Class XII Electrochemistry - Nernst equation.Arunesh Gupta
Introduction, application of electrochemistry, metallic conduction & electrolytic conduction, electrolytes, electrochemical cell & electrolytic cell, Galvanic cell (Daniell cell), Standard reduction & oxidation potential, SHE as reference electrode, Standard emf of a cell or standard cell potential, Electrochemical series & its application, Nernst equation, Relationship between (i) Standard cell potential & equilibrium constant (ii) standard cell potential & standard Gibbs energy, some numerical problems.
Class XII Electrochemistry - Nernst equation.Arunesh Gupta
Introduction, application of electrochemistry, metallic conduction & electrolytic conduction, electrolytes, electrochemical cell & electrolytic cell, Galvanic cell (Daniell cell), Standard reduction & oxidation potential, SHE as reference electrode, Standard emf of a cell or standard cell potential, Electrochemical series & its application, Nernst equation, Relationship between (i) Standard cell potential & equilibrium constant (ii) standard cell potential & standard Gibbs energy, some numerical problems.
GROUP THEORY
CONSTRUCTING CHARACTER TABLE IS FOLLOWED BY 4 STEPS through orthogonality rule
STEP 1 : FIND THE NUMBER OF IRRs
Number of IRs = Number of classes.- In C3v
there is 3 classes so Г1,Г2 Г3
STEP 2: FIND OUT THE DIMENSIONS
Sum of the squares of the dimensions of IRRs = Order of the Group
We have to identify a set of 3 positive integers (I1 I2 I3 dimensions of IRRs) which satisfy this condition
The only value of I which satisfy this condition are 1,1,2 so that I12 = I22
SO 3 IRRs of C3v ,two are 1-D and one is 2-D
STEP 3 : FIND character of two 1-D IRRs
In every point group is 1-D IRR who characters are equal to 1 .this IRRs is called totally symmetric IRR
Thus we have
Which satisfy the rule sum of the square of the characters of all operations in any IRR is equal to the order of the group
FIND characters of another 1-D IRRsConditions
All the characters of this IRRs equal to +1 or -1
Also IRR must be Orthogonal to Г1
Г1 has six +1 as characters of the sym operations 1 for E ; 2 (1) for C3 ; 3 (1) for σv
The characters of Г2 is Orthogonal to Г1 so it has three +1 and three -1
For E in 1-D is +1 ; for 2 C3 in 1-D is +1 ; FOR 3 σV is -1
Lesson Plan for Calss XII Chemistry of NCERT Lesson CBSE This lesson plan covers Nucleophilic substitution reaction, SN1 and SN2 reaction,Factors affecting nucleophilic substitution reaction, sterric hindrance,
Aplicando la ecuación de Nernst, calcular el potencial de cada semicélula y la fuerza electromotriz de la pila
Pt | Cr2+ (0,001M), Cr3+(1M) || V3+(0,01M), V2+ (1M) | Pt
Datos: E0 (V3+/V2+) = – 0,26 V; E0(Cr3+/Cr2+) = – 0,41 V
R= 8,314 JK-1 mol-1; F= 96485 Cmol-1; T= 25 ºC = 298 K
IA on effect of temperature of NaOH on the rate of hydrogen production, and f...Lawrence kok
IA on effect of temperature of NaOH on the rate of hydrogen production, and finding Ea for reaction between aluminium and sodium hydroxide measured using a pressure sensor.
GROUP THEORY
CONSTRUCTING CHARACTER TABLE IS FOLLOWED BY 4 STEPS through orthogonality rule
STEP 1 : FIND THE NUMBER OF IRRs
Number of IRs = Number of classes.- In C3v
there is 3 classes so Г1,Г2 Г3
STEP 2: FIND OUT THE DIMENSIONS
Sum of the squares of the dimensions of IRRs = Order of the Group
We have to identify a set of 3 positive integers (I1 I2 I3 dimensions of IRRs) which satisfy this condition
The only value of I which satisfy this condition are 1,1,2 so that I12 = I22
SO 3 IRRs of C3v ,two are 1-D and one is 2-D
STEP 3 : FIND character of two 1-D IRRs
In every point group is 1-D IRR who characters are equal to 1 .this IRRs is called totally symmetric IRR
Thus we have
Which satisfy the rule sum of the square of the characters of all operations in any IRR is equal to the order of the group
FIND characters of another 1-D IRRsConditions
All the characters of this IRRs equal to +1 or -1
Also IRR must be Orthogonal to Г1
Г1 has six +1 as characters of the sym operations 1 for E ; 2 (1) for C3 ; 3 (1) for σv
The characters of Г2 is Orthogonal to Г1 so it has three +1 and three -1
For E in 1-D is +1 ; for 2 C3 in 1-D is +1 ; FOR 3 σV is -1
Lesson Plan for Calss XII Chemistry of NCERT Lesson CBSE This lesson plan covers Nucleophilic substitution reaction, SN1 and SN2 reaction,Factors affecting nucleophilic substitution reaction, sterric hindrance,
Aplicando la ecuación de Nernst, calcular el potencial de cada semicélula y la fuerza electromotriz de la pila
Pt | Cr2+ (0,001M), Cr3+(1M) || V3+(0,01M), V2+ (1M) | Pt
Datos: E0 (V3+/V2+) = – 0,26 V; E0(Cr3+/Cr2+) = – 0,41 V
R= 8,314 JK-1 mol-1; F= 96485 Cmol-1; T= 25 ºC = 298 K
IA on effect of temperature of NaOH on the rate of hydrogen production, and f...Lawrence kok
IA on effect of temperature of NaOH on the rate of hydrogen production, and finding Ea for reaction between aluminium and sodium hydroxide measured using a pressure sensor.
IA on effect of inhibitor concentration copper on enzyme catalase (yeast extr...Lawrence kok
IA on effect of inhibitor concentration copper on enzyme catalase (yeast extract) on the rate of decomposition of H2O2 measured using a pressure sensor.
Master Thesis Total Oxidation Over Cu Based Catalystsalbotamor
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and isothermal reduction with propane and hydrogen.
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transmission mode at the Cu K edge and Ce LIII edge, as well as online mass spectrometry
(MS) at the outlet of the reactor.
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IA on effectiveness of different types of catalysts MnO2 vs Fe(NO3)3 on the rate of decomposition of H2O2 measured using a pressure sensor.
1. Pt/Pd surface
Catalytic Properties of Transition metal
• Variable oxidation state - lose and gain electron easily.
• Use 3d and 4s electrons to form weak bond.
• Act as Homogeneous or Heterogenous catalyst – lower activation energy
• Homogeneous catalyst – catalyst and reactant in same phase/state
• Heterogeneous catalyst – catalyst and reactant in diff phase/state
• Heterogenous catalyst- Metal surface provide active site (lower Ea )
• Surface catalyst bring molecule together (close contact) -bond breaking/making easier
Transition metal as catalyst with diff oxidation states
2H2O2 + Fe2+ → 2H2O+O2+Fe3+
H2O2+Fe2+→H2O + O2 + Fe3+
Fe3+ + I - → Fe2+ + I2
Fe2+ ↔ Fe3+
Rxn slow if only I- is added H2O2 + I- → I2 + H2O + O2
Rxn speed up if Fe2+/Fe3+ added
Fe2+ change to Fe3+ and is change back to Fe2+ again
recycle
molecule adsorp on
surface catalyst
Pt/Pd surface
Bond break
Bond making
3+
CH2 = CH2 + H2 → CH3 - CH3
Nickel catalyst
Without
catalyst, Ea
CH2= CH2 + H2 CH3 - CH3
Surface of catalyst for adsorption
With catalyst, Ea
adsorption
H2
adsorption
C2H4
bond breaking
making
desorption
C2H6
Fe2+ catalyst
How catalyst work ?
Activation energy
Comparing MnO2 vs Fe(NO3)3 on the rate of decomposition of H2O2 measured using a pressure sensor.
2. Across period
Cr - 4s13d5
• half filled more stable
Cu - 4s13d10
• fully filled more stable
Ca
4s2
K
4s1
Transition metal have partially fill 3d orbital
• 3d and 4s electron can be lost easily
• electron fill from 4s first then 3d
• electron lost from 4s first then 3d
• 3d and 4s energy level close together (similar in energy)
Filling electron- 4s level lower, fill first Losing electron- 4s higher, lose first
3d
4s
Comparing MnO2 vs Fe(NO3)3 on the rate of decomposition of H2O2 measured using a pressure sensor.
3. Solid vs solution catalyst will be compared. Assuming same surface area
Fe3+ solution and MnO2 solid
Same amount were used – 0.00005mol
5% H2O2 used. Pressure sensor to measure O2 released.
Diff temp used to find Ea
Reaction mechanism
Procedure:
0.00005mol of each catalyst added to H2O2
H2O2 and Fe(NO3)3 solution immersed in water bath at various temp.
Diff temp used (17, 30, 35, 38, 42C)
Mass of MnO2 is 0.00435g – (0.00005mol)
Ex: 1g of Fe(NO3)3 added to 100ml water – conc is = 0.0247M
To transfer 0.00005mol Fe3+ to H2O2, the vol needed will be 2ml
2ml Fe(NO3)3 was added to 1ml 5% H2O2 in a boiling tube
Pressure sensor attached.
1. Comparing homogenous solution (diff transition metal) against solid MnO2
2. Which transition metal works best (same amt of catalyst added, 0.0005mol)
3. Measure Ea value for diff transition metal and compared to MnO2 which is 41kJmol-1
4. Will Ea higher/lower for heterogenous catalyst (MnO2) compared to homogenous catalyst
like CuSO4, FeSO4, FeCI3
Research Questions
Hydrogen peroxide decomposition – O2 production
2H2O2→ 2H2O + O2
Fe3+ and MnO2
Comparing MnO2 vs Fe(NO3)3 on the rate of decomposition of H2O2 measured using a pressure sensor.
4. Expt done at diff temp for Fe(NO3)3
Slope/gradient taken over 15s
Rate decomposition increases exponential with temp.
Temp/
C
Temp/
K
Rate
kPa/s
1/T k ln k
17 290 0.06074 0.00344 0.06074 -2.801
30 303 0.5423 0.00330 0.5423 -0.611
35 308 1.277 0.00324 1.277 0.2445
38 311 1.505 0.00321 1.505 0.409
42 315 2.588 0.00317 2.588 0.951
Assuming rate constant, k. = Rate of decomposition.
y = 0.0051e0.1518x
R² = 0.989
0
0.5
1
1.5
2
2.5
3
3.5
0 10 20 30 40 50
Rate
decomposition
temp/C
Rate decomposition vs temp/C
Comparing MnO2 vs Fe(NO3)3 on the rate of decomposition of H2O2 measured using a pressure sensor.
Fe(NO3)3 catalyst at diff temp
5. Temp/
C
Temp/
K
Rate
kPa/s
1/T k ln k
17 290 0.06074 0.00344 0.06074 -2.801
30 303 0.5423 0.00330 0.5423 -0.611
35 308 1.277 0.00324 1.277 0.2445
38 311 1.505 0.00321 1.505 0.409
42 315 2.588 0.00317 2.588 0.951
Arrhenius Eqn
Ea from its gradient
Arrhenius Eqn - Ea by graphical Method
RT
Ea
e
A
k
.
.
T
R
E
A
k a 1
ln
ln
Plot ln k vs 1/T
ln both sides
-Ea/R
Gradient = -Ea/R
Gradient = - 13953
-13953 = -Ea/R
Ea = 13953 x 8.314
= 116 kJmol-1
ln k
1/T
Arrhenius plot to find Ea for Fe(NO3)3
y = -13953x + 45.294
R² = 0.9913
-3
-2.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
0.00315 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345 0.0035
lnk
1/T
lnk vs 1/T
Assuming rate constant, k. = Rate of decomposition.
6. Expt done at diff temp for MnO2 Slope/gradient taken over 15s
MnO2 catalyst at diff temp.
Rate decomposition increases exponential with temp.
Temp/
C
Temp/
K
Rate
kPa/s
1/T k ln k
17 290 0.4098 0.00344 0.4098 -0.892
30 303 0.7466 0.00330 0.7466 -0.2922
35 308 0.8274 0.00324 0.8274 -0.1894
38 311 1.353 0.00321 1.353 0.3023
42 315 1.445 0.00317 1.445 0.368
Assuming rate constant, k. = Rate of decomposition.
y = 0.1644e0.0514x
R² = 0.9426
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
0 10 20 30 40 50
Rate
decomposition
temp/C
Rate decomposition vs temp/C
Comparing MnO2 vs Fe(NO3)3 on the rate of decomposition of H2O2 measured using a pressure sensor.
7. Arrhenius Eqn
Ea from its gradient
Arrhenius Eqn - Ea by graphical Method
RT
Ea
e
A
k
.
.
T
R
E
A
k a 1
ln
ln
Plot ln k vs 1/T
ln both sides
-Ea/R
Gradient = -Ea/R
Gradient = - 4703
-4703 = -Ea/R
Ea = 4703 x 8.314
= 39 kJmol-1
ln k
1/T
Arrhenius plot to find Ea for MnO2
Temp/
C
Temp/
K
Rate
kPa/s
1/T k ln k
17 290 0.4098 0.00344 0.4098 -0.892
30 303 0.7466 0.00330 0.7466 -0.2922
35 308 0.8274 0.00324 0.8274 -0.1894
38 311 1.353 0.00321 1.353 0.3023
42 315 1.445 0.00317 1.445 0.368
y = -4703.1x + 15.248
R² = 0.9375
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.00315 0.0032 0.00325 0.0033 0.00335 0.0034 0.00345 0.0035
lnk
1/T
lnk vs 1/T
Assuming rate constant, k. = Rate of decomposition.
0
20
40
60
80
100
120
140
Fe(NO3)3 MnO2
Activation
energy
Type of catalyst
Type of catalyst vs activation energy
Comparison between solid catalyst (MnO2) vs solution (FeNO3)3
Lit value Ea.for MnO2 = (41 kJmol-1).
% 𝐞𝐫𝐫𝐨𝐫 =
(𝐋𝐢𝐭 𝐯𝐚𝐥𝐮𝐞−𝐞𝐱𝐩𝐭 𝐯𝐚𝐥𝐮𝐞)
𝐋𝐢𝐭 𝐯𝐚𝐥𝐮𝐞
x 100%
% error =
𝟒𝟏−𝟑𝟗
𝟒𝟏
𝐱 𝟏𝟎𝟎% = 𝟓%
MnO2 system has a lower Ea compared to Fe(NO3)3 system.
8. Method 1 Method 2
Time Time
Volume Pressure
• Rate = Δ vol O2 over time
• Volume recorded
• Rate = Δ pressure O2 over time
• Pressure recorded
Procedure
2H2O2 → O2 + 2H2O
Rxn: H2O2 with diff (catalyst) measured using TWO diff methods
• 2H2O2 → O2 + 2H2O
(H2O2 limiting, KI excess)
• Pipette 1ml 1.0M KI to 20ml of 1.5% H2O2
• Vol O2 released recorded at 1 min interval
• Repeated using 3% H2O2 conc
Time/m Vol O2
(H2O2 1.5%)
Vol O2
(H2O2 3.0%)
0 0.0 0.0
1 8.5 14.0
2 15.0 26.5
3 21.0 34.0
4 26.0 39.0
Volume O2
Time
3 %
1.5 %
Comparing MnO2 vs Fe(NO3)3 on the rate of decomposition of H2O2 measured using a pressure sensor.
9. • 2H2O2 → O2 + 2H2O
(H2O2 limiting, KI excess)
• Pipette 1ml 1.0M KI to 20ml of 1.5% H2O2
• Pressure O2 released recorded at 1 min interval
• Repeat using 3% H2O2 conc
Method 1 Method 2
Time Time
Volume Pressure
• Rate = Δ vol O2 over time
• Volume recorded
• Rate = Δ pressure O2 over time
• Pressure recorded
Procedure
2H2O2 → O2 + 2H2O
Time
3 %
1.5 %
Time/m Pressure O2
(H2O2 1.5%)
Pressure O2
(H2O2 3%)
0 101.3 101.3
1 102.4 103.4
2 103.5 105.6
3 110.3 115.2
4 113.5 118.2
Pressure O2
Comparing MnO2 vs Fe(NO3)3 on the rate of decomposition of H2O2 measured using a pressure sensor.