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Electrode Potential
For any electrode ⟶ oxidation potential = - Reduction potential
Ecell = R.P of cathode – R.P of anode
Ecell = R.P of cathode + O.P of anode
Ecell is always a + ve quantity & Anode will be electrode of low R.P
Ecell = SRP of cathode – R.P of anode
∎ Greater the SRP value greater will be oxidising power.
Gibbs Free Energy change:
∆G = - nFEcell
∆G 0 = -nFE o cell
Where, n ⟶ no. e
- involved in cell (electrode) reaction
F ⟶ Faraday's constant = 96500 C ≈ (96485 C)
= charge on 1 mole of e
19 23 (1.6 10 C ) 6.022 10 96500 C
− × × × =
Note : 1.E.M.F. is an intensive properties (not dependent on mass or size) so we can not add or
subtract.
E
o value of two reaction to calculate E
o of any other reaction.
- ∆G is an, extensive properties we can add or subtract. ∆G value of two reaction.
Nernst Equation: (Effect of concentration and temp of an emf of cell)
0 ⇒ ∆ G = ∆ G + RT nQ (where Q is reaction quotient)
0 ∆ G = − RT nK eq
0 cell cell
RT
E E nQ nF
0 2.^
cell cell log
RT
E E Q
nF
At 298 K
Ecell Ecell log Q n
= − [At 298 K]
At chemical equilibrium
∆G = 0 ; Ecell = 0.
0
log
cell eq
nE K =
nEcell log Keq n
Note- Nernst equation for an actual reaction
( ) ( )
n M aq ne M s
( )
log
o RP SRP (^) n aq
E E
n M
For an electrode M(s)/M n+ .
96
Summary
1 / /
log [ ]
n n
o M M M M n
RT
E E
nF M
+ =^ + − +
Concentration Cell: A cell in which both the electrods are made up of same material.
For all concentration Cell:
0 Ecell = 0
(a) Electrolyte Concentration Cell:
Concentration cell in which the electrodes are of same material, but they are in contact with the
different
concentration of their ions
eg.
2 2 Zn s ( ) / Zn ( c 1 (^) ) || Zn ( c 2 ) / Zn s ( )
1
log 2
C
E
C
For spontaneous cell reaction Ecell > 0 ⇒ C 1 (^) < C 2
i.e. concentration of anodic compartment < cathodic compartment
(B) Diff erent Types Of Electrodes:
- Metal-Metal ion Electrode ( ) /
n M s M
. ( )
n M ne M s
0.0591 (^1) log[ ]
o n E E M n
= +
- Gas-ion Electrode pt / H (^) 2 ( Patm ) / H ( XM )
as a reduction electrode (^2)
H aq e H Patm
2
1 2
0.0591log [ ]
o H
P
E E
H
- Oxidation-reduction Electrode
2 3 Pt / Fe , Fe
as a reduction electrode
3 2 Fe e Fe
2
3
[ ]
0.0591 log [ ]
o Fe E E Fe
- Metal – Metal insoluble salt electrode e.g. Ag/AgCl, Cl-
as a reduction electrode AgCl s ( ) e Ag s ( ) Cl
− −
0 / / / /
Cl AgCl Ag Cl AgCl Ag
E − = E − −
- 0.0591log[Cl ]
This electrode has a fixed value of reduction potential at a given concentration of anion, hence
can be used as a reference electrode
- Mercury - Mercuric oxide - Hydroxide ion half cell
HgO s ( ) H O 2 ( ) 2 e Hg ( ) 2 OH
− −
/ / / /
[ ]
o OH HgO Hg OH HgO Hg
RT
E E n OH F
− =^ − −^
- Mercury- Mercurous chloride - chloride ion half cell
Hg Cl 2 2 (^) ( ) s 2 e 2 Hg ( ) 2 Cl ( aq )
−
97
w α q w = zq w = Z it Z = electrochemical equivalent of
substance
Second Law: When same quantity of charge is passed through different electrolytes, then the
masses of different substances deposited at respective electrodes will be in the proportion of their
equivalent weights.
w α q
W
constant E
1 2
1 2
W W
E E
When W = E, then change q = 96500 coulomb = 1 Faradays
or No of equivalents of substances deposited or evolved
= No of faradays of charge (used in the electrolysis)
W i t current efficiency factor
E
× ×
Current Efficiency =
actual mass deposited produced
Theoritical mass deposited produced
×
Condition for Simultaneous Deposition of Cu & Fe at Cathode
(^2) / 2 2 / 2
log log 2 2
o o Cu Cu Fe Fe
E E
Cu Fe
+ −^ + =^ + − +
Condition for the simultaneous deposition of Cu & Fe on cathode.
