All Exams >
Chemistry >
Physical Chemistry >
All Questions
All questions of Electrochemistry for Chemistry Exam
When a conductance cell was filled with a 0.0025 M solution of K2SO4, its resistance was 326 Ω. If cell constant is 0.2281 cm-1, the specific conductance (Ω-1cm-1) of K2SO4 solution is:- a)4.997 × 10-4
- b)5.997 × 10-4
- c)6.997 × 10-4
- d)7.997 × 10-4
Correct answer is option 'C'. Can you explain this answer?
When a conductance cell was filled with a 0.0025 M solution of K2SO4, its resistance was 326 Ω. If cell constant is 0.2281 cm-1, the specific conductance (Ω-1cm-1) of K2SO4 solution is:
a)
4.997 × 10-4
b)
5.997 × 10-4
c)
6.997 × 10-4
d)
7.997 × 10-4
|
Pooja Choudhury answered |
The standard reduction potential at 298 K for the following half cells are given:
Which is the strongest reducing agent:- a)Zn(s)
- b)Cr(s)
- c)H2(g)
- d)Fe2+ (aq)
Correct answer is option 'A'. Can you explain this answer?
The standard reduction potential at 298 K for the following half cells are given:
Which is the strongest reducing agent:
a)
Zn(s)
b)
Cr(s)
c)
H2(g)
d)
Fe2+ (aq)
|
|
Pooja Choudhury answered |
Zn has minimum reduction potential it means Zn is strong reducing agent. Hence A is correct.
The unit of molar conductivity is:- a)Ω-1
- b)S m-1 mol−1
- c)m2 mol−1
- d)S m2 mol−1
Correct answer is option 'D'. Can you explain this answer?
The unit of molar conductivity is:
a)
Ω-1
b)
S m-1 mol−1
c)
m2 mol−1
d)
S m2 mol−1
|
|
Pooja Choudhury answered |
- The molar conductivity of an electrolytic solution is the conductance of the volume of the solution containing a unit mole of electrolyte that is placed between two electrodes of unit area cross-section or at a distance of one-centimeter apart.
- The unit of molar conductivity is S⋅m2⋅mol-1.
The cell potential for the following electrochemical system at 25°C is:Al(s) | Al3+ (0.01 M) || Fe2+ (0.1 M) | Fe (s)Given: Standard reduction potential of Al3+ + 3e– → Al is –1.66 V at 25°C
Standard reduction potential of Fe2+ + 2e– → Fe is –0.44 V at 25°C- a)1.23 V
- b)1.21 V
- c)1.22 V
- d)–2.10 V
Correct answer is option 'A'. Can you explain this answer?
The cell potential for the following electrochemical system at 25°C is:
Al(s) | Al3+ (0.01 M) || Fe2+ (0.1 M) | Fe (s)
Given: Standard reduction potential of Al3+ + 3e– → Al is –1.66 V at 25°C
Standard reduction potential of Fe2+ + 2e– → Fe is –0.44 V at 25°C
Standard reduction potential of Fe2+ + 2e– → Fe is –0.44 V at 25°C
a)
1.23 V
b)
1.21 V
c)
1.22 V
d)
–2.10 V
|
Raksha Pillai answered |
Solution:
The cell potential (Ecell) of an electrochemical cell is given by the difference between the standard reduction potentials of the half-cells.
Ecell = E°(reduction at cathode) - E°(reduction at anode)
Given: E°(Al3+ + 3e- → Al) = 1.66 V
E°(Fe2+ + 2e- → Fe) = 0.44 V
We need to determine the cell potential for the following electrochemical system at 25C:
Al(s) | Al3+ (0.01 M) || Fe2+ (0.1 M) | Fe(s)
The cell diagram represents the two half-cells in the electrochemical cell. The two vertical lines represent the phase boundary between the two half-cells, and the double vertical line represents the salt bridge.
The half-cell reactions for the given electrochemical system are:
Al3+ + 3e- → Al (reduction at cathode)
Fe2+ → Fe + 2e- (oxidation at anode)
Step 1: Write the balanced cell reaction:
Al(s) + Fe2+ (0.1 M) → Al3+ (0.01 M) + Fe(s)
Step 2: Determine the reduction potential for the cathode:
E°(Al3+ + 3e- → Al) = 1.66 V
Step 3: Determine the oxidation potential for the anode:
E°(Fe2+ → Fe + 2e-) = -0.44 V (Note: The oxidation potential is the negative of the reduction potential.)
Step 4: Calculate the cell potential:
Ecell = E°(reduction at cathode) - E°(reduction at anode)
Ecell = 1.66 V - (-0.44 V)
Ecell = 2.10 V
However, this value is incorrect as the reduction potential for iron in the given electrochemical system is not standard. We need to use the Nernst equation to calculate the actual cell potential.
Step 5: Calculate the actual cell potential using the Nernst equation:
Ecell = E° - (RT/nF) ln(Q)
where,
E° = standard cell potential
R = gas constant = 8.314 J/mol K
T = temperature in Kelvin
n = number of electrons transferred
F = Faraday constant = 96,485 C/mol
Q = reaction quotient
At equilibrium, Q = K, the equilibrium constant.
K = [Al3+] / [Fe2+]
At 25°C, the equilibrium constant for the reaction is:
K = [Al3+] / [Fe2+]
K = (0.01 M) / (0.1 M)
K = 0.1
Substituting the values in the Nernst equation:
Ecell = E° - (RT/nF) ln(Q)
Ecell = 1.66 V - (0.0257 V) ln(0.1)
Ecell = 1.23 V
Therefore, the cell potential for the given electrochemical system at 25°C is 1.23 V.
Answer: Option (a) 1.23 V.
The cell potential (Ecell) of an electrochemical cell is given by the difference between the standard reduction potentials of the half-cells.
Ecell = E°(reduction at cathode) - E°(reduction at anode)
Given: E°(Al3+ + 3e- → Al) = 1.66 V
E°(Fe2+ + 2e- → Fe) = 0.44 V
We need to determine the cell potential for the following electrochemical system at 25C:
Al(s) | Al3+ (0.01 M) || Fe2+ (0.1 M) | Fe(s)
The cell diagram represents the two half-cells in the electrochemical cell. The two vertical lines represent the phase boundary between the two half-cells, and the double vertical line represents the salt bridge.
The half-cell reactions for the given electrochemical system are:
Al3+ + 3e- → Al (reduction at cathode)
Fe2+ → Fe + 2e- (oxidation at anode)
Step 1: Write the balanced cell reaction:
Al(s) + Fe2+ (0.1 M) → Al3+ (0.01 M) + Fe(s)
Step 2: Determine the reduction potential for the cathode:
E°(Al3+ + 3e- → Al) = 1.66 V
Step 3: Determine the oxidation potential for the anode:
E°(Fe2+ → Fe + 2e-) = -0.44 V (Note: The oxidation potential is the negative of the reduction potential.)
Step 4: Calculate the cell potential:
Ecell = E°(reduction at cathode) - E°(reduction at anode)
Ecell = 1.66 V - (-0.44 V)
Ecell = 2.10 V
However, this value is incorrect as the reduction potential for iron in the given electrochemical system is not standard. We need to use the Nernst equation to calculate the actual cell potential.
Step 5: Calculate the actual cell potential using the Nernst equation:
Ecell = E° - (RT/nF) ln(Q)
where,
E° = standard cell potential
R = gas constant = 8.314 J/mol K
T = temperature in Kelvin
n = number of electrons transferred
F = Faraday constant = 96,485 C/mol
Q = reaction quotient
At equilibrium, Q = K, the equilibrium constant.
K = [Al3+] / [Fe2+]
At 25°C, the equilibrium constant for the reaction is:
K = [Al3+] / [Fe2+]
K = (0.01 M) / (0.1 M)
K = 0.1
Substituting the values in the Nernst equation:
Ecell = E° - (RT/nF) ln(Q)
Ecell = 1.66 V - (0.0257 V) ln(0.1)
Ecell = 1.23 V
Therefore, the cell potential for the given electrochemical system at 25°C is 1.23 V.
Answer: Option (a) 1.23 V.
In the electrolytic cell, flow of electrons is from:- 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
Correct answer is option 'C'. Can you explain this answer?
In the electrolytic cell, flow of electrons is from:
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
|
Edurev.iitjam answered |
Flow of Electrons in an Electrolytic Cell
In an electrolytic cell, the flow of electrons is from: Anode to Cathode through Internal Supply
This means that option B is correct. Here's why:
In an electrolytic cell, the flow of electrons is from: Anode to Cathode through Internal Supply
This means that option B is correct. Here's why:
- Anode: This is where oxidation takes place in an electrolytic cell. During oxidation, a substance loses electrons. This means that the anode is the source of electrons.
- Cathode: This is where reduction takes place in an electrolytic cell. During reduction, a substance gains electrons, meaning that the cathode is where electrons are received.
- Flow of electrons: Since electrons are produced at the anode (through oxidation) and consumed at the cathode (through reduction), the flow of electrons is from the anode to the cathode.
- Internal supply: In an electrolytic cell, the power supply is connected to the anode and cathode, creating an electric current within the cell. This current forces the electrons to move from the anode to the cathode, hence the term "through internal supply".
In summary, in an electrolytic cell, the flow of electrons is from anode to cathode through internal supply as a result of the oxidation and reduction reactions taking place at the anode and cathode, respectively.
The standard reduction potentials E°, for the half reactions are as:
The emf for the cell reaction,
- a)–0.35 V
- b)+0.35 V
- c)+1.17 V
- d)–1.17 V
Correct answer is option 'B'. Can you explain this answer?
The standard reduction potentials E°, for the half reactions are as:
The emf for the cell reaction,
a)
–0.35 V
b)
+0.35 V
c)
+1.17 V
d)
–1.17 V
|
Asf Institute answered |
Fe2++2e−⟶Fe; 20mmE∘=−0.41V
20mm Zn⟶Zn2++2e−; E∘=+0.76V
⇒Fe2++Zn⟶Zn2++Fe; E∘=+0.35V
20mm Zn⟶Zn2++2e−; E∘=+0.76V
⇒Fe2++Zn⟶Zn2++Fe; E∘=+0.35V
Which of the following expressions is correct:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
Which of the following expressions is correct:
a)
b)
c)
d)
|
|
Pooja Choudhury answered |
According to kholaraus law,
NH4OH = NH4 + OH
i.e. NH4 + Cl + Na + OH – Na – Cl = NH4Cl + NaOH - NaCl
NH4OH = NH4 + OH
i.e. NH4 + Cl + Na + OH – Na – Cl = NH4Cl + NaOH - NaCl
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 g. The valencies of A, B and C respectively:- a)3, 1 and 2
- b)1, 3 and 2
- c)3, 1 and 3
- d)2, 3 and 2
Correct answer is option 'B'. Can you explain this answer?
Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolyzed under identical condition using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the weights of B and C deposited were 2.7 g. The valencies of A, B and C respectively:
a)
3, 1 and 2
b)
1, 3 and 2
c)
3, 1 and 3
d)
2, 3 and 2
|
|
Pooja Choudhury answered |
According to faraday's law:
W = ZQ = EQ/96500
For A:
2.1=(7/x)Q/96500
For B:
2.7=(27/y)Q/96500
For C:
7.2=(48/z)Q/96500
x: y:z ::(7/2.1):(27/2.7):(48/7.2)
W = ZQ = EQ/96500
For A:
2.1=(7/x)Q/96500
For B:
2.7=(27/y)Q/96500
For C:
7.2=(48/z)Q/96500
x: y:z ::(7/2.1):(27/2.7):(48/7.2)
= 3.33:10:6.66
=1:3:2
by solving these equations,
x=1, y=3, z=2
by solving these equations,
x=1, y=3, z=2
Galvanization is applying a coating of:- a)Pb
- b)Cr
- c)Cu
- d)Zn
Correct answer is option 'D'. Can you explain this answer?
Galvanization is applying a coating of:
a)
Pb
b)
Cr
c)
Cu
d)
Zn
|
Akash Kulkarni answered |
Galvanization is the process of coating a metal object with a layer of zinc to protect it from corrosion. The correct answer is option 'D' which is Zinc.
Zinc as a coating material:
Zinc is a highly reactive metal and has a strong affinity towards oxygen. When exposed to air, it reacts with oxygen to form a thin layer of zinc oxide on its surface. This layer acts as a barrier between the metal and the air, preventing further oxidation. Zinc is also very ductile and can be easily shaped and molded to fit any object. These properties make it an ideal coating material for metal objects.
Galvanization process:
The galvanization process involves coating a metal object with a layer of zinc to protect it from corrosion. The process can be carried out using one of two methods: hot-dip galvanizing or electroplating.
Hot-dip galvanizing:
In hot-dip galvanizing, the metal object is first cleaned and then dipped into a bath of molten zinc. The high temperature of the zinc bath causes the zinc to react with the surface of the metal, forming a layer of zinc-iron alloy. The object is then removed from the bath and allowed to cool, forming a layer of pure zinc on its surface.
Electroplating:
In electroplating, the metal object is first cleaned and then placed in a solution containing zinc ions. A direct current is then passed through the solution, causing the zinc ions to be deposited onto the surface of the metal object. The object is then removed from the solution and rinsed to remove any excess zinc.
Advantages of galvanization:
Galvanization provides several benefits, including:
1. Corrosion resistance: Zinc is a highly corrosion-resistant material that protects the underlying metal from rust and other forms of corrosion.
2. Longevity: Galvanized objects have a long lifespan and can last for decades without needing to be replaced.
3. Low maintenance: Galvanized objects require very little maintenance, making them a cost-effective choice for many applications.
Conclusion:
In conclusion, galvanization is the process of coating a metal object with a layer of zinc to protect it from corrosion. Zinc is an ideal coating material due to its high reactivity, ability to form a protective oxide layer, and ductility. Galvanization provides several benefits, including corrosion resistance, longevity, and low maintenance.
Zinc as a coating material:
Zinc is a highly reactive metal and has a strong affinity towards oxygen. When exposed to air, it reacts with oxygen to form a thin layer of zinc oxide on its surface. This layer acts as a barrier between the metal and the air, preventing further oxidation. Zinc is also very ductile and can be easily shaped and molded to fit any object. These properties make it an ideal coating material for metal objects.
Galvanization process:
The galvanization process involves coating a metal object with a layer of zinc to protect it from corrosion. The process can be carried out using one of two methods: hot-dip galvanizing or electroplating.
Hot-dip galvanizing:
In hot-dip galvanizing, the metal object is first cleaned and then dipped into a bath of molten zinc. The high temperature of the zinc bath causes the zinc to react with the surface of the metal, forming a layer of zinc-iron alloy. The object is then removed from the bath and allowed to cool, forming a layer of pure zinc on its surface.
Electroplating:
In electroplating, the metal object is first cleaned and then placed in a solution containing zinc ions. A direct current is then passed through the solution, causing the zinc ions to be deposited onto the surface of the metal object. The object is then removed from the solution and rinsed to remove any excess zinc.
Advantages of galvanization:
Galvanization provides several benefits, including:
1. Corrosion resistance: Zinc is a highly corrosion-resistant material that protects the underlying metal from rust and other forms of corrosion.
2. Longevity: Galvanized objects have a long lifespan and can last for decades without needing to be replaced.
3. Low maintenance: Galvanized objects require very little maintenance, making them a cost-effective choice for many applications.
Conclusion:
In conclusion, galvanization is the process of coating a metal object with a layer of zinc to protect it from corrosion. Zinc is an ideal coating material due to its high reactivity, ability to form a protective oxide layer, and ductility. Galvanization provides several benefits, including corrosion resistance, longevity, and low maintenance.
The conductance at infinite dilution follow the order- a)Li+ > Na+ > K+
- b)Na+ > Li+ > K+
- c)K+ > Li+ > Na+
- d)K+ > Na+ > Li+
Correct answer is option 'D'. Can you explain this answer?
The conductance at infinite dilution follow the order
a)
Li+ > Na+ > K+
b)
Na+ > Li+ > K+
c)
K+ > Li+ > Na+
d)
K+ > Na+ > Li+
|
|
Pooja Choudhury answered |
- Smaller the ion, the greater will be the solvation, and hence lower will be the molar conductance.
An electrochemical cell consists of two half-cell reactions.
The mass of copper (in grams) dissolved on passing 0.5 A current for 1h is [Given, atomic mass of Cu is 63.6, F = 96500 C mol–1]- a)0.88
- b)1.18
- c)0.29
- d)0.59
Correct answer is option 'D'. Can you explain this answer?
An electrochemical cell consists of two half-cell reactions.
The mass of copper (in grams) dissolved on passing 0.5 A current for 1h is [Given, atomic mass of Cu is 63.6, F = 96500 C mol–1]
a)
0.88
b)
1.18
c)
0.29
d)
0.59
|
Veda Institute answered |
m/E = Q/F
m/(63.6/2) = (0.5 x 60 x 60)/96500
m = (1800 x 63.6)/(96500 x 2)
m = 0.59
m/(63.6/2) = (0.5 x 60 x 60)/96500
m = (1800 x 63.6)/(96500 x 2)
m = 0.59
(in S m2 mol–1) by the relation (M is molarity in mol m–3)
Which of the following ion is expected to have least value of molar conductivity at infinite dilution in an aqueous solution:- a)Na+
- b)K+
- c)Rb+
- d)Cs+
Correct answer is option 'A'. Can you explain this answer?
Which of the following ion is expected to have least value of molar conductivity at infinite dilution in an aqueous solution:
a)
Na+
b)
K+
c)
Rb+
d)
Cs+
|
Sagarika Patel answered |
Na+ does not have the lowest molar conductance of all ions. It has lower molar conductance than the ions that are greater in size than Na such as K+, Cs+ as the molar conductance is inversely proportional to solvation of ions. Smaller the ion, greater will be the solvation and hence lower will be the molar conductance.
.
The electrochemical cell shown below is a concentration cell.
M|M2+ (saturated solut ion of a sparingly so luble salt, MX2 || M2+ (0.001 mol dm–3)| M.
The emf of the cell depends on the difference in concentration of M2+ ions at the two electrodes. The emf of the cell at 298 is 0.059 V.Q. The value of ΔG (kJ mol-1) for the given cell is (take 1 F = 96500 C mol-1)- a)–5.7
- b)5.7
- c)11.4
- d)–11.4
Correct answer is option 'D'. Can you explain this answer?
The electrochemical cell shown below is a concentration cell.
M|M2+ (saturated solut ion of a sparingly so luble salt, MX2 || M2+ (0.001 mol dm–3)| M.
The emf of the cell depends on the difference in concentration of M2+ ions at the two electrodes. The emf of the cell at 298 is 0.059 V.
M|M2+ (saturated solut ion of a sparingly so luble salt, MX2 || M2+ (0.001 mol dm–3)| M.
The emf of the cell depends on the difference in concentration of M2+ ions at the two electrodes. The emf of the cell at 298 is 0.059 V.
Q.
The value of ΔG (kJ mol-1) for the given cell is (take 1 F = 96500 C mol-1)
a)
–5.7
b)
5.7
c)
11.4
d)
–11.4
|
|
Pooja Choudhury answered |
At cathode:M+2(aq)+2e−→M(s)
At anode:M(s)+2X−(aq)→MX2(aq)+2e−
n - factor of the cell reaction is 2.
At anode:M(s)+2X−(aq)→MX2(aq)+2e−
n - factor of the cell reaction is 2.
ΔG=−nFEcell
= −2×96500×0.059
= −11.4 kJmole−1
Molar conductivity is defined as:Where k = conductivity, G = conductance, l = distance between two electrodes and Vm = volume of solution containing 1 mol of electrolyte- a)
- b)KVm
- c)Gl
- d)Kl2
Correct answer is option 'B'. Can you explain this answer?
Molar conductivity is defined as:
Where k = conductivity, G = conductance, l = distance between two electrodes and Vm = volume of solution containing 1 mol of electrolyte
a)
b)
KVm
c)
Gl
d)
Kl2
|
Vedika Singh answered |
- The molar conductivity of an electrolytic solution is the conductance of the volume of the solution containing a unit mole of electrolyte that is placed between two electrodes of unit area cross-section or at a distance of one-centimeter apart.
Where C = concentration of electrolyte
If x is the specific resistance of the electrolyte solution and y is he molarity of the solution, then 
is given by:
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
If x is the specific resistance of the electrolyte solution and y is he molarity of the solution, then is given by:
is given by:a)
b)
c)
d)
|
|
Shubham Rane answered |
Molar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of electrolyte.
Choose the correct statement:
a)Conductance is an extensive propertyb)Specific conductance is an extensive propertyc)Equivalent conductivity is an intensive property d)Molar conductivity is an extensive propertyCorrect answer is option 'A,C'. Can you explain this answer?
|
Vipin Singh answered |
C should not be ans it is intensive prpt.
The conductivity of N/10 KCl solution at 20°C is 0.0212 S cm–1 and the resistance of cell containing this solution at 20°C is 55 ohm. The cell constant is:
- a)2.173 cm–1
- b)1.166 cm–1
- c)4.616 cm–1
- d)3.324 cm–1
Correct answer is option 'B'. Can you explain this answer?
The conductivity of N/10 KCl solution at 20°C is 0.0212 S cm–1 and the resistance of cell containing this solution at 20°C is 55 ohm. The cell constant is:
a)
2.173 cm–1
b)
1.166 cm–1
c)
4.616 cm–1
d)
3.324 cm–1
|
Athul Chaudhary answered |
Given information:
- Conductivity of N/10 KCl solution at 20°C = 0.0212 S cm^(-1)
- Resistance of cell containing the solution at 20°C = 55 ohm
To find:
Cell constant
Solution:
The conductivity of a solution is given by the equation:
Conductivity = (1 / Resistance) x Cell constant
We are given the conductivity and resistance, so we can rearrange the equation to solve for the cell constant:
Cell constant = Conductivity x Resistance
Substituting the given values into the equation, we get:
Cell constant = 0.0212 S cm^(-1) x 55 ohm
Calculating the product, we find:
Cell constant = 1.166 S cm^(-1)
Therefore, the correct answer is option 'B' (1.166 cm^(-1)).
Summary:
The cell constant is calculated using the equation Cell constant = Conductivity x Resistance. Substituting the given values, we find the cell constant to be 1.166 cm^(-1).
- Conductivity of N/10 KCl solution at 20°C = 0.0212 S cm^(-1)
- Resistance of cell containing the solution at 20°C = 55 ohm
To find:
Cell constant
Solution:
The conductivity of a solution is given by the equation:
Conductivity = (1 / Resistance) x Cell constant
We are given the conductivity and resistance, so we can rearrange the equation to solve for the cell constant:
Cell constant = Conductivity x Resistance
Substituting the given values into the equation, we get:
Cell constant = 0.0212 S cm^(-1) x 55 ohm
Calculating the product, we find:
Cell constant = 1.166 S cm^(-1)
Therefore, the correct answer is option 'B' (1.166 cm^(-1)).
Summary:
The cell constant is calculated using the equation Cell constant = Conductivity x Resistance. Substituting the given values, we find the cell constant to be 1.166 cm^(-1).
According to the Nernst equation, the potential of an electrode changes by 59.2 mV whenever the ratio of the oxidized and the reduced species changed by a factor of 10 at 25°C. What would be the corresponding change in the electrode potential if the experiment is carried out at 30°C:
- a)60.2 mV
- b)71.0 mV
- c)59.2 mV
- d)None of these.
Correct answer is option 'C'. Can you explain this answer?
According to the Nernst equation, the potential of an electrode changes by 59.2 mV whenever the ratio of the oxidized and the reduced species changed by a factor of 10 at 25°C. What would be the corresponding change in the electrode potential if the experiment is carried out at 30°C:
a)
60.2 mV
b)
71.0 mV
c)
59.2 mV
d)
None of these.
|
|
Edurev.iitjam answered |
The cell constant of a conductivity cell is defined as:Where l = distance between electrodes and A = area of cross-section of each of the electrodes- a)
- b)K = lA
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The cell constant of a conductivity cell is defined as:
Where l = distance between electrodes and A = area of cross-section of each of the electrodes
a)
b)
K = lA
c)
d)
|
|
Asf Institute answered |
- The cell constant is a multiplier constant specific to a conductivity sensor.
- The measured current is multiplied by the cell constant to determine the electrical conductivity of the solution.
- The cell constant, known as K, refers to a theoretical electrode consisting of two 1 cm square plates 1 cm apart.
- A cell constant has units of 1/cm (per centimeter), where the number refers to the ratio of the distance between the electrode plates to the surface area of the plate.
