Electrochem - Copy.pptx

Galvanic (Voltaic cell) Electrolytic cell
Chemical to electrical Electrical to chemical
Electronic & electrolytic conductors

Galvanic cell – Daniel cell
https://youtu.be/xDITrdbajAs

Representation of galvanic cell.
• Anode Representation:
Zn│Zn2+
(1M) or Zn ; Zn2+
(1M)
Zn │ ZnSO4 (1M) or Zn ; ZnSO4 (1M)
• Cathode Representation:
Cu2+
(1M) │ Cu or Cu2+
(1M) ;Cu
Cu2+
(1M) ; Cu or CuSO4(1M) │ Cu
• Cell Representation:
Zn │ ZnSO4 (1M)║ CuSO4(1M) │Cu

Anode:
• Electrode at which oxidation occurs
• is where electrons are produced
• has a negative sign
Cathode:
* Electrode at which reduction occurs
• is where electrons are consumed
• has a positive sign
Oxdn. Is Losing Electrons, Rdn. Is Gaining Electrons
Anode vs Cathode in Galvanic cell

external circuit
internal circuit
Electrolytic Cell

Electrolytic cell
9
https://youtu.be/HQ9Fhd7P_HA

Comparison between Galvanic and
Electrolytic Cell
Galvanic Cell
• Cell reaction is spontaneous
• Converts Chemical Energy to
Electrical Energy
• Anode is negative terminal and
cathode is positive terminal
• Have two electrodes and two
electrolytes
• Used as portable source of
electric energy.
• Salt bridge is used
• Eg: Daniel Cell
Electrolytic Cell
• Cell reaction is non spontaneous
• Converts Electrical Energy into
Chemical Energy
• Anode is positive terminal and
cathode is negative terminal
• Have two electrodes and single
electrolyte
• Used in Electrolysis apparatus.
• Salt bridge is not used
• Eg : Electroplating.

Comparison…….contd..
•Similarities:
•Involves oxidation at anode & reduction at cathode
• oxidation & reduction in the separate regions
•Electrons flow from anode to cathode in the external
circuit
•Reactions occur on electrode surface only

12
Porous diaphragm Salt bridge

Liquid Junction Potential (LJP)
• 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).
 Same electrolytes of
different concentrations.
With Porous disc

Salt Bridge - Elimination of LJP
A device that permits electrical contact between two solutions,
while preventing direct reaction between the reactants/mixing.
Cell: Zn/Zn2+(0.01M) // Cu2+(0.1M)/Cu

 It maintains electrical neutrality in the two half cells
 It provides electrical contact between the two electrolyte solutions
of a cell
 It minimizes LJP in galvanic cells containing two dissimilar
solutions in contact
Functions of SB
salts used in salt
bridge: potassium nitrate,
potassium chloride, ammonium
nitrate etc.

Origin of single electrode potential
Zinc ions moves into
solution leaving
behind electrons
making it electron rich
Copper ions gets deposited as
copper leaving behind free
negatively charged sulfate ions in
solution makes the electrode
electron poor

The rate of the reaction depends on the
• nature of metal
• temperature
• concentration of the metal ions in the solution
Helmholtz electrical double layer

Measurement of electrode
potential
• It is impossible to determine the absolute half cell potential.
• We can only measure the difference in potential between two
electrodes potentiometrically, by combining them to form a
complete cell.
Platinum
SHE
https://www.google.com/search?q=measuring+single+electrode+potential+of+zinc+electrode+w
hen+copuled+with+SHE&rlz=1C1CHBF_enIN922IN922&source=lnms&tbm=isch&sa=X&ved=2ahU
KEwjb-rD-5MDsAhVD4jgGHQX8AwwQ_AUoAXoECBUQAw&biw=1366&bih=625#imgrc=sQH-
GZEvHbrCjM

Zn electrode
Cu electrode
Measurement of electrode
potential
Ecell = Ecathode- Eanode

Standard electrode potential: E0 is the electrode potential
when electrode is in contact with a solution of unit concentration at
298 K involving pure solids and liquids

Nernst Equation
An expression of a quantitative relationship between electrode
potential/cell potential and concentration of the electrolyte
species in an electro-chemical reaction
Mn+
aq) + ne- M(s)
Vant Hoff reaction isotherm: R = 8.314J/mol/K T in kelvin
F = 96500 C
K
ln
RT
o
G
G 



]
[M
[M]
ln
RT
G
G n
o





]
[M
[M]
ln
RT
nFE
nFE n
o





]
[M
1
ln
nF
RT
E
E n
o



]
[M
1
log
nF
RT
2.303
E
E n
o



K,
T
At 298

]
[M
1
log
n
0.0592
E
E n
o



]
[M
log
n
0.0592
E
E n
o 



Nernst Equation
.
From Nernst equation,
If concentration of solution (Mn+) and temperature is increased, the
electrode potential increases and vice versa.
 
  







anode
at
species
of
Conc
Cathode
at
species
of
Conc
log
nF
RT
303
.
2
E
E 0
cell
cell
For Daniel cell
 
 






 

2
2
A
o
C
0
cell
zn
Cu
log
nF
RT
303
.
2
E
-
E
E

Significance of the Nernst equation
• To calculate the potential of a cell that operates under non-standard
conditions.
• To measure the equilibrium constant for a reaction, when the overall
cell potential for the reaction is zero
At equilibrium the overall cell potential for the reaction is zero. i.e. E=0
Nernst equation,
E=E° - (RT/nF)lnKc
0 = Eo - RT/nF lnKc
Eo = RT/nF lnKc
RTlnKc = nFE°
Kc = e nFE° / RT

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
ECell 3 factors:
*The nature/composition of the electrodes
*Temperature E α T
*Conc. of the electrolyte solns E α C.
Ecell is always +ve
ΔG = - nFE

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.
The emf cannot be measured accurately using ordinary voltmeter
• part of the cell current is drawn to deflect the needle
• part of the emf is used to overcome the internal resistance of the
cell
Measured emf < actual emf of cell

The potentiometric measurement of emf of a cell:
Poggendorff’s compensation method
AB- Potentiometric wire
ES- Standard cell
Ex- unknown cell
G- Galvanometer
J- Sliding contact.
C-Rh – adjustable resistance
S- Storage battery
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 ═ AD
Es AD1
Ex = AD x Es
AD1
+ -
G
J
S
C
Rh
B
A
Ex
Es
D D’

Energetics of Cell Reactions
• Net electrical work performed by the cell reaction of a galvanic cell:
W = QE ------ (1)
Q is the quantity of electrical charge in coulombs produced by the reaction and E is
the emf of the cell in volts
Charge on 1mole of electrons = F Coulombs (96,500 )
When ‘n’ moles of electrons are involved in the cell reaction,
the total charge on ‘n’ moles of electrons = nF coulombs
Q = nF
Substituting for Q , W = nFE
The cell does net work at the expense of the decrease in free energy change (ΔG)
accompanying the cell reaction
i.e., Net electrical work = Decrease in free energy
ΔG = - nFE

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.
Platinum

• 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).
Limitations
 Construction and working is difficult.
 Pt is susceptible for poisoning.
 Cannot be used in the presence of oxidizing agents.

Applications
• Primary reference electrode: To determine electrode potential of
other unknown electrodes
• Electrode potential of Hydrogen Electrode is given as follows:
• H+ + e- 1/2 H2(g)
• E = Eo - 2.303 RT/nF log [H2]1/2/[H+]
E = 0 - 0.0591 log 1/[H+]
E = - 0.0591pH
To determine the pH of a solution. Cell Scheme: Pt,H2,H+
(x)// SHE
• The emf of the cell is determined.
• E (cell) = E (C) – E(A)
E (cell) = 0 – (- 0.0592 pH)
E (cell) = 0.0592 pH
pH = E(cell)/ 0.0592

• Consists of a tube, in the
bottom of which is a layer
of mercury, over which is
placed a paste of mercury
and mercurous chloride.
• Remaining portion of the
cell is filled with a solution
of normal/decinormal/
saturated solution of KCl.
• A platinum wire sealed at its
end fixed into the main tube
dipping into the mercury
layer for electrical contact.
Calomel electrode

• Representation: Hg; Hg2Cl2 / KCl
2Hg → Hg2
2+ + 2e- Hg2
2+ + 2Cl- → Hg2Cl2
As anode: 2Hg + 2Cl- → Hg2Cl2 + 2e-
As Cathode: Hg2Cl2 + 2e- → 2Hg + 2 Cl-
E= Eo - 0.0591 log [Cl-]2 at 298 K
2
Nernst equation E= 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 0.2422/2444 V
Calomel electrode
[Reactant]
[product]
log
n
0.0592
E
E o



To determine the EMF of a cell and pH of a solution.
Pt|H2|H+
(X) ||KCl|Hg2Cl2|Hg
Ecell= 0.2422-(- 0.0592pH)
pH = (Ecell – 0.2422) / 0.0592
Calomel electrode

Advantages
• It is very simple to construct.
• It can be used for a long time without much attention.
• Electrode potential is stable over a long period.
• It has low temperature coefficient of emf.
• It is less prone to contamination.
Disadvantages
• Calomel electrodes should not be used above 50oC .
• Calomel electrode should be used with proper precaution as
mercury compounds are toxic.
Calomel electrode

40
The emf of a cell consisting of a hydrogen and the
normal calomel is 0.664 V at 25 C. Calculate the pH
of the solution containing the hydrogen electrode.
Ecell= Ecal (normal) –( −0.0591pH)
0.664 = 0.2810+0.0591 pH
0.383=0.0591 pH
pH= 6.48

Ion Selective Electrode
The electrode which is sensitive to a specific ion
present in an electrolyte whose potential
depends upon the activity of specific ion in the
electrolyte is called ion selective electrode.
The magnitude of potential of this electrode is an
indicator of the activity of the specific ion in the
electrolyte. Example for this type of electrode is
glass electrode.

The electrode consists of a thin glass membrane
(about 50 micrometer thick), sealed onto one end
of a heavy–walled glass tube.
A special variety of glass (corning 0l5 glass with
approximate composition 20% Na2O, 6% CaO &
72% SiO2) is used
It has low melting point and high electrical
resistance.
The glass bulb is filled with a solution of constant
pH (0.1 M HCl).
42

A silicate glass used for membranes consists of an
infinite 3D- network of SiO4
4- groups
There are sufficient cations to balance the negative
charge of the silicate groups within the interstices
of this structure.
Singly charged cations such as sodium and lithium
are mobile in the lattice and are responsible for
electrical conduction within the membrane.
The glass is a partially hydrated aluminosilicate
containing sodium or calcium ions.
43

The silica glass structure is shaped in such a way
that it allows Na+ ions some mobility. The metal
cations in the hydrated gel diffuse out of the
glass and into the solution while H+ from
solution can diffuse into the hydrated gel.
44

Silver/silver chloride reference electrode which
is connected to one of the terminals of a
potential measuring device.
Note that internal reference electrode is a part
of the glass electrode and it is not the pH
sensing element.
Only the potential that occurs between the
outer surface of the glass bulb and the test
solution responds to pH changes.

Glass electrode
Thin walled glass
membrane
Solution of
unknown pH
Glass Membrane
Outer Inner
E1 E2
HCl, 0.1 mol dm-3(Reference solution of dilute HCl)
Electrical contact
Protective sleeve
Ag/AgCl wire
H+ + Na+ ↔ Na+ + H+
(soln.) (glass) (soln.) (glass)

Schematic diagram of the
structure of glass, which consist of an irregular
network of SiO4 tetrahedra connected through
Oxygen atoms.
Cations are coordinated to the oxygen atoms.

Electrode Potential of glass electrode.
The overall potential of the glass electrode is given by:
Eg = Eb + Eref. + Easy.
It has three components:
 The boundary potential Eb,
 Internal reference electrode potential Eref.
 Asymmetric potential Easy

• 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.
= Eo
g + (RT/nF) ln C1
= Eo
g + 0.0592 log [H+]
Eg = Eo
g – 0.0592 pH

Application of glass electrode
Determination of pH:
Cell: SCE │Test solution ║ GE
E cell = Eg – Ecal.
E cell = Eo
g – 0.0591 pH – 0.2422
pH = Eo
g -Ecell – Ecal. / 0.0591

53
A glass electrode dipped in a soln. of pH = 4 offered an
emf of 0.2066 V with SCE at 298 K. When dipped in a
soln. of unknown pH at the same temperature, the
recorded emf was 0.1076 V. Calculate the pH of the
soln. [ESCE = 0.2412 V].
pH= (Eg
o− Ecell − Ecal(decinormal) ) / 0.0591
4=( Eg
o− 0.2066− 0.3335) /0.0591
Eg
o= 0.7765 V
pH= ( 0.7765− 0.1076−0.3335) / 0.0591
pH= 5.67

Q-1
KCl solution is used to make salt bridge because
• KCl is highly soluble in water
• To increase liquid junction potential
• Mobilities of K+ and Cl- are nearly the same
• Mobility of K+ is greater than Cl-
Which of the following statements is incorrect?
• An external power supply is needed in an electrolytic cell.
• In a galvanic cell difference between the electrode potentials of two
electrodes is responsible for the flow of electrons between the two half
cells.
• The electrode with a higher standard electrode potential will undergo
reduction with respect to other electrode.
• In a galvanic cell, the passage of a current through the electrolytes
drives a redox reaction.
Which of the following factors will not affect appreciably the emf of the cell?
• 1. Nature of the electrode
• 2. Concentration of the electrolyte
• 3. Temperature
• 4. Liquid junction potential 54

Q-2
The EMF of the following cell is found to be 0.20 V at 298 K,
Cd(s)/Cd2+(aq)(?)// Ni2+ (aq) (2.0 M)/Ni (s)
What is the molar concentration of Cd2+ ions in solution?
• 1. 4.00 M -0.25 + 0.40 = 0.15 V
• 2. 0.040 M
• 3. 0.400 M Eo
Cd2+/Cd = − 0.40 V
• 4. 0.004 M Ni2+/ Ni = - 0.25 V
Ans:
55
 
 






R
P
log
nF
RT
303
.
2
-
E
E 0
cell
cell
 
 






2
x
log
2
/
0591
.
0
-
0.15
20
.
0
 
 






2
x
log
2
/
0591
.
0
-
05
.
0
 
log2
-
[logx
0591
.
0
-
1
.
0 
 
log2
-
[logx
692
.
1 

 
[logx
3010
.
0
692
.
1 


 
[logx
391
.
1 

 
M
0.04

X

If E1, E2 and E3 are the emf values of the following three galvanic cells
respectively
Mg|Mg2+(0.02 M)||Cu2+(0.2 M)|Cu
Mg|Mg2+(0.2 M)||Cu2+(0.2 M)|Cu
Mg|Mg2+(0.1 M)||Cu2+(0.2 M)|Cu
Which one of the following is true?
• E3>E2>E1
• E1>E3>E2
• E1>E2>E3
• E2>E3>E1
Q-2
 
 






R
P
log
nF
RT
303
.
2
-
E
E 0
cell
cell
 
 






0.2
0.2
log
0.0295
-
[2.71]
E2
 
(-1)
log
0.0295
-
2.71
E1   
0.0295
2.71
E1 

 
2.7395
E1 
 
  






0.2
0.02
log
0.0591/2
-
(-2.37)]
-
[0.34
E1
 
2.71
E2 
 
 






0.2
0.1
log
0.0295
-
[2.71]
E3
 
(-0.3010)
log
0.0295
-
2.71
E3 
 
2.7188
E3 

57
Q-2
Which among the following combinations of electrodes in 1M concentration of itꞌs
salt solution at 298K produce maximum emf?
(Given Eo of Fe3+ = 0.77 V, Cu2+ = 0.34 V, Sn2+ = -0.14 V and Mg2+ = -2.37 V)
1. Fe3+ and Sn2+
2. Cu2+ and Mg2+
3. Fe3+ and Mg2+
4. Cu2+ and Sn2+
A standard cell has a
1) 0.0V potential at 298K
2) Negligible temp. coeff. Of emf
3) 1.0V potential at 298K
4) high temp. coeff. Of emf
 
0.91
0.14)
(-
-
0.77
E 

 
2.71
2.37)
(-
-
0.34
E 

 
3.14
2.37)
(-
-
0.77
E 

 
0.48
0.14)
(-
-
0.34
E 


Q-2
What is EoNi2+/Ni at 298 K if the emf of Zn(s)/Zn2+(1M)││Ni2+(1M)/Ni(s) is 0.51V and
Zn(s)/Zn2+(1M)││SCE is 1.002V (ESCE = 0.2422V).
1. 0.7568
2. -1.2698
3. 1.2442
4. -0.2498
 
Zn
calomel
cell E
-
E
E 
 
Zn
E
-
0.2422
1.002 
 
0.7598
-
EZn 
Ecell = Ec-EA
0.51 = Ec- (- 0.7598)
0.51 = Ec+ 0.7598
-0.2498 = ENi2+/Ni
ENi2+/Ni = E0Ni2+/Ni -0.059 /2 log 1/1
= -0.2498V

59
The standard free energy change in joules of the following cell at 300 K.
Cu(s) / Cu2+(0.1M)║Ag+(0.25M) / Ag(s); Eocell = 0.46 V
1. -88,780
2. -44,390
3. 88,780
4. 44,390
Q-2
ΔG° = -nFE°
= -2×96500×0.46 = -88,780 J

60
Saturated calomel electrode is called as a secondary reference electrode because
1. Its potential depends on the concentration of KCl used
2. Its potential is fixed as zero volt
3. Its potential is constant and measured with respect to primary reference
electrode
4. Mercurous chloride decompose above 50° C
Calomel electrode is an example of
1. Redox electrode
2. Metal insoluble salt electrode
3. Ion selective electrode
4. Gas electrode
Q-2

61
Calculate the free energy of Cu electrode with Cu2+ concentration 0.015 M.
1. -44816 J
2. -55217 J
3. -65620 J
4. -65260 J
Q-2
E = E° - 0.0591/2 log1/[Cu2+)
= 0.34-0.0591/2 log 1/0.015 = 0.2861 V
ΔG = -nFE = -2×96500×0.2861
= -55217 J
The emf of a cell, Pt | H2 | H+ (x) || KC l |Hg2Cl2 | Hg is 0.83V at 298K.
Then the pH of the unknown solution is
1) 8.9
2) 9.9
3) 4.11
4) 5.02
pH = (Ecell – 0.2422)/0.0592
= (0.83-0.2422)/0.0592
= 9.9

Q3
Calculate the standard emf in volts for a cell
containing Sn2+ / Sn and Br2 / Br - electrodes.
[ Eo ( Sn2+ / Sn) = − 0.14 V, Eo ( Br2 / Br -) = 1.08 V]
E o cell = E o cathode − Eo anode
Because reduction potential of Eo ( Br2 / Br-) is
higher, it is cathodic
E o cell = Eo ( Br2 / Br -) - Eo ( Sn2+ / Sn)
= 1.08 – (− 0.14)
= 1.22 V
62

Q3
Using the electrochemical series, calculate the emf in volts for the cell
Fe(s) | Fe2+(0.1 M) || Cd2+(0.2 M) | Cd at 298 K. Write the cell reactions.
From the series we have;
Eo Cd2+/Cd = − 0.40 V ; Eo Fe2+/Fe = − 0.44 V
At anode Fe →Fe2+ + 2 e−
At Cathode Cd2+ + 2 e− → Cd
Net reaction: Fe + Cd2+→ Fe2+ + Cd
EMF of the cell at 298 K is given by
Eocell = Eocathode − Eo anode
= − 0.40 − (− 0.44)
= 0.04 V
Ecell = Eocell − (0.0591 / n) log [ Fe2+ ] / [Cd2+]
= 0.04 −( 0.0591/ 2 ) log [0.1] / [0.2]
= 0.0488 V
63

Q3
Find the molar concentration of Cd2+ ions in the given electrochemical cell.
Zn / Zn2+ (0.1 M) // Cd2+(M)/ Cd
Given Eo Zn2+/Zn = − 0.76 V; Eo Cd2+/Cd = − 0.40 V ; and Ecell = 0.3305 V at
298 K
Cell representation
Zn│Zn2+(1M)││Ag+(10M)│Ag
Cell reaction:
Zn + 2Ag+ → Zn2+ + 2Ag
Eo
cell = Eo
cathode − Eo
anode
= 0.80 − ( −0.76) = 1.56 V
Ecell = E°cell − (0.0592/ 2) log [1]/[10]2
= 1.56 – (0.0591 /2) log [1] / [10]2
= 1.6192 V
64

Q3
Write the cell reactions and calculate the EMF in volts for the following
cell at 298K.
Mg/ Mg2+ (0.001M) // Cu2+ ( 0.0001M) / Cu .
Given Eo Cu2+/Cu = 0.34 V and Eo Mg2+/Mg = − 2.37V
At anode Mg → Mg2+ + 2 e−
At cathode Cu2+ + 2 e− → Cu
Net reaction Mg + Cu2+ → Mg 2+ + Cu
Ecell = Eocell − (0.0591/n ) log [ Mg2+]/ [Cu2+]
Eocell = Eocathode − Eo anode
= 0.34 – [ −2.37]
= 2.71V
Ecell = 2.71 – (0.0591 /2) log [0.001]/ [0.0001]
= 2.6805 V
65

Q3
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.
∆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
∆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 = nF(δE / δT) P
= 2 x 96,500 x (0-00004) = -7.72JK-1
66

Q4
The emf of a cell consisting of a hydrogen and
the normal calomel is 0.664 V at 25 ºC. Calculate
the pH of the solution containing the hydrogen
electrode.
Ecell= Ecal (normal) – (−0.0591pH)
0.664 = 0.2810+0.0591 pH
0.383=0.0591 pH
pH= 6.48
68

Q4
A glass electrode dipped in a soln. of pH = 4 offered an
emf of 0.2066 V with decinormal calomel electrode at
298 K. When dipped in a soln. of unknown pH at the
same temperature, the recorded emf was 0.1076 V.
Calculate the pH of the soln. [CE = 0.2412 V].
pH= (Eg
o− Ecell − Ecal(decinormal) ) / 0.0591
4=( Eg
o− 0.2066− 0.3335) /0.0591
Eg
o= 0.7765 V
pH= ( 0.7765− 0.1076−0.3335) / 0.0591
pH= 5.67
69

Q4
Write the cell scheme and determine the electrode
potential of zinc immersed in 0.1 M ZnSO4. Given
E.M.F. of cell =1.0022 V and Eo (Calomel electrode)
=0.2422V.
Zn / ZnSO4( 0.1M) // KCl; Hg2Cl2;Hg
Ecell = E cathode – E anode
1.0022 = 0.2422 – E Zn2+ / Zn
E Zn2+ / Zn = 0.2422 – 1.0022
= - 0.76 V
70

Electrochem - Copy.pptx

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