Section (C) – Conductance:
Conductance =
Resistance
Specific conductance of conductivity:
(Reciprocal of specific resistance)
K
= K = specific conductance
Equivalent Conductance: The conductance of all the ions produced one gram equivalent of
an electrolyte in a given solution.
1000 eq
K
Normality
×
= unit: - ohm
2 eq
Molar conductance: The conductance of all the ions produced by ionization of 1 gm mole
of an electrolyte when present in v ml solution.
eq
K
Molarity
×
= unit: - ohm
2 mole
Specific conductance = conductance
a
×
, ℓ = distance between electrodes of conductivity cell.
99
Kohlrausch’s Law:
Variation of λ (^) eq / λ (^) mof a solution with concentration:
(i) Strong electrolyte:
These solution are found to follow debye huckle onsagar equation at low concentrations.
o
λ m λ m b c
∞ = −
(ii) Weak electrolytes:
Kohlrausch Law: λ n λ n λ
∞ ∞ ∞ =^ + + +^ − − Where λ is the molar conductivity
n+ = No. of cations obtained after dissociation per formula unit
n- = No. of anions obtained after dissociation per formula unit
Application of Kohlrausch Law:
- Calculation of
o λ m of weal electrolytes:
( 3 ) ( 3 ) ( ) ( )
o o o o
λ M CH COOH = λ M CH COONa + λ M HCl −λ M NaCl
- To calculate degree of dissociation of a weak electrolyte
0
0
m
m
2
eq
c K
- Solubility (s) of sparingly soluble salt & their Ksp.
m m k^ solubility
∞ = = ×
Ksp = S
2 .
Note: (i) Conductance of mixture of two electrolytes
Ctotal = ∑ Celectrolytes + C (^) water ; K (^) total = ∑ K (^) electrolytes + Kwater
(ii) All the electrolytes will be parallel b/w two electrodes.
(iii) C or K is proportional to concentration of the solution [for any strong electrolyte (100%
dissolvated)]
Ionic Mobility: It is the distance travelled by the ion per second under the potential
gradient of 1 volts per cm. It’s unit is cm 2 S
Absolute ionic mobility:
Ionic mobility at infinite dilution is called absolute ionic mobility and represented by
0
μ c
or
0
μ a
or
Speed of the ion at infinite dilution under unit potential gradient (in cm
2 sec
0 λ c (^) ∝ μ c ;
0 λ a (^) ∝μ a
0 o
λ c = F μ c ;
0 o
λ a = F ×μ a
100
- Weak acid with weak base: CH 3 COOH+NH 4 OH→CH 3 COONH 4 +H 2 O
- Mixture of strong acid and weak acid vs strong or weak base:
(a) (HCl+CH 3 COOH) + NaOH (b) (HCl + CH 3 COOH) + NH 4 OH
102
- How long (approximately) should water be electrolysed by passing through 100 amperes current so
that the oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)
(2018)
(a) 1.6 hours
(b) 6.4 hours
(c) 0.8 hours
(d) 3.2 hours
- When an electric current is passed through acidified water, 112 mL of hydrogen gas at N.T.P. was
collected at the cathode in 965 seconds. The current passed, in ampere, is: (2018)
(a) 0.
(b) 0.
(c) 1.
(d) 2.
- When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount
of p-aminophenol produced is : (2018)
(a) 109.0 g
(b) 98.1 g
(c) 9.81 g
(d) 10.9 g
- Given, 𝐸𝐸𝐶𝐶𝑙𝑙 2 /𝐶𝐶𝑙𝑙^
− ° = 1.36𝑉𝑉, 𝐸𝐸𝐶𝐶𝑟𝑟 3+/𝐶𝐶𝑟𝑟
° = −0.74𝑉𝑉 (2017)
𝐸𝐸 𝐶𝐶𝑟𝑟 2 𝑂𝑂 7 2− /𝐶𝐶𝑟𝑟 3+
° = 1.33, 𝐸𝐸𝑀𝑀𝑀𝑀𝑂𝑂 4
−/𝑀𝑀𝑀𝑀2+
° = 1.51𝑉𝑉
Among the following, the strongest reducing agent is
(a) Cr
(b) Mn 2+
(c) Cr 3+
(d) Cl
103
Practice Questions
- Electrolysis of dilute aqueous NaCI solution was carried out by passing 10 mA current. The time
required to liberate 0.01 mole of H 2 gas at the cathode is (1F = 96500 C mol
(a) 9.65 × 10 4 s
(b) 19.3 × 10 4 s
(c) 28.95 × 10 4 s
(d) 38.6 × 10 4 s
- In the electrolytic cell, flow of electrons is from (2003)
(a) cathode to anode in solution
(b) cathode to anode through external supply
(c) cathode to anode through internal supply
(d) anode to cathode through internal supply
- Standard electrode potential data are useful for understanding the suitability of an oxidant in a
redox titration. Some half-cell reactions and their standard potentials are given below : (2002)
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
3 - 2 4
2 3 2 7
/ MnO /
/ /
3 2
2
Cr Cr Mn
Cr O Cr Cl Cl
E V E V
E V E V
MnO aq H aq e Mn aq H O l E V
Cr O aq H aq e Cr aq H O l E V
Fe aq e Fe aq E V
Cl g e Cl aq E V
− + −
° °
° °
− + − +
− −
Identify the incorrect statement regarding the quantitative estimation of aqueous Fe(NO 3 ) (^2)
(a)
MnO 4 can be used in aqueous HCI
(b)
2- Cr O 2 7 can be used in aqueous HCI
(c)
MnO 4 can be used in aqueous H 2 SO 4
(d)
2- Cr O 2 7 can be used in aqueous H 2 SO 4
105
- Saturated solution of KNO 3 is used to make 'salt-bridge' because (2001) (a)
velocity of K
is greater than that of
NO 3
(b) velocity of
NO 3 is greater than that of K
(c) velocities of both K
and
NO 3 are nearly the same
(d) KNO 3 is highly soluble in water
The gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y
and I MZ - at 25°C. If
the order of reduction potential is Z> Y > X, then (1999)
(a) Y will oxidise X and not Z
(b) Y will oxidise Z and not X
(c) Y will oxidise both X and Z
(d) Y will reduce both X and Z
- The standard reduction potential values of three metallic cations, X, Y, Z are 0.52, — 3.03 and — 1.
V respectively. The order of reducing power of the corresponding metals is ( 1998)
(a) Y > Z > X
(b) X > Y > Z
(c) Z > Y > X
(d) Z > X > Y
- The standard reduction potentials E 0 , for the half reactions are as (1989)
Zn = Zn 2+
Fe = Fe 2+
The emf for the cell reaction, Fe 2+
(a) - 0.35 V
(b) + 0.35 V
(c) +1.17 V
(d) - 1.17 V
106
- Faraday's laws of electrolysis are related to the (1983)
(a) atomic number of the cation
(b) atomic number of the anion
(c) equivalent weight of the electrolyte
(d) speed of the cation
- The standard reduction potentials at 298K for the following half cells are given : (1981)
( ) ( )
( ) ( )
2
3
Zn aq e Zn s E V
Cr aq e Cr s E V
2 H (^) ( aq (^) ) 2 e H (^) 2 ( g (^) ); E 0.000 V
( ) ( )
3 2 Fe aq e Fe aq ; E 0.770 V
Which is the strongest reducing agent?
(a) Zn(s)
(b) Cr(s)
(c) H 2 (g)
(d) Fe(s)
- For the following cell, Zn (s)|ZnSO 4 (aq)) || CuSO 4 (aq )|Cu(s) when the concentration of Zn 2+ is 10
times the concentration of Cu 2+ , the expression for ∆G (in J mol
[F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1V] (2017)
(a)2.303 RT+1.1F
(b) 1.1 F
(c) 2.303 RT – 2.2F
(d) -2.2F
- Galvanisation is applying a coating of (2017)
(a) Cr
108
(b) Cu
(c) Zn
(d) Pb
- Given below are the half-cell reactions (2014)
Mn 2+
2(Mn 3+
The E° for 3Mn 2+ → Mn + 2Mn 3+ will be
(a) - 2.69 V; the reaction will not occur
(b) - 2.69 V; the reaction will occur
(c) - 0.33 V; the reaction will not occur
(d) - 0.33 V; the reaction will occur
- The equivalent conductance of NaCI at concentration C and at infinite dilution are λc and λ∞
respectively. The correct relationship between λc and λ∞ is given as (where, the constant B is positive)
(2014)
(a) λC = λ∞ + (B)C
(b) λC = λ∞ - (B)C
(c) λC = λ∞ - (B) C
(d) λC = λ∞ + (B) C
- Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution of 0.
M solution of same electrolyte is 1.4 S m
- and resistance of same solution of the same electrolyte is 280
Ω. The molar conductivity of 0.5 M solution of the electrolyte in Sm 2 mol
(2014)
(a) 5 x 10
(b) 5 x 10
(c) 5 x 10 3
(d) 5 x 10 2
109
- Consider the following cell reaction,
2Fe(s)+ O 2 (g) + 4H
(aq) → 2Fe 2+ (aq) + 2H 2 O(l), E° = 1.67 V
At [Fe 2+ ] = 10
- M, P(O 2 ) = 0.1 atm and pH = 3, the cell potential at 25°C is (2011)
(a) 1.47 V
(b) 1.77 V
(c) 1.87 V
(d) 1.57 V
- AgNO 3 (aqueous) was added to an aqueous KCl solution gradually and the conductivity of the
solution was measured. The plot of conductance (Λ) versus the volume of AgNO 3 is (2011)
(a) P
111
(b) Q
(c) R
(d) S
- The half cell reactions for rusting of iron are: (2005)
∆G° (in kJ) for the reaction is
(a) – 76
(b) – 322
(c) – 122
(d) – 176
- Zn|Zn 2+ (a = 0.1M) || Fe 2+ (a = 0.01 M) | Fe (2004)
The emf of the above cell is 0.2905 V. Equilibrium constant for the cell reaction is
(a) 10 0.32/0.
(b) 10 0.32/0.
(c) 10 0.26/0.
(d) 10 0.32/0.
- The correct order of equivalent conductance at infinite dilution of LiCI, NaCI and KCl is
(2001)
(a) LiCI > NaCI > KCI
(b) KCl > NaCI > LiCI
(c) NaCl > KCI > LiCl
(d) LiCI > KCl > NaCI
112
- Two aqueous solutions A and B containing solute CuSO 4 and NaBr respectively were electrolysed
using platinum electrodes. The pH of the resulting solutions will show a/an :
(a) Increase in both the solutions
(b) Decrease in both the solutions
(c) Increase in A and decrease in B
(d) Decrease in A and increase in B
- What is the time (in sec) required for depositing all the silver present in 125 mL of 1 M AgNO (^3)
solution by passing a current of 241.25 A? (1 F = 96500 C)
(a) 10
(b) 50
(c) 1000
(d) 100
- The products formed when an aqueous solution of NaBr is electrolysed in a cell having inert
electrodes are:
(a) Na and Br (^2)
(b) Na and O 2
(c) H 2 , Br 2 and NaOH
(d) H 2 and 0 (^2)
- (^3)
0 Al / Al E (^) + = - 1.66 V and Ksp of Al(OH) 3 = 1.0 × 10
- . Reduction potential of the above couple at pH =
14 is:
(a) – 2.31 V
(b) + 2.
(c) – 1.01 V
(d) + 1.01 V
- The standard electrode potential for the reactions,
114
Ag
(aq) + e
Sn 2+ (aq) + 2e
- → Sn(s) at 25°C are 0.80 volt and -0.14 volt, respectively. The emf of the cell Sn|Sn 2+ (
M) || Ag
(1 M)|Ag is:
(a) 0.66 volt
(b) 0.80 volt
(c) 1.08 volt
(d) 0.94 volt
- The standard reduction potential of Cu 2+ / Cu and Cu 2+ / Cu + are 0.337 and 0.153 respectively. The
standard electrode potential of Cu
/ Cu half-cell is:
(a) 0.184 V
(b) 0.827V
(c) 0.521 V
(d) 0.490V
- (^3 2 )
0 0 / /
Fe Fe Fe Fe E (^) + + = + V E (^) + = − V What is (^2)
0 Fe Fe / E (^) + and is Fe
stable to disproportionation in
aqueous solution under standard conditions
(a) + 0.44 V, yes
(b) – 0.44 V, No
(c) + 0.44 V, No
(d) – 0.44 V, yes
- E° for F 2 + 2e
2F^
- is 2.8 V, E° for ½ F 2 + e - = F - is
(a) 2.8 V
(b) 1.4 V
(c) – 2.8 V
(d) – 1.4 V
115
