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Molar conductivity for a compound AB is 145.0 Scm2mol-1 and for CB is 110.1 Scm2mol-1. Limiting molar conductivity for A+ is 73.5 Scm2mol-1. What is limiting molar conductivity for C+?- a)326.6 S cm2 mol-1
- b)38.6 S cm2 mol-1
- c)181.6 S cm2 mol-1
- d)90.8 S cm2 mol-1
Correct answer is option 'B'. Can you explain this answer?
Molar conductivity for a compound AB is 145.0 Scm2mol-1 and for CB is 110.1 Scm2mol-1. Limiting molar conductivity for A+ is 73.5 Scm2mol-1. What is limiting molar conductivity for C+?
a)
326.6 S cm2 mol-1
b)
38.6 S cm2 mol-1
c)
181.6 S cm2 mol-1
d)
90.8 S cm2 mol-1
|
Subhankar Choudhary answered |
Given:
Molar conductivity of AB = 145.0 Scm²mol⁻¹
Molar conductivity of CB = 110.1 Scm²mol⁻¹
Limiting molar conductivity of A = 73.5 Scm²mol⁻¹
To find: Limiting molar conductivity of C
Formula used:
Kohlrausch law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of the molar conductivity of its cation and anion, each multiplied by its own migration constant. That is,
Λ°(AB) = λ⁺(A) + λ⁻(B)
Λ°(CB) = λ⁺(C) + λ⁻(B)
where, λ⁺ and λ⁻ are the molar conductivities of the cation and anion, respectively, and the migration constants are represented by Greek letters.
Calculation:
Using Kohlrausch law,
λ⁺(A) + λ⁻(B) = 73.5 Scm²mol⁻¹
λ⁺(C) + λ⁻(B) = 110.1 Scm²mol⁻¹
Subtracting the first equation from the second equation,
(λ⁺(C) + λ⁻(B)) - (λ⁺(A) + λ⁻(B)) = 110.1 Scm²mol⁻¹ - 73.5 Scm²mol⁻¹
λ⁺(C) - λ⁺(A) = 36.6 Scm²mol⁻¹
Therefore, the limiting molar conductivity of C can be calculated as:
λ⁺(C) = λ⁺(A) + 36.6 Scm²mol⁻¹
λ⁺(C) = 73.5 Scm²mol⁻¹ + 36.6 Scm²mol⁻¹
λ⁺(C) = 110.1 Scm²mol⁻¹
Hence, the limiting molar conductivity of C is 110.1 Scm²mol⁻¹.
Therefore, the correct option is (B) 38.6 S cm²mol⁻¹.
Molar conductivity of AB = 145.0 Scm²mol⁻¹
Molar conductivity of CB = 110.1 Scm²mol⁻¹
Limiting molar conductivity of A = 73.5 Scm²mol⁻¹
To find: Limiting molar conductivity of C
Formula used:
Kohlrausch law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of the molar conductivity of its cation and anion, each multiplied by its own migration constant. That is,
Λ°(AB) = λ⁺(A) + λ⁻(B)
Λ°(CB) = λ⁺(C) + λ⁻(B)
where, λ⁺ and λ⁻ are the molar conductivities of the cation and anion, respectively, and the migration constants are represented by Greek letters.
Calculation:
Using Kohlrausch law,
λ⁺(A) + λ⁻(B) = 73.5 Scm²mol⁻¹
λ⁺(C) + λ⁻(B) = 110.1 Scm²mol⁻¹
Subtracting the first equation from the second equation,
(λ⁺(C) + λ⁻(B)) - (λ⁺(A) + λ⁻(B)) = 110.1 Scm²mol⁻¹ - 73.5 Scm²mol⁻¹
λ⁺(C) - λ⁺(A) = 36.6 Scm²mol⁻¹
Therefore, the limiting molar conductivity of C can be calculated as:
λ⁺(C) = λ⁺(A) + 36.6 Scm²mol⁻¹
λ⁺(C) = 73.5 Scm²mol⁻¹ + 36.6 Scm²mol⁻¹
λ⁺(C) = 110.1 Scm²mol⁻¹
Hence, the limiting molar conductivity of C is 110.1 Scm²mol⁻¹.
Therefore, the correct option is (B) 38.6 S cm²mol⁻¹.
Chemical used in salt bridge isa. KOHb. KCIc. KNO2d. KBrCorrect answer is option 'B'. Can you explain this answer?
|
Anand Saha answered |
KCl is used as salt bridge because it provides positive K+ ions and negative Cl- ions as the salt bridge needs to maintain the neutrality in the system by providing enough negative ions equal to the positive ions during oxidation.
The reduction potential of an element A is -2.71V.What can be concluded from this?- a)A will be a good oxidising agent
- b)A will accept electrons easily
- c)A will undergo reduction easily
- d)A will undergo oxidation easily
Correct answer is 'D'. Can you explain this answer?
The reduction potential of an element A is -2.71V.What can be concluded from this?
a)
A will be a good oxidising agent
b)
A will accept electrons easily
c)
A will undergo reduction easily
d)
A will undergo oxidation easily
|
Avantika Dasgupta answered |
Reduction Potential of Element A
The reduction potential of an element A is -2.71V. This means that the element A has a strong tendency to undergo oxidation and lose electrons. The more negative the reduction potential value, the easier it is for the element to undergo oxidation and lose electrons.
Explanation of Options
a) A will be a good oxidising agent - This statement is incorrect. A good oxidizing agent is one that accepts electrons from other species and undergoes reduction. But, since the reduction potential of element A is negative, it indicates that the element A is likely to undergo oxidation and lose electrons, so it is not a good oxidizing agent.
b) A will accept electrons easily - This statement is incorrect. The reduction potential value is negative, which means that the element A has a strong tendency to lose electrons and undergo oxidation, rather than accepting electrons and undergoing reduction.
c) A will undergo reduction easily - This statement is incorrect. The reduction potential value is negative, which means that the element A has a strong tendency to undergo oxidation and lose electrons, rather than undergoing reduction and gaining electrons.
d) A will undergo oxidation easily - This statement is correct. The reduction potential value of element A is negative, which means that the element A has a strong tendency to undergo oxidation and lose electrons. The more negative the reduction potential value, the easier it is for the element to undergo oxidation and lose electrons. Therefore, element A will undergo oxidation easily.
Conclusion
The correct answer is option 'D'. The reduction potential value of an element indicates its tendency to undergo oxidation or reduction. A negative reduction potential value indicates a strong tendency to undergo oxidation and lose electrons, while a positive reduction potential value indicates a strong tendency to undergo reduction and gain electrons.
The reduction potential of an element A is -2.71V. This means that the element A has a strong tendency to undergo oxidation and lose electrons. The more negative the reduction potential value, the easier it is for the element to undergo oxidation and lose electrons.
Explanation of Options
a) A will be a good oxidising agent - This statement is incorrect. A good oxidizing agent is one that accepts electrons from other species and undergoes reduction. But, since the reduction potential of element A is negative, it indicates that the element A is likely to undergo oxidation and lose electrons, so it is not a good oxidizing agent.
b) A will accept electrons easily - This statement is incorrect. The reduction potential value is negative, which means that the element A has a strong tendency to lose electrons and undergo oxidation, rather than accepting electrons and undergoing reduction.
c) A will undergo reduction easily - This statement is incorrect. The reduction potential value is negative, which means that the element A has a strong tendency to undergo oxidation and lose electrons, rather than undergoing reduction and gaining electrons.
d) A will undergo oxidation easily - This statement is correct. The reduction potential value of element A is negative, which means that the element A has a strong tendency to undergo oxidation and lose electrons. The more negative the reduction potential value, the easier it is for the element to undergo oxidation and lose electrons. Therefore, element A will undergo oxidation easily.
Conclusion
The correct answer is option 'D'. The reduction potential value of an element indicates its tendency to undergo oxidation or reduction. A negative reduction potential value indicates a strong tendency to undergo oxidation and lose electrons, while a positive reduction potential value indicates a strong tendency to undergo reduction and gain electrons.
The solubility of [Co(NH3)4Cl2] CIO4_________ if the 
= 50, 
= 70, and the measured resistance was 33.5Ω in a cell with cell constant of 0.20 is ____.- a)59.7 mmol/L
- b)49.7 mmol/L
- c)39.7 mmol/L
- d)29.7 mmol/L
Correct answer is option 'B'. Can you explain this answer?
The solubility of [Co(NH3)4Cl2] CIO4_________ if the = 50, = 70, and the measured resistance was 33.5Ω in a cell with cell constant of 0.20 is ____.
= 50, = 70, and the measured resistance was 33.5Ω in a cell with cell constant of 0.20 is ____.a)
59.7 mmol/L
b)
49.7 mmol/L
c)
39.7 mmol/L
d)
29.7 mmol/L
|
Ambition Institute answered |
The correct answer is option B
Given,
λCo(NH3)4Cl2+=50 λClo-4 =70
λ∞= λCo(NH3)4Cl2+ + λClo-4
λ∞= 50 + 70
λ∞=120
(x) Cell constant = 1/A
0.02 = l/A
Resistance(R) =33.5Ω
K =c.x (x = is cell constant)
S =49.7 mol/L
Given,
λCo(NH3)4Cl2+=50 λClo-4 =70
λ∞= λCo(NH3)4Cl2+ + λClo-4
λ∞= 50 + 70
λ∞=120
(x) Cell constant = 1/A
0.02 = l/A
Resistance(R) =33.5Ω
K =c.x (x = is cell constant)
S =49.7 mol/L
A half cell reaction A- → A + e- has a large negative reduction potential. It follows that :- a)A is easily reduced
- b)A – is easily reduced
- c)A – is easily oxidised
- d)A is easily oxidised
Correct answer is option 'C'. Can you explain this answer?
A half cell reaction A- → A + e- has a large negative reduction potential. It follows that :
a)
A is easily reduced
b)
A – is easily reduced
c)
A – is easily oxidised
d)
A is easily oxidised
|
Kalyan Chavan answered |
Can you please provide more details or context about the half-cell reaction you are referring to?
Hydrogen gas is not liberated when the following metal is added to dil. HCl.- a)Mg
- b)Sn
- c)Ag
- d)Zn
Correct answer is option 'C'. Can you explain this answer?
Hydrogen gas is not liberated when the following metal is added to dil. HCl.
a)
Mg
b)
Sn
c)
Ag
d)
Zn
|
Nikita Singh answered |
The metals, present below hydrogen in the electrochemical series, cannot liberate hydrogen from the dilute acids.
Among the given metals only Ag is present below hydrogen in electrochemical series, so it does not evolve hydrogen with dil HCl.
Ag−I−dilHCl ⟶ No reaction
Among the given metals only Ag is present below hydrogen in electrochemical series, so it does not evolve hydrogen with dil HCl.
Ag−I−dilHCl ⟶ No reaction
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.
Can you explain the answer of this question below:Nernst equation for an electrode is based on the variation of electrode potential of an electrode with:
- A:
temperature only
- B:
Concentration of electrolyte only
- C:
Both a and b
- D:
Density of the electrodes
The answer is c.
Nernst equation for an electrode is based on the variation of electrode potential of an electrode with:
temperature only
Concentration of electrolyte only
Both a and b
Density of the electrodes
|
|
Nikita Singh answered |
Nernst equation for an electrode is based on the variation of electrode potential of an electrode with temperature and concentration of electrolyte.
Temperature for the measurement of standard electrode potential is- a)298K
- b)300K
- c)30?C
- d)310K
Correct answer is option 'A'. Can you explain this answer?
Temperature for the measurement of standard electrode potential is
a)
298K
b)
300K
c)
30?C
d)
310K
|
Anaya Patel answered |
The standard electrode potentials are customarily determined at solute concentrations of 1 Molar, gas pressures of 1 atmosphere, and a standard temperature which is usually 25°C i.e, 298 K.
Cell reaction is spontaneous when- a)EθRed is positive
- b)ΔGθ is positive
- c)ΔGθ is negative
- d)EθRed is negative.
Correct answer is option 'C'. Can you explain this answer?
Cell reaction is spontaneous when
a)
EθRed is positive
b)
ΔGθ is positive
c)
ΔGθ is negative
d)
EθRed is negative.
|
|
Nandini Patel answered |
ΔGdegree must be negative for the reaction to be spontaneous.
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.
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
|
|
Vikram Kapoor 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
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
Salt bridge is indicated in the cell representation by :- a)I
- b)!!
- c)((
- d)II
Correct answer is option 'D'. Can you explain this answer?
Salt bridge is indicated in the cell representation by :
a)
I
b)
!!
c)
((
d)
II
|
|
Khushi Pandey answered |
Indicàted by Twø parallel linés (||)
. In the construction of a salt bridge, saturated solution of KNO3 is used because:- a)Velocity of K+ and NO3– are same
- b)Velocity of NO3– is greater than that of K+
- c)Velocity of K+ is greater than that of NO3–
- d)KNO3 is highly soluble in water
Correct answer is option 'A'. Can you explain this answer?
. In the construction of a salt bridge, saturated solution of KNO3 is used because:
a)
Velocity of K+ and NO3– are same
b)
Velocity of NO3– is greater than that of K+
c)
Velocity of K+ is greater than that of NO3–
d)
KNO3 is highly soluble in water
|
|
Riya Agarwal answered |
Velocities of both should be same to balance the amount of both ions in the soln. if the vel of any of them is more...then its ions will release more
Conductivity is represented by the symbol- a)P
- b)Δ
- c)λ
- d)κ
Correct answer is option 'D'. Can you explain this answer?
Conductivity is represented by the symbol
a)
P
b)
Δ
c)
λ
d)
κ
|
Rajat Patel answered |
Units of Electrical Conductivity. Electrical conductivity is denoted by the symbol σ and has SI units of siemens per meter (S/m). In electrical engineering, the Greek letter κ is used. Sometimes the Greek letter γ represents conductivity.
The free energy change for the following cell reaction is given as :
2Au3+ (aq) + 3Cu (s) → 2Au (s) + 3Cu2+ (aq)
- a)6 FE°cell
- b)3 FE°cell
- c)-2 FE°cell
- d)-6 FE°cell
Correct answer is option 'D'. Can you explain this answer?
The free energy change for the following cell reaction is given as :
2Au3+ (aq) + 3Cu (s) → 2Au (s) + 3Cu2+ (aq)
2Au3+ (aq) + 3Cu (s) → 2Au (s) + 3Cu2+ (aq)
a)
6 FE°cell
b)
3 FE°cell
c)
-2 FE°cell
d)
-6 FE°cell
|
|
Preeti Iyer answered |
The correct answer is Option D.
EO = EOCa2+/ Ca - EOAu2+/ Au
= -2.87 - (1.50)
= -2.87 - 1.50
= -4.37 V
rGO = -nFEO
= -6 FEO
= -2.87 - (1.50)
= -2.87 - 1.50
= -4.37 V
rGO = -nFEO
= -6 FEO
Select the correct statement(s).- a)
(molar conductance at infinite dilution) of strong electrolytes as well as weak electrolytes can be obtained by Kohlrausch’s law. 
- b) of strong as well as weak electrolytes can be obtained by extrapolation of graph between
- c)
can be obtained by known value of 
- d)
Correct answer is option 'A,C,D'. Can you explain this answer?
Select the correct statement(s).
a)
(molar conductance at infinite dilution) of strong electrolytes as well as weak electrolytes can be obtained by Kohlrausch’s law. b)
of strong as well as weak electrolytes can be obtained by extrapolation of graph between c)
can be obtained by known value of d)
|
|
Anaya Patel answered |
For strong electrolyte, variation in graph is uniform, hence
can be extrapolated to 
In case of weak electrolyte, variation of
is rapid in lower concentration range (higher dilution), hence graph is not uniform and 
cannot be obtained by extrapolation to
Thus, (b) is incorrect.
Thus, correct based on Kohlrausch’s law.
Thus, it is also correct based on Kohlrausch's law.
Comprehension TypeDirection : This section contains 2 paragraphs, each describing theory, experiments, data, etc. Four questions related to the paragraphs have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).Passage IConsider the following solutions of an electrolyte
Q. Conductivity.of 0.2 M solution is- a)2.60 Sm-1
- b)0.25 Sm-1
- c)4.00 Sm-1
- d)0.50 Sm-1
Correct answer is option 'B'. Can you explain this answer?
Comprehension Type
Direction : This section contains 2 paragraphs, each describing theory, experiments, data, etc. Four questions related to the paragraphs have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d).
Passage I
Consider the following solutions of an electrolyte
Q.
Conductivity.of 0.2 M solution is
a)
2.60 Sm-1
b)
0.25 Sm-1
c)
4.00 Sm-1
d)
0.50 Sm-1
|
|
Krishna Iyer answered |
For 0.1 M solution, k = 1.30 Sm-1
Conductivity (k) = Conductance x Cell constant
Conductivity (k) = Conductance x Cell constant
Consider the following reaction which of the following statement is true for this cell reaction.
(Zn + Cu2+ → Zn2+ + Cu)- a)Zn2+ ions are oxidized to Zn
- b)Zn is oxidized to Zn2+ ions
- c)Zn is reduced to Zn2+ ions
- d)Cu2+ ions are oxidized to Cu
-
Correct answer is option 'B'. Can you explain this answer?
Consider the following reaction which of the following statement is true for this cell reaction.
(Zn + Cu2+ → Zn2+ + Cu)
(Zn + Cu2+ → Zn2+ + Cu)
a)
Zn2+ ions are oxidized to Zn
b)
Zn is oxidized to Zn2+ ions
c)
Zn is reduced to Zn2+ ions
d)
Cu2+ ions are oxidized to Cu
|
Geetika Shah answered |
- For the reaction Zn + Cu2+ → Zn2+ + Cu, Zn is oxidized to Zn2+ while Cu2+ is reduced to Cu
- In a redox reaction, the reactant that loses electrons (is oxidized) causes a reduction and is called a reducing agent. In the example above, zinc metal is the reducing agent; it loses two electrons (is oxidized) and becomes Zn2+ ion.
The reduction potential of an element A is -2.71V.What can be concluded from this?- a)A will be a good oxidising agent
- b)A will accept electrons easily
- c)A will undergo reduction easily
- d)A will undergo oxidation easily
Correct answer is option 'D'. Can you explain this answer?
The reduction potential of an element A is -2.71V.What can be concluded from this?
a)
A will be a good oxidising agent
b)
A will accept electrons easily
c)
A will undergo reduction easily
d)
A will undergo oxidation easily
|
Dr Manju Sen answered |
Reduction potential means to accept electrons to reduce oneself.
A + e- → A- ∆Ereduction = +ve value
Since, the reduction potential is negative, it means that the reaction will reverse to make ∆E value +ve. So the reaction becomes,
A → A+ + e-
This becomes oxidation of A. So oxidation of A will be easy.
A + e- → A- ∆Ereduction = +ve value
Since, the reduction potential is negative, it means that the reaction will reverse to make ∆E value +ve. So the reaction becomes,
A → A+ + e-
This becomes oxidation of A. So oxidation of A will be easy.
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 reduction potential of an element A is 1.71 V. What can be concluded from this?- a)A will lose electrons easily
- b)A will undergo reduction easily
- c)A will undergo oxidation easily
- d)A will be a good reducing agent
Correct answer is option 'B'. Can you explain this answer?
The reduction potential of an element A is 1.71 V. What can be concluded from this?
a)
A will lose electrons easily
b)
A will undergo reduction easily
c)
A will undergo oxidation easily
d)
A will be a good reducing agent
|
|
Rajeev Saxena answered |
The standard reduction potential is the tendency for a chemical species to be reduced, and is measured in volts at standard conditions. The more positive the potential is the more likely it will be reduced. Hence, A will undergo reduction easily.
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.
Resistance of 0.2 M soluton of an electrolyte is 50?. The specific conductance of solution is 1.3 Sm-1. If resistance of the 0.4 M solution of the same electrolyte is 260Ω, its molar conductivity is[AlEEE 2011]- a)6250 S m2 mol-1
- b)6.25 x 10-4 S m2 mol-1
- c)62.5 x 10-4 S m2 mol-1
- d)62.5 S m2 mol-1
Correct answer is option 'B'. Can you explain this answer?
Resistance of 0.2 M soluton of an electrolyte is 50?. The specific conductance of solution is 1.3 Sm-1. If resistance of the 0.4 M solution of the same electrolyte is 260Ω, its molar conductivity is
[AlEEE 2011]
a)
6250 S m2 mol-1
b)
6.25 x 10-4 S m2 mol-1
c)
62.5 x 10-4 S m2 mol-1
d)
62.5 S m2 mol-1
|
|
Gaurav Kumar answered |
Specific conductance (k) of 0.2 M solution
= Conductance x Cell constant
Specific conductance of 0.4M solution
= Conductance x Cell constant
Specific conductance of 0.4M solution
The gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y- and 1 M Z- at 298 K.If the standard reduction potential
then,- 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)Z- will reduce both X and Y
Correct answer is option 'D'. Can you explain this answer?
The gas X at 1 atm is bubbled through a solution containing a mixture of 1 M Y- and 1 M Z- at 298 K.If the standard reduction potential
then,
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)
Z- will reduce both X and Y
|
|
Nikita Singh answered |
In ECS, pair with more negative values of E°red reducing agent is above oxidising agent.
Thus, Z/Z- is the best reducing agent
Thus, Z- will reduce both X and Y and itself will be oxidised to Z .
At 300K molar conductivity of solution A is 350 units, and at infinite dilution the molar conductivity of the same sample is 480 unit. Predict the percentage dissociation of the electrolyte.- a)73.0%
- b)37.0%
- c)63.0%
- d)137.0%
Correct answer is option 'A'. Can you explain this answer?
At 300K molar conductivity of solution A is 350 units, and at infinite dilution the molar conductivity of the same sample is 480 unit. Predict the percentage dissociation of the electrolyte.
a)
73.0%
b)
37.0%
c)
63.0%
d)
137.0%
|
Gunjan Lakhani answered |
Stronger the oxidizing agent, greater is the:- a)Reactivity
- b)Ionic behaviour
- c)Oxidation potential
- d)Reduction potential
Correct answer is option 'D'. Can you explain this answer?
Stronger the oxidizing agent, greater is the:
a)
Reactivity
b)
Ionic behaviour
c)
Oxidation potential
d)
Reduction potential
|
|
Rajeev Saxena answered |
Lithium is strongest Reducing agent because of lowest standard reduction potential. When something is oxidized, it reduces another substance, becoming a reducing agent. Hence lithium is the strongest reducing agent. remember, Li is the strongest reducing agent and F is the strongest oxidizing agent!
Which of the following relations are not correct?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Which of the following relations are not correct?
a)
b)
c)
d)
|
|
Nandini Iyer answered |
The correct answer is Option B
k =G*l/A (Since, G = 1/R)
= l/AR
Therefore, kR = l/A
So, l/A = kR not k/R
k =G*l/A (Since, G = 1/R)
= l/AR
Therefore, kR = l/A
So, l/A = kR not k/R
EMF of the following cell is 0.2905 V 
The equilibrium constant for the cell reaction is [IIT JEE 2004]- a)100.32/0.059
- b)100.32/0.0295
- c)100.26/0.0295
- d)100.32/0.295
Correct answer is option 'B'. Can you explain this answer?
EMF of the following cell is 0.2905 V
The equilibrium constant for the cell reaction is
[IIT JEE 2004]
a)
100.32/0.059
b)
100.32/0.0295
c)
100.26/0.0295
d)
100.32/0.295
|
Dilip Chaurasiya answered |
E0 cell = .0591/n log k. k= equilibrium constant
Which of the following statements is incorrect about cell constant?- a)It depends on the distance between the electrodes and their area of cross section
- b)It is represented by the symbol G*
- c)It has the dimensions length-2
- d)It is calculated by the expression G* = l/A
Correct answer is option 'C'. Can you explain this answer?
Which of the following statements is incorrect about cell constant?
a)
It depends on the distance between the electrodes and their area of cross section
b)
It is represented by the symbol G*
c)
It has the dimensions length-2
d)
It is calculated by the expression G* = l/A
|
Om Desai answered |
The correct answer is Option C.
Cell constant can be defined as the ratio of the distance between conductance-titration electrodes to the area of the electrodes, measured from the determined resistance of a solution of known specific conductance. It has dimension length-1
Cell constant can be defined as the ratio of the distance between conductance-titration electrodes to the area of the electrodes, measured from the determined resistance of a solution of known specific conductance. It has dimension length-1
For the cell, 
and for the cell Pt(H2) | H+ (1M)| Ag, 
Thus Ecell for theAg|Ag+ (0.1M) || Zn2+ (0.1M) | Zn is ....................and cell reaction is...............- a)1.44 V ,spontaneous
- b)0.4 V ,spontaneous
- c)-1.44 V ,non-spontaneous
- d)-1.53 V ,non-spontaneous
Correct answer is option 'D'. Can you explain this answer?
For the cell, and for the cell Pt(H2) | H+ (1M)| Ag,
and for the cell Pt(H2) | H+ (1M)| Ag, Thus Ecell for the
Ag|Ag+ (0.1M) || Zn2+ (0.1M) | Zn is ....................and cell reaction is...............
a)
1.44 V ,spontaneous
b)
0.4 V ,spontaneous
c)
-1.44 V ,non-spontaneous
d)
-1.53 V ,non-spontaneous
|
|
Geetika Shah answered |
Ecell < 0, hence reaction is non-spontaneous.
In the equation, ΔG° = – nF E° cell ; F is:- a)Boltzmann constant
- b)Faraday’s constant
- c)Gas constant
- d)Universal gas constant
Correct answer is option 'B'. Can you explain this answer?
In the equation, ΔG° = – nF E° cell ; F is:
a)
Boltzmann constant
b)
Faraday’s constant
c)
Gas constant
d)
Universal gas constant
|
|
Nandini Iyer answered |
The correct answer is Option B.
The relationship between ΔGo and Eo is given by the following equation: ΔGo=−nFEo. Here, n is the number of moles of electrons and F is the Faraday constant.
The relationship between ΔGo and Eo is given by the following equation: ΔGo=−nFEo. Here, n is the number of moles of electrons and F is the Faraday constant.
Consider the cell reaction:
Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0 m) | Cu (s)
If we wish to make a cell with more positive voltage using the same substances, we should:
- a)Increase [Cd2+] as well as [Cu2+] to 2.0 M
- b)Increase only [Cu2+] to 2.0 M
- c)Reduce only [Cd2+] to 0.1 M
- d)Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0M
Correct answer is option 'D'. Can you explain this answer?
Consider the cell reaction:
Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0 m) | Cu (s)
If we wish to make a cell with more positive voltage using the same substances, we should:
Cd(s) | Cd2+ (1.0 M) || Cu2+ (1.0 m) | Cu (s)
If we wish to make a cell with more positive voltage using the same substances, we should:
a)
Increase [Cd2+] as well as [Cu2+] to 2.0 M
b)
Increase only [Cu2+] to 2.0 M
c)
Reduce only [Cd2+] to 0.1 M
d)
Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0M
|
|
Geetika Shah answered |
The correct answer is Option D.
Redox reaction:
Cd(s)→Cd2++2e
Cu2++2e→Cu(s)
Ecell = E°cell − (0.059/2) log ([Cd2+]/ [Cu2+])
Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0M
Cd(s)→Cd2++2e
Cu2++2e→Cu(s)
Ecell = E°cell − (0.059/2) log ([Cd2+]/ [Cu2+])
Decreases [Cd2+] to 0.1M and increases [Cu2+] to 1.0M
The electrode potential measures the :- a)tendency of a cell reaction to occur
- b)current carried by an elelctrode
- c)tendency of the electrode to gain or lose electrons
- d)difference in the ionisation of electrode and metal ion
Correct answer is option 'C'. Can you explain this answer?
The electrode potential measures the :
a)
tendency of a cell reaction to occur
b)
current carried by an elelctrode
c)
tendency of the electrode to gain or lose electrons
d)
difference in the ionisation of electrode and metal ion
|
|
Nandini Patel answered |
The tendency of an electrode to lose or gain electrons when it is in contact with its own ions in solution is called electrode potential.
Since the tendency to lose electrons means also the tendency to get oxidised, this tendency is called oxidation potential. Similarly, the tendency to gain electrons means the tendency to get reduced. Hence this tendency is called reduction potential.
Kohlrausch’s Law shows that:- a)at infinite dilution the ionic conductivity of ions is additive.
- b)at infinite dilution the ionic conductivity of all the ions of the electrolyte become equal.
- c)at infinite dilution the concentration of the electrolyte becomes unity.
- d)at infinite dilution the concentration of ions increases.
Correct answer is option 'A'. Can you explain this answer?
Kohlrausch’s Law shows that:
a)
at infinite dilution the ionic conductivity of ions is additive.
b)
at infinite dilution the ionic conductivity of all the ions of the electrolyte become equal.
c)
at infinite dilution the concentration of the electrolyte becomes unity.
d)
at infinite dilution the concentration of ions increases.
|
|
Preeti Khanna answered |
The correct answer is option A
Kohlrausch's law states that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the anions and cations. If a salt is dissolved in water, the conductivity of the solution is the sum of the conductances of the anions and cations.
Hence, at infinite dilution the ionic conductivity of ions is additive.
Hence, at infinite dilution the ionic conductivity of ions is additive.
In an electrochemical cell, the electrode having a lower reduction potential will act as:- a)Salt bridge
- b)Electrolyte
- c)Anode
- d)Cathode
Correct answer is option 'C'. Can you explain this answer?
In an electrochemical cell, the electrode having a lower reduction potential will act as:
a)
Salt bridge
b)
Electrolyte
c)
Anode
d)
Cathode
|
Sargam Singh answered |
A substance with lower reduction potential has more tendency to oxidize .in a electrochemical cell anode performs oxidation reaction hence the electrode will function as a anode
Gibbs free energy change for a cell reaction is positive what does it indicates?- a)cell will discharge easily
- b)Cell reaction is spontaneous
- c)Cell reaction is non spontaneous
- d)Cell will work under standard conditions
Correct answer is option 'C'. Can you explain this answer?
Gibbs free energy change for a cell reaction is positive what does it indicates?
a)
cell will discharge easily
b)
Cell reaction is spontaneous
c)
Cell reaction is non spontaneous
d)
Cell will work under standard conditions
|
|
Rajeev Saxena answered |
No, reaction cannot be spontaneous (continue to happen) when the change in gibbs free energy is positive. ... For a spontaneous process to happen , the change in Gibbs free energy must be negative. A roaring bonfire is an example of a spontaneous reaction.
Three cell A, B and C has equilibrium constant in the ratio 1:4 : 9 respectively. Arrange the following cells in the order of increasing Gibbs free energy.- a)A>B>C
- b)Cannot be answered
- c)B>C>A
- d)A
Correct answer is option 'A'. Can you explain this answer?
Three cell A, B and C has equilibrium constant in the ratio 1:4 : 9 respectively. Arrange the following cells in the order of increasing Gibbs free energy.
a)
A>B>C
b)
Cannot be answered
c)
B>C>A
d)
A
|
|
Nandini Patel answered |
A>B>C
Smaller the value of equilibrium constant (k) larger will be value of Gibbs free energy.
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 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
During electrolysis, the reaction that takes place at cathode is:- a)Hydrolysis
- b)Reduction
- c)Oxidation
- d)Neutralization
Correct answer is option 'B'. Can you explain this answer?
During electrolysis, the reaction that takes place at cathode is:
a)
Hydrolysis
b)
Reduction
c)
Oxidation
d)
Neutralization
|
|
Rajat Patel answered |
The electrode at which oxidation takes place is known as the anode, while the electrode at which reduction take place is called the cathode. If you see galvanic cell reduction take place at the left electrode, so the left one is the cathode. Oxidation takes place at the right electrode, so the right one is the anode.
For the following cell with hydrogen electrodes at two different pressure p1 and p2 
emf is given by- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
For the following cell with hydrogen electrodes at two different pressure p1 and p2
emf is given by
a)
b)
c)
d)
|
|
Krishna Iyer answered |
For SHE E°SHE = 0.00 V
Oxidation at anode (left)
Reduction at cathode (right)
Net
Oxidation at anode (left)
Reduction at cathode (right)
Net
This is the type of the cell in which electrodes at different pressures are dipped in same electrolyte and connectivity is made by a salt-bridge.
Reaction Quotient (Q) 
∵
During electrolysis, the reaction that takes place at anode is:- a)Reduction
- b)Neutralization
- c)Hydrolysis
- d)Oxidation
Correct answer is option 'D'. Can you explