + RT/nF ln C1 + EAg/AgCl + Easy.
= Eog + RT/nF ln C1
= Eog + 0.0592 log [H+]
Eg = Eog – 0.0592 pH.
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• Advantages:
1. It can be used without interference in solutions containing strong oxidants, strong reductants, proteins, viscos fluids and gases as the glass is chemically robust.
2. It can be used for solutions having pH values 2 to 10. With some special glass (by incorporation of Al2O3 or B2O3) measurements can be extended to pH values up to 12.
3. It is immune to poisoning and is simple to operate
4. The equilibrium is reached quickly & the response is rapid
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5. It can be used for very small quantities of the solutions. Small electrodes can be used for pH measurement in one drop of solution in a tooth cavity or in the sweat of the skin (micro determinations using microelectrodes)
6. If recently calibrated, the glass electrode gives an accurate response.
7. The glass electrode is much more convenient to handle than the inconvenient hydrogen gas electrode.
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Disadvantages:
The bulb of this electrode is very fragile and has to be used with great care.
The alkaline error arises when a glass electrode is employed to measure the pH of solutions having pH values in the 10-12 range or greater. In the presence of alkali ions, the glass surface becomes responsive to both hydrogen and alkali ions. Low pH values arise as a consequence and thus the glass pH electrode gives erroneous results in highly alkaline solutions.
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The acid error results in highly acidic solutions (pH less than zero)Measured pH values are high.
Dehydration of the working surface may cause erratic electrode performance. It is crucial that the pH electrode be sufficiently hydrated before being used. When not in use, the electrode should be stored in an aqueous solution because once it is dehydrated, several hours are required to rehydrate it fully.
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As the glass membrane has a very high electrical resistance (50 to 500 mΩ), the ordinary potentiometer cannot be used for measurement of the potential of the glass electrode. Thus special electronic potentiometers are used which require practically no current for their operation.
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Standardization has to be carried out frequently because asymmetry potential changes gradually with time. Because of an asymmetry potential, not all glass electrodes in a particular assembly have the same value of Eo
G . For this reason, it is best to determine Eo
G for each electrode before use.
The commercial verson is moderately expensive
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Limitations.
The bulb is very fragile and has to be used with great care.
In the presence of alkali ions, the glass surface becomes responsive to both hydrogen and alkali ions. Measured pH values are low.
In highly acidic solutions (pH less than zero) measured pH values are high.
When not in use, the electrode should be stored in an aqueous solution.
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Applications.
• Determination of pH:
Cell: SCE ║Test solution / GE
E cell = Eg – Ecal.
E cell = Eog – 0.0592 pH – 0.2422
pH = Eog -Ecell – Ecal. / 0.0592
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Problems
The cell SCE ΙΙ (0.1M) HCl Ι AgCl(s) /Ag
gave emf of 0.24 V and 0.26 V with buffer having pH value 2.8 and unknown pH value respectively. Calculate the pH value of unknown buffer solution. Given ESCE= 0.2422 V
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Eog= 0.0592pH +Ecell + Ecal.
= 0.0592x2.8 +0.24 + 0.2422
=0.648 V
pH = Eog -Ecell – Ecal. / 0.0592
= 0.648 -0.26-0.2422/0.0592
= 2.46
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CONCENTRATION CELLS.
Two electrodes of the same metal are in contact with solutions of different concentrations.
Emf arises due to the difference in concentrations.
Cell Representation:
M/ Mn+[C1] ║ Mn+/M[C2]
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Construction.
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• At anode: Zn →Zn2+(C1) + 2e-
• At cathode: Zn2+(C2) + 2e-→ Zn
• Ecell = EC-EA
= E0 + (2.303RT/ nF)logC2- [E0+(2.303RT/nF)logC1]
• Ecell = (0.0592/n) log C2/C1
Ecell is positive only if C2 > C1
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• Anode - electrode with lower electrolyte
concentration.
• Cathode – electrode with higher electrolyte
concentration.
• Higher the ratio [C2/C1] higher is the emf.
• Emf becomes zero when [C1] = [C2].
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Problems
• Zn/ZnSO4(0.001M)||ZnSO4(x)/Zn is 0.09V
at 25˚C. Find the concentration of the
unknown solution.
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Ecell = 0.0592/n log C2/C1
0.09 =(0.0592/2) log ( x / 0.001)
x =1.097M
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2. Calculate the valency of mercurous ions
with the help of the following cell.
Hg/ Mercurous || Mercurous /Hg
nitrate (0.001N) nitrate (0.01N) when the emf observed at 18˚ C is 0.029 V
Ecell=(2.303 RT/nF) log C2/C1
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Ecell=(2.303 RT/nF) log C2/C1
0.029 = 2.303RT/n) log (0.01/0.001)
0.029 =0.057 x 1/ n
n = 0.057/0.029 =` 2
Valency of mercurous ions is 2, Hg2 2+
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Assignment
Answer the following questions: 1.Distinguish between electrolytic and galvanic
cells.2.Explain the origin of electrode potential. What are
the sign conventions for electrode potential?3.Give reasons for the following. i) The glass electrode changes its emf over a
period of time. ii) KCl is preferred instead of NaCl as an
electrolyte in the preparation of salt bridge4. What is meant by a standard cell? Give an
example
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5. Quote any four limitations of glass electrode6.Define liquid junction potential. How it can be
eliminated or minimized?7.Derive Nernst equation for the single electrode
potential.8.Describe potentiometric determination of emf of a
cell.9.Writ e construction and working of Calomel
Electrode10.What are concentration cells? Show that emf of
concentration cell becomes zero at a certain point of its working.