Which of the following ion has highest molar conductivity at infinite dilution in an aqueous solution:- a)F–
- b)Cl–
- c)Br–
- d)HO–
Correct answer is option 'D'. Can you explain this answer?
Which of the following ion has highest molar conductivity at infinite dilution in an aqueous solution:
a)
F–
b)
Cl–
c)
Br–
d)
HO–
|
Rahul Chatterjee answered |
OH−
OH− ions will have highest molar conductivity amongst the given anions.
Which solution will conduct the electricity:
- a)Sugar in water
- b)Sugar in ethanol
- c)Iodine in ethanol
- d)MgCl2 in water
Correct answer is option 'D'. Can you explain this answer?
Which solution will conduct the electricity:
a)
Sugar in water
b)
Sugar in ethanol
c)
Iodine in ethanol
d)
MgCl2 in water
|
|
Sarthak Chavan answered |
**Conductivity of Solutions**
**Introduction:**
Conductivity refers to the ability of a substance to conduct electricity. In solutions, the presence of charged particles (ions) enables them to conduct electricity. Substances that dissolve in water and dissociate into ions are known as electrolytes and are conductive. On the other hand, substances that do not dissociate into ions and do not conduct electricity are called non-electrolytes.
**Explanation of Options:**
a) Sugar in Water:
Sugar (sucrose) is a covalent compound that does not dissociate into ions when dissolved in water. Therefore, it does not conduct electricity.
b) Sugar in Ethanol:
Similar to sugar in water, sugar in ethanol also does not dissociate into ions. Ethanol is a covalent compound, and therefore, the solution does not conduct electricity.
c) Iodine in Ethanol:
Iodine is a nonpolar covalent compound that does not dissociate into ions when dissolved in ethanol. Hence, the solution of iodine in ethanol does not conduct electricity.
d) MgCl2 in Water:
MgCl2 is an ionic compound composed of magnesium ions (Mg2+) and chloride ions (Cl-). When this compound is dissolved in water, it dissociates into its constituent ions. This dissociation process allows the solution to conduct electricity. The magnesium ions (Mg2+) and chloride ions (Cl-) are free to move in the solution and carry electric charge, enabling the conduction of electricity.
**Conclusion:**
Among the given options, only the solution of MgCl2 in water conducts electricity. This is because MgCl2 is an ionic compound that dissociates into ions when dissolved in water. The presence of these ions allows the solution to conduct electricity. The other options (sugar in water, sugar in ethanol, and iodine in ethanol) do not dissociate into ions and therefore do not conduct electricity.
**Introduction:**
Conductivity refers to the ability of a substance to conduct electricity. In solutions, the presence of charged particles (ions) enables them to conduct electricity. Substances that dissolve in water and dissociate into ions are known as electrolytes and are conductive. On the other hand, substances that do not dissociate into ions and do not conduct electricity are called non-electrolytes.
**Explanation of Options:**
a) Sugar in Water:
Sugar (sucrose) is a covalent compound that does not dissociate into ions when dissolved in water. Therefore, it does not conduct electricity.
b) Sugar in Ethanol:
Similar to sugar in water, sugar in ethanol also does not dissociate into ions. Ethanol is a covalent compound, and therefore, the solution does not conduct electricity.
c) Iodine in Ethanol:
Iodine is a nonpolar covalent compound that does not dissociate into ions when dissolved in ethanol. Hence, the solution of iodine in ethanol does not conduct electricity.
d) MgCl2 in Water:
MgCl2 is an ionic compound composed of magnesium ions (Mg2+) and chloride ions (Cl-). When this compound is dissolved in water, it dissociates into its constituent ions. This dissociation process allows the solution to conduct electricity. The magnesium ions (Mg2+) and chloride ions (Cl-) are free to move in the solution and carry electric charge, enabling the conduction of electricity.
**Conclusion:**
Among the given options, only the solution of MgCl2 in water conducts electricity. This is because MgCl2 is an ionic compound that dissociates into ions when dissolved in water. The presence of these ions allows the solution to conduct electricity. The other options (sugar in water, sugar in ethanol, and iodine in ethanol) do not dissociate into ions and therefore do not conduct electricity.
Saturated solution of KNO3 is used to make ‘salt-bridge’ because:- a)Velocity of K+ is greater than that of NO3-
- b)Velocity NO3- is greater than that of K+
- c)Velocities of both K+ and NO3- are nearly the same
- d)KNO3 is highly soluble in water
Correct answer is option 'C'. Can you explain this answer?
Saturated solution of KNO3 is used to make ‘salt-bridge’ because:
a)
Velocity of K+ is greater than that of NO3-
b)
Velocity NO3- is greater than that of K+
c)
Velocities of both K+ and NO3- are nearly the same
d)
KNO3 is highly soluble in water
|
|
Tanishq Goyal answered |
A homemade hand warmer. When the KNO3 is dissolved in water, it absorbs heat, making the solution cold. However, when the solution is exposed to air, the water evaporates, leaving behind solid KNO3 crystals and releasing heat. This exothermic reaction creates a heat source that can be used to warm hands or other body parts.
To make the hand warmer, you will need:
- KNO3 (potassium nitrate)
- Water
- Two sealable plastic bags (one larger and one smaller)
- A measuring cup
- A spoon
- A thermometer
Instructions:
1. Measure out 100 grams of KNO3 and pour it into the larger plastic bag.
2. Add 50 milliliters of water to the bag and seal it tightly.
3. Use your hands to mix the KNO3 and water together until the KNO3 is completely dissolved.
4. Place the bag in the refrigerator and let it chill for at least 30 minutes.
5. Once the solution is cold, remove it from the refrigerator and use the thermometer to check the temperature. It should be around 0°C (32°F).
6. Pour the solution into the smaller plastic bag and seal it tightly.
7. Place the smaller bag inside the larger bag and seal it tightly as well.
8. Massage the bags with your hands to mix the solution and activate the reaction.
9. The bags will start to warm up quickly and provide heat for up to an hour.
Caution: Potassium nitrate can be dangerous if ingested or inhaled. Use caution when handling and storing the chemicals, and do not puncture or open the plastic bags.
To make the hand warmer, you will need:
- KNO3 (potassium nitrate)
- Water
- Two sealable plastic bags (one larger and one smaller)
- A measuring cup
- A spoon
- A thermometer
Instructions:
1. Measure out 100 grams of KNO3 and pour it into the larger plastic bag.
2. Add 50 milliliters of water to the bag and seal it tightly.
3. Use your hands to mix the KNO3 and water together until the KNO3 is completely dissolved.
4. Place the bag in the refrigerator and let it chill for at least 30 minutes.
5. Once the solution is cold, remove it from the refrigerator and use the thermometer to check the temperature. It should be around 0°C (32°F).
6. Pour the solution into the smaller plastic bag and seal it tightly.
7. Place the smaller bag inside the larger bag and seal it tightly as well.
8. Massage the bags with your hands to mix the solution and activate the reaction.
9. The bags will start to warm up quickly and provide heat for up to an hour.
Caution: Potassium nitrate can be dangerous if ingested or inhaled. Use caution when handling and storing the chemicals, and do not puncture or open the plastic bags.
Based on the data given blow strongest oxidizing agent will be:
- a)Cl
- b)Cr3+
- c)Mn2+
- d)MnO4–
Correct answer is option 'D'. Can you explain this answer?
Based on the data given blow strongest oxidizing agent will be:
a)
Cl
b)
Cr3+
c)
Mn2+
d)
MnO4–
|
|
Pooja Choudhury answered |
As MnO4- has highest reduction potential among them so it is strongest oxidizing agent.
The concentration of K+ ion inside a biological cell is 20 times higher than outside. The magnitude of potential difference between the two sides is [Given: 2.303 RT/F = 59 mV]- a)0 mV
- b)26 mV
- c)77 mV
- d)177 mV
Correct answer is option 'C'. Can you explain this answer?
The concentration of K+ ion inside a biological cell is 20 times higher than outside. The magnitude of potential difference between the two sides is [Given: 2.303 RT/F = 59 mV]
a)
0 mV
b)
26 mV
c)
77 mV
d)
177 mV
|
|
Pooja Choudhury answered |
This is a concentration cell,
Ecell=0.059log(C1/C2)
C1=20C2
Ecell=0.059log(20)
Ecell=0.0767V
=76.7mV
Which of the following are correct:- a)Conductivity of a solution increases with dilution
- b)Molar conductivity of a solution increases with dilution
- c)Conductivity of a solution decreases with dilution
- d)Molar conductivity of a solution decreases with dilution
Correct answer is option 'B,C'. Can you explain this answer?
Which of the following are correct:
a)
Conductivity of a solution increases with dilution
b)
Molar conductivity of a solution increases with dilution
c)
Conductivity of a solution decreases with dilution
d)
Molar conductivity of a solution decreases with dilution
|
Isha Bose answered |
**Conductivity and Molar Conductivity**
Conductivity is a measure of a solution's ability to conduct electricity. It depends on the concentration of ions present in the solution. Molar conductivity, on the other hand, is a measure of the conductivity of a solution per unit concentration of solute. It is calculated by dividing the conductivity of the solution by its molar concentration.
**Effect of Dilution on Conductivity and Molar Conductivity**
1. **Conductivity of a Solution Increases with Dilution (Incorrect)**
When a solution is diluted by adding more solvent, the concentration of ions in the solution decreases. As a result, the number of charge carriers responsible for conducting electricity also decreases. Therefore, the conductivity of a solution generally decreases with dilution, not increases.
2. **Molar Conductivity of a Solution Increases with Dilution (Correct)**
Molar conductivity is the ratio of conductivity to molar concentration. When a solution is diluted, the molar concentration decreases. However, the conductivity decreases at a slower rate than the molar concentration. This is because the decrease in concentration of ions is offset by the increase in the average distance between the ions, which reduces the effect of their mutual repulsion. As a result, the molar conductivity of a solution generally increases with dilution.
3. **Conductivity of a Solution Decreases with Dilution (Correct)**
As mentioned earlier, when a solution is diluted, the concentration of ions decreases. Since conductivity is directly proportional to the concentration of ions, the conductivity of a solution decreases with dilution. This is the correct statement.
4. **Molar Conductivity of a Solution Decreases with Dilution (Incorrect)**
As explained earlier, the molar conductivity of a solution increases with dilution. Therefore, this statement is incorrect.
In conclusion, the correct statements are:
b) Molar conductivity of a solution increases with dilution
c) Conductivity of a solution decreases with dilution
Conductivity is a measure of a solution's ability to conduct electricity. It depends on the concentration of ions present in the solution. Molar conductivity, on the other hand, is a measure of the conductivity of a solution per unit concentration of solute. It is calculated by dividing the conductivity of the solution by its molar concentration.
**Effect of Dilution on Conductivity and Molar Conductivity**
1. **Conductivity of a Solution Increases with Dilution (Incorrect)**
When a solution is diluted by adding more solvent, the concentration of ions in the solution decreases. As a result, the number of charge carriers responsible for conducting electricity also decreases. Therefore, the conductivity of a solution generally decreases with dilution, not increases.
2. **Molar Conductivity of a Solution Increases with Dilution (Correct)**
Molar conductivity is the ratio of conductivity to molar concentration. When a solution is diluted, the molar concentration decreases. However, the conductivity decreases at a slower rate than the molar concentration. This is because the decrease in concentration of ions is offset by the increase in the average distance between the ions, which reduces the effect of their mutual repulsion. As a result, the molar conductivity of a solution generally increases with dilution.
3. **Conductivity of a Solution Decreases with Dilution (Correct)**
As mentioned earlier, when a solution is diluted, the concentration of ions decreases. Since conductivity is directly proportional to the concentration of ions, the conductivity of a solution decreases with dilution. This is the correct statement.
4. **Molar Conductivity of a Solution Decreases with Dilution (Incorrect)**
As explained earlier, the molar conductivity of a solution increases with dilution. Therefore, this statement is incorrect.
In conclusion, the correct statements are:
b) Molar conductivity of a solution increases with dilution
c) Conductivity of a solution decreases with dilution
Choose the correct options:
- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
Choose the correct options:
a)
b)
c)
d)
|
Chirag Verma answered |
Correct Answer :- c
Explanation : molar conductance = Valency × equivalent conductance.
Al2(SO4)3 is commonly known as aluminium sulphate. There is two aluminium atom in aluminium sulphate, and because of this, the total valency of the formula is 6+
AM = 6eq for Al2(SO4)3
Which of the following solutions will show the minimum value of transference number of Cl–1 ions:
- a)0.1 M NaCl solut ion
- b)0.1 M KCl solut ion
- c)0.1 M HCl solution
- d)0.1 M MgCl2 solution
Correct answer is option 'C'. Can you explain this answer?
Which of the following solutions will show the minimum value of transference number of Cl–1 ions:
a)
0.1 M NaCl solut ion
b)
0.1 M KCl solut ion
c)
0.1 M HCl solution
d)
0.1 M MgCl2 solution
|
Anagha Chauhan answered |
Because 
has maximum value of transference no. as co mpared to other cations. Hence 
will be minimum in 0.1 M HCl solution
has maximum value of transference no. as co mpared to other cations. Hence will be minimum in 0.1 M HCl solutionAt 20°C, the standard EMF of a certain cell is +0.2699 V, and at 30°C it is + 0.2669 V. What can you say about the standard entropy of this reaction? Assume that the standard ΔH° and ΔS° are independent of temperature:- a)ΔS° = 0
- b)ΔS° = + ve
- c)ΔS° = -ve
- d)None of these.
Correct answer is option 'C'. Can you explain this answer?
At 20°C, the standard EMF of a certain cell is +0.2699 V, and at 30°C it is + 0.2669 V. What can you say about the standard entropy of this reaction? Assume that the standard ΔH° and ΔS° are independent of temperature:
a)
ΔS° = 0
b)
ΔS° = + ve
c)
ΔS° = -ve
d)
None of these.
|
|
Pooja Choudhury answered |
As per temperature coefficient of cell, dS= nF[dE/dT] at constant pressure Here we can se that as temp. Increase E is decrease it means if we take T as positive then E will be negative. So, here [dE/dT] this ratio will be negative which is directly proportional to the entropy change. Hence, we can say that dS= (-ve)
Which of the following will have same value of molar conductivity and equivalent conductivity:- a)0.1 M NaCl solution
- b)0.1 M Na2SO4 solution
- c)0.1 M KCl solution
- d)0.1 M MgCl2 solution
Correct answer is option 'A,C'. Can you explain this answer?
Which of the following will have same value of molar conductivity and equivalent conductivity:
a)
0.1 M NaCl solution
b)
0.1 M Na2SO4 solution
c)
0.1 M KCl solution
d)
0.1 M MgCl2 solution
|
|
Anagha Bajaj answered |
**Explanation:**
To understand why the molar conductivity and equivalent conductivity are the same for option A and C, let's first define what these terms mean.
**Molar Conductivity:** Molar conductivity (Λm) is a measure of the ability of a solution to conduct electricity. It is defined as the conductivity of a solution containing 1 mole of electrolyte dissolved in 1 liter of solution.
**Equivalent Conductivity:** Equivalent conductivity (Λeq) is a measure of the conductivity of an electrolyte solution containing 1 equivalent of electrolyte dissolved in 1 liter of solution. An equivalent is the amount of an electrolyte that provides 1 mole of charge.
Now, let's analyze the given options and determine the molar conductivity and equivalent conductivity for each solution.
a) 0.1 M NaCl solution:
- NaCl dissociates into Na+ and Cl- ions in solution.
- Since NaCl is a strong electrolyte, it completely dissociates into ions.
- The molar conductivity and equivalent conductivity of NaCl solution will be the same because 1 mole of NaCl provides 1 equivalent of charge.
b) 0.1 M Na2SO4 solution:
- Na2SO4 dissociates into 2 Na+ ions and 1 SO4- ion in solution.
- Since Na2SO4 is a strong electrolyte, it completely dissociates into ions.
- The molar conductivity of Na2SO4 solution will be higher than the equivalent conductivity because 1 mole of Na2SO4 provides 2 equivalents of charge.
c) 0.1 M KCl solution:
- KCl dissociates into K+ and Cl- ions in solution.
- Since KCl is a strong electrolyte, it completely dissociates into ions.
- The molar conductivity and equivalent conductivity of KCl solution will be the same because 1 mole of KCl provides 1 equivalent of charge.
d) 0.1 M MgCl2 solution:
- MgCl2 dissociates into 1 Mg2+ ion and 2 Cl- ions in solution.
- Since MgCl2 is a strong electrolyte, it completely dissociates into ions.
- The molar conductivity of MgCl2 solution will be higher than the equivalent conductivity because 1 mole of MgCl2 provides 2 equivalents of charge.
Therefore, the options with the same value of molar conductivity and equivalent conductivity are A (0.1 M NaCl solution) and C (0.1 M KCl solution).
To understand why the molar conductivity and equivalent conductivity are the same for option A and C, let's first define what these terms mean.
**Molar Conductivity:** Molar conductivity (Λm) is a measure of the ability of a solution to conduct electricity. It is defined as the conductivity of a solution containing 1 mole of electrolyte dissolved in 1 liter of solution.
**Equivalent Conductivity:** Equivalent conductivity (Λeq) is a measure of the conductivity of an electrolyte solution containing 1 equivalent of electrolyte dissolved in 1 liter of solution. An equivalent is the amount of an electrolyte that provides 1 mole of charge.
Now, let's analyze the given options and determine the molar conductivity and equivalent conductivity for each solution.
a) 0.1 M NaCl solution:
- NaCl dissociates into Na+ and Cl- ions in solution.
- Since NaCl is a strong electrolyte, it completely dissociates into ions.
- The molar conductivity and equivalent conductivity of NaCl solution will be the same because 1 mole of NaCl provides 1 equivalent of charge.
b) 0.1 M Na2SO4 solution:
- Na2SO4 dissociates into 2 Na+ ions and 1 SO4- ion in solution.
- Since Na2SO4 is a strong electrolyte, it completely dissociates into ions.
- The molar conductivity of Na2SO4 solution will be higher than the equivalent conductivity because 1 mole of Na2SO4 provides 2 equivalents of charge.
c) 0.1 M KCl solution:
- KCl dissociates into K+ and Cl- ions in solution.
- Since KCl is a strong electrolyte, it completely dissociates into ions.
- The molar conductivity and equivalent conductivity of KCl solution will be the same because 1 mole of KCl provides 1 equivalent of charge.
d) 0.1 M MgCl2 solution:
- MgCl2 dissociates into 1 Mg2+ ion and 2 Cl- ions in solution.
- Since MgCl2 is a strong electrolyte, it completely dissociates into ions.
- The molar conductivity of MgCl2 solution will be higher than the equivalent conductivity because 1 mole of MgCl2 provides 2 equivalents of charge.
Therefore, the options with the same value of molar conductivity and equivalent conductivity are A (0.1 M NaCl solution) and C (0.1 M KCl solution).
For HCl solution at 25ºC, the equivalent conductivity at infinite dilution is 425 Ω-1 cm2 eq-1. The specific conductance of a solution of HCl is 3.825 Ω-1 cm-1. If the apparent degree of dissociat ion is 90%, the normality of solution is- a)0.9
- b)10.00
- c)1.1
- d)1.2
Correct answer is option 'B'. Can you explain this answer?
For HCl solution at 25ºC, the equivalent conductivity at infinite dilution is 425 Ω-1 cm2 eq-1. The specific conductance of a solution of HCl is 3.825 Ω-1 cm-1. If the apparent degree of dissociat ion is 90%, the normality of solution is
a)
0.9
b)
10.00
c)
1.1
d)
1.2
|
|
Vandana Chopra answered |
In a typical Conductometric titration of a strong acid with a weak base, the curve resembles:- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In a typical Conductometric titration of a strong acid with a weak base, the curve resembles:
a)
b)
c)
d)
|
Pie Academy answered |
Here consider a strong acid as HCL and weak base as Ammonium hydroxide that is NH4OH. Suppose the acid has high concentration of H positive ions due to which it show high conductance, when a weak base is added to it, the cation of weak base combines with Cl Negative ion to form ammonium chloride precipitate thus decreasing the conductivity by utilizing H positive ion to form water molecule due to this decrease in conductivity the graph shows on negative fall in conductance after all the edge positive and Cl negative ions are combine end point is reached where no free ions are possible for conduction after this when the basis added to form NH4 positive and was OH negative since it is a weak base it has low conductivity and the graph increases very slowly towards conduction minute changes in increasing conduction can be seen when a weak bases are added
plot a) show high conductivity which is not possible in weak bases
plot c) and d) involve 3 mixture components which is not the case
The perfect diagram would look like this and B is the closest.
plot c) and d) involve 3 mixture components which is not the case
The perfect diagram would look like this and B is the closest.
The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is:- a)LiCl > NaCl > KCl
- b)KCl > NaCl > LiCl
- c)NaCl > KCl > LiCl
- d)LiCl > KCl > NaCl
Correct answer is option 'B'. Can you explain this answer?
The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is:
a)
LiCl > NaCl > KCl
b)
KCl > NaCl > LiCl
c)
NaCl > KCl > LiCl
d)
LiCl > KCl > NaCl
|
|
Stuti Patel answered |
The correct order of equivalent conductance at infinite dilution is KCl > NaCl > LiCl.
Anion is same (chloride ion) for all the species. Larger is the size of the cation, greater is the equivalent conductance at infinite dilution and vice versa.
The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:
- a)201.28 S cm2 mol–1
- b)698.28 S cm2 mol–1
- c)390.71 S cm2 mol–1
- d)540.48 S cm2 mol–1
Correct answer is option 'C'. Can you explain this answer?
The molar conductivity of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 S cm2 mol–1 respectively. The molar conductivity of CH3COOH at infinite dilution is:
a)
201.28 S cm2 mol–1
b)
698.28 S cm2 mol–1
c)
390.71 S cm2 mol–1
d)
540.48 S cm2 mol–1
|
|
Asf Institute answered |
Which one of the following solutions has lowest conducting power:
- a)0.1 M CH3COOH
- b)0.1 M NaCl
- c)0.1 M KNO3
- d)0.1 M HCl
Correct answer is option 'A'. Can you explain this answer?
Which one of the following solutions has lowest conducting power:
a)
0.1 M CH3COOH
b)
0.1 M NaCl
c)
0.1 M KNO3
d)
0.1 M HCl
|
|
Aryan Choudhary answered |
**Answer:**
To determine the solution with the lowest conducting power, we need to consider the dissociation of the solutes in water. The greater the degree of dissociation, the higher the conducting power of the solution.
**Dissociation of the solutes:**
a) CH3COOH (acetic acid):
CH3COOH ⇌ CH3COO- + H+
b) NaCl (sodium chloride):
NaCl ⇌ Na+ + Cl-
c) KNO3 (potassium nitrate):
KNO3 ⇌ K+ + NO3-
d) HCl (hydrochloric acid):
HCl ⇌ H+ + Cl-
**Factors affecting the degree of dissociation:**
1. Nature of the solute: Strong acids and strong bases dissociate completely in water, while weak acids and weak bases dissociate to a lesser extent.
2. Concentration of the solution: Higher concentrations tend to increase the degree of dissociation.
**Explanation:**
a) 0.1 M CH3COOH (acetic acid):
Acetic acid is a weak acid and does not dissociate completely in water. It only partially ionizes into CH3COO- and H+. Therefore, the conducting power of this solution will be lower compared to the others.
b) 0.1 M NaCl (sodium chloride):
Sodium chloride is an ionic compound and dissociates completely into Na+ and Cl- ions in water. This solution will have a higher conducting power compared to acetic acid.
c) 0.1 M KNO3 (potassium nitrate):
Potassium nitrate is also an ionic compound and dissociates completely into K+ and NO3- ions in water. This solution will have a higher conducting power compared to acetic acid.
d) 0.1 M HCl (hydrochloric acid):
Hydrochloric acid is a strong acid and dissociates completely into H+ and Cl- ions in water. This solution will have a higher conducting power compared to acetic acid.
Therefore, the solution with the lowest conducting power is 0.1 M CH3COOH (acetic acid) due to its weak acid nature and partial dissociation in water.
To determine the solution with the lowest conducting power, we need to consider the dissociation of the solutes in water. The greater the degree of dissociation, the higher the conducting power of the solution.
**Dissociation of the solutes:**
a) CH3COOH (acetic acid):
CH3COOH ⇌ CH3COO- + H+
b) NaCl (sodium chloride):
NaCl ⇌ Na+ + Cl-
c) KNO3 (potassium nitrate):
KNO3 ⇌ K+ + NO3-
d) HCl (hydrochloric acid):
HCl ⇌ H+ + Cl-
**Factors affecting the degree of dissociation:**
1. Nature of the solute: Strong acids and strong bases dissociate completely in water, while weak acids and weak bases dissociate to a lesser extent.
2. Concentration of the solution: Higher concentrations tend to increase the degree of dissociation.
**Explanation:**
a) 0.1 M CH3COOH (acetic acid):
Acetic acid is a weak acid and does not dissociate completely in water. It only partially ionizes into CH3COO- and H+. Therefore, the conducting power of this solution will be lower compared to the others.
b) 0.1 M NaCl (sodium chloride):
Sodium chloride is an ionic compound and dissociates completely into Na+ and Cl- ions in water. This solution will have a higher conducting power compared to acetic acid.
c) 0.1 M KNO3 (potassium nitrate):
Potassium nitrate is also an ionic compound and dissociates completely into K+ and NO3- ions in water. This solution will have a higher conducting power compared to acetic acid.
d) 0.1 M HCl (hydrochloric acid):
Hydrochloric acid is a strong acid and dissociates completely into H+ and Cl- ions in water. This solution will have a higher conducting power compared to acetic acid.
Therefore, the solution with the lowest conducting power is 0.1 M CH3COOH (acetic acid) due to its weak acid nature and partial dissociation in water.
The velocity of Li+ ion in water is 2 × 10–2 cm/sec when 100 V is applied between two electrodes separated by 2 cm. The mobility of Li+ ion in water is,
- a)4 × 10–4 cm2 s–1 V–1
- b)1 × 10–4 s–1 V–1
- c)4 V cm2 s–1
- d)2.5 × 105 V s cm–2
Correct answer is option 'A'. Can you explain this answer?
The velocity of Li+ ion in water is 2 × 10–2 cm/sec when 100 V is applied between two electrodes separated by 2 cm. The mobility of Li+ ion in water is,
a)
4 × 10–4 cm2 s–1 V–1
b)
1 × 10–4 s–1 V–1
c)
4 V cm2 s–1
d)
2.5 × 105 V s cm–2
|
|
Edurev.iitjam answered |
For the following cell,
Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s)
When the concentration of Zn2+ is 10 times the concentration of Cu2+, the expression for ΔG (in J mol-1) is [F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1 V]
- a)2.303 RT – 2.2 F
- b)2.303 RT + 1.1 F
- c)1.1 F
- d)–2.2 F
Correct answer is option 'A'. Can you explain this answer?
For the following cell,
Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s)
When the concentration of Zn2+ is 10 times the concentration of Cu2+, the expression for ΔG (in J mol-1) is [F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1 V]
a)
2.303 RT – 2.2 F
b)
2.303 RT + 1.1 F
c)
1.1 F
d)
–2.2 F
|
|
Vedika Singh answered |
The molality of (NH4)2SO4 solution that has the same ionic strength as 1 mol kg–1 solution of KCl is:- a)1/3 mol kg–1
- b)1/2 mol kg–1
- c)2/5 mol kg–1
- d)3/5 mol kg–1
Correct answer is option 'A'. Can you explain this answer?
The molality of (NH4)2SO4 solution that has the same ionic strength as 1 mol kg–1 solution of KCl is:
a)
1/3 mol kg–1
b)
1/2 mol kg–1
c)
2/5 mol kg–1
d)
3/5 mol kg–1
|
|
Hrishikesh Verma answered |
Calculation of Ionic Strength:
Ionic strength is the measure of the concentration of ions in a solution. It is calculated using the formula:
Ionic strength (I) = 1/2 ∑ ci zi^2
Where,
ci = concentration of the ith ion
zi = charge on the ith ion
For 1 mol kg^-1 solution of KCl, the concentration of K+ and Cl- ions is 1 mol kg^-1 each. Therefore,
Ionic strength (I) = 1/2 [ (1 mol kg^-1) (1^2) + (1 mol kg^-1) (1^2) ]
I = 1 mol kg^-1
Calculation of Molality of (NH4)2SO4 solution:
To find the molality of (NH4)2SO4 solution that has the same ionic strength as 1 mol kg^-1 solution of KCl, we need to first calculate the concentration of ammonium and sulfate ions in the solution.
(NH4)2SO4 dissociates to give 2 NH4+ ions and 1 SO4^2- ion. Therefore, the total concentration of ions in the solution is:
Concentration of ions = 2[NH4+] + [SO4^2-]
Let x be the molality of (NH4)2SO4 solution. Then,
Concentration of NH4+ ions = 2x mol kg^-1
Concentration of SO4^2- ion = x mol kg^-1
Using the formula for ionic strength, we can write:
I = 1/2 [ (2x mol kg^-1) (1^2) + (x mol kg^-1) (2^2) ]
1 = 2x + 2x
x = 1/3 mol kg^-1
Therefore, the molality of (NH4)2SO4 solution that has the same ionic strength as 1 mol kg^-1 solution of KCl is 1/3 mol kg^-1.
Ionic strength is the measure of the concentration of ions in a solution. It is calculated using the formula:
Ionic strength (I) = 1/2 ∑ ci zi^2
Where,
ci = concentration of the ith ion
zi = charge on the ith ion
For 1 mol kg^-1 solution of KCl, the concentration of K+ and Cl- ions is 1 mol kg^-1 each. Therefore,
Ionic strength (I) = 1/2 [ (1 mol kg^-1) (1^2) + (1 mol kg^-1) (1^2) ]
I = 1 mol kg^-1
Calculation of Molality of (NH4)2SO4 solution:
To find the molality of (NH4)2SO4 solution that has the same ionic strength as 1 mol kg^-1 solution of KCl, we need to first calculate the concentration of ammonium and sulfate ions in the solution.
(NH4)2SO4 dissociates to give 2 NH4+ ions and 1 SO4^2- ion. Therefore, the total concentration of ions in the solution is:
Concentration of ions = 2[NH4+] + [SO4^2-]
Let x be the molality of (NH4)2SO4 solution. Then,
Concentration of NH4+ ions = 2x mol kg^-1
Concentration of SO4^2- ion = x mol kg^-1
Using the formula for ionic strength, we can write:
I = 1/2 [ (2x mol kg^-1) (1^2) + (x mol kg^-1) (2^2) ]
1 = 2x + 2x
x = 1/3 mol kg^-1
Therefore, the molality of (NH4)2SO4 solution that has the same ionic strength as 1 mol kg^-1 solution of KCl is 1/3 mol kg^-1.
Which statement is not correct:- a)Conductance of an electrolytic solution increases with dilution
- b)Conductance of an electrolytic solution decreases with dilution
- c)Equivalent conductivity solution decreases solution increases with dilution
- d)none
Correct answer is option 'B'. Can you explain this answer?
Which statement is not correct:
a)
Conductance of an electrolytic solution increases with dilution
b)
Conductance of an electrolytic solution decreases with dilution
c)
Equivalent conductivity solution decreases solution increases with dilution
d)
none
|
|
Sagarika Yadav answered |
Conductance and Equivalent Conductivity of Electrolytic Solutions
Conductance and equivalent conductivity are two important properties of electrolytic solutions. Let's understand them in detail.
Conductance of Electrolytic Solutions
Conductance is the ability of an electrolytic solution to conduct electricity. It is expressed in units of Siemens per centimeter (S/cm). The conductance of an electrolytic solution depends on various factors such as:
- The concentration of the electrolyte
- The nature of the electrolyte
- The temperature of the solution
- The area of the electrodes in the solution
Dilution and Conductance
The conductance of an electrolytic solution increases with dilution. This is because, when an electrolytic solution is diluted, the concentration of ions decreases, and the distance between the ions increases. As a result, there are more free ions that can conduct electricity, which leads to an increase in the conductance of the solution.
Therefore, option 'A' is correct.
Equivalent Conductivity of Electrolytic Solutions
Equivalent conductivity is another important property of electrolytic solutions. It is defined as the conductance of a solution containing one gram equivalent of the electrolyte between two electrodes placed one centimeter apart. It is expressed in units of Siemens per centimeter per gram equivalent (S/cm/eq).
Dilution and Equivalent Conductivity
The equivalent conductivity of an electrolytic solution decreases with dilution. This is because, when an electrolytic solution is diluted, the concentration of ions decreases, and the distance between the ions increases. As a result, the number of ions that contribute to the conductivity of the solution decreases, which leads to a decrease in the equivalent conductivity of the solution.
Therefore, option 'C' is correct.
Conclusion
In summary, the statement that is not correct is option 'B'. The conductance of an electrolytic solution increases with dilution, while the equivalent conductivity of an electrolytic solution decreases with dilution.
Conductance and equivalent conductivity are two important properties of electrolytic solutions. Let's understand them in detail.
Conductance of Electrolytic Solutions
Conductance is the ability of an electrolytic solution to conduct electricity. It is expressed in units of Siemens per centimeter (S/cm). The conductance of an electrolytic solution depends on various factors such as:
- The concentration of the electrolyte
- The nature of the electrolyte
- The temperature of the solution
- The area of the electrodes in the solution
Dilution and Conductance
The conductance of an electrolytic solution increases with dilution. This is because, when an electrolytic solution is diluted, the concentration of ions decreases, and the distance between the ions increases. As a result, there are more free ions that can conduct electricity, which leads to an increase in the conductance of the solution.
Therefore, option 'A' is correct.
Equivalent Conductivity of Electrolytic Solutions
Equivalent conductivity is another important property of electrolytic solutions. It is defined as the conductance of a solution containing one gram equivalent of the electrolyte between two electrodes placed one centimeter apart. It is expressed in units of Siemens per centimeter per gram equivalent (S/cm/eq).
Dilution and Equivalent Conductivity
The equivalent conductivity of an electrolytic solution decreases with dilution. This is because, when an electrolytic solution is diluted, the concentration of ions decreases, and the distance between the ions increases. As a result, the number of ions that contribute to the conductivity of the solution decreases, which leads to a decrease in the equivalent conductivity of the solution.
Therefore, option 'C' is correct.
Conclusion
In summary, the statement that is not correct is option 'B'. The conductance of an electrolytic solution increases with dilution, while the equivalent conductivity of an electrolytic solution decreases with dilution.
The fraction of the total current carried by an ion is known as:
- a)Transport number of that ion
- b)Conductance of that ion
- c)Both a and b
- d)None of these.
Correct answer is option 'A'. Can you explain this answer?
The fraction of the total current carried by an ion is known as:
a)
Transport number of that ion
b)
Conductance of that ion
c)
Both a and b
d)
None of these.
|
|
Abhijeet Majumdar answered |
Transport number of an ion:
The transport number of an ion refers to the fraction of the total current carried by that specific ion in an electrolytic solution or a cell. It is denoted by the symbol "t" and is expressed as a decimal or fraction.
Conductance of an ion:
Conductance is a measure of an ion's ability to conduct electric current. It represents the ease with which ions move through an electrolyte solution. The conductance of an ion depends on its concentration, mobility, and charge.
Explanation:
In an electrolytic solution or a cell, multiple ions may be present. When an electric current is passed through the solution, the ions move towards the electrodes, carrying the charge. The transport number of an ion indicates the proportion of the total current that is carried by that specific ion.
The transport number of an ion can be determined experimentally using various techniques, such as the Hittorf's method or the moving boundary method. These methods allow for the measurement of changes in the concentrations of ions at the electrodes and help in calculating the transport number.
The conductance of an ion is related to its transport number. The conductance is influenced by factors such as ion concentration, ion mobility, and charge. However, it does not directly represent the fraction of total current carried by an ion.
Conclusion:
In conclusion, the fraction of the total current carried by an ion is known as the transport number of that ion. It is an important parameter in understanding the contribution of individual ions to the overall current flow in an electrolytic solution or a cell. The conductance of an ion, on the other hand, represents its ability to conduct the electric current and is influenced by various factors. Therefore, the correct answer is option 'A' - Transport number of that ion.
The transport number of an ion refers to the fraction of the total current carried by that specific ion in an electrolytic solution or a cell. It is denoted by the symbol "t" and is expressed as a decimal or fraction.
Conductance of an ion:
Conductance is a measure of an ion's ability to conduct electric current. It represents the ease with which ions move through an electrolyte solution. The conductance of an ion depends on its concentration, mobility, and charge.
Explanation:
In an electrolytic solution or a cell, multiple ions may be present. When an electric current is passed through the solution, the ions move towards the electrodes, carrying the charge. The transport number of an ion indicates the proportion of the total current that is carried by that specific ion.
The transport number of an ion can be determined experimentally using various techniques, such as the Hittorf's method or the moving boundary method. These methods allow for the measurement of changes in the concentrations of ions at the electrodes and help in calculating the transport number.
The conductance of an ion is related to its transport number. The conductance is influenced by factors such as ion concentration, ion mobility, and charge. However, it does not directly represent the fraction of total current carried by an ion.
Conclusion:
In conclusion, the fraction of the total current carried by an ion is known as the transport number of that ion. It is an important parameter in understanding the contribution of individual ions to the overall current flow in an electrolytic solution or a cell. The conductance of an ion, on the other hand, represents its ability to conduct the electric current and is influenced by various factors. Therefore, the correct answer is option 'A' - Transport number of that ion.
FAD is a redox active molecule which takes part in many important biological reactions. the redox potential of FAD at pH 7.0 is given below:
Calculate the redox potential when the media is acidified to pH0- a)0 V
- b)0.24 V
- c)0.12 V
- d)None of these.
Correct answer is option 'B'. Can you explain this answer?
FAD is a redox active molecule which takes part in many important biological reactions. the redox potential of FAD at pH 7.0 is given below:
Calculate the redox potential when the media is acidified to pH0
a)
0 V
b)
0.24 V
c)
0.12 V
d)
None of these.
|
Satyabrata Sahoo answered |
Which of the following ion is expected to have highest value of molar conductivity at infinite dilution in an aqueous solution:- a)Na+
- b)K+
- c)H+
- d)
Correct answer is option 'C'. Can you explain this answer?
Which of the following ion is expected to have highest value of molar conductivity at infinite dilution in an aqueous solution:
a)
Na+
b)
K+
c)
H+
d)
|
|
Athul Chaudhary answered |
H+ ions will have the maximum molar conductivity amongst the given cations
The mean ionic activity coefficient of 0.0005 mol kg–1 CaCl2 in water at 25°C is:- a)0.98
- b)0.67
- c)0.81
- d)0.91
Correct answer is option 'D'. Can you explain this answer?
The mean ionic activity coefficient of 0.0005 mol kg–1 CaCl2 in water at 25°C is:
a)
0.98
b)
0.67
c)
0.81
d)
0.91
|
|
Vikram Kapoor answered |
Correct Answer :- d
Explanation : log¥ = - 0.509|Z+Z-|√I
I = 0.0015
log¥= -0. 509|2×-1|√0.0015
= - 0.0394
¥ = (10) ^-0.0394
= 0.91
The unit of A in Debye-Huckel-Onsager equation for strong electrolytes is:- a)Ω-1cm2 M-1/ 2
- b)Ω-1cm-2 M1/ 2
- c)Ω-1cm2 M-1
- d)M-1/ 2
Correct answer is option 'D'. Can you explain this answer?
The unit of A in Debye-Huckel-Onsager equation for strong electrolytes is:
a)
Ω-1cm2 M-1/ 2
b)
Ω-1cm-2 M1/ 2
c)
Ω-1cm2 M-1
d)
M-1/ 2
|
|
Pooja Choudhury answered |
The influence of ion-ion interactions on the conductivity of strong electrolytes was studied by Debye and Huckel.
They considered that each ion is surrounded by an ionic atmosphere of opposite sign, and derived an expression relating the molar conductance of strong electrolytes with the concentration by assuming complete dissociations.
It was further developed by Onsager. For a uni-univalent electrolyte the Debye Huckel and Onsager equation is given below.
They considered that each ion is surrounded by an ionic atmosphere of opposite sign, and derived an expression relating the molar conductance of strong electrolytes with the concentration by assuming complete dissociations.
It was further developed by Onsager. For a uni-univalent electrolyte the Debye Huckel and Onsager equation is given below.
Where A and B are constants which depend only in the nature of the solvent and temperature.
Chapter doubts & questions for Electrochemistry - Physical Chemistry 2025 is part of Chemistry exam preparation. The chapters have been prepared according to the Chemistry exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Chemistry 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Electrochemistry - Physical Chemistry in English & Hindi are available as part of Chemistry exam.
Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
Physical Chemistry
83 videos|142 docs|67 tests
|
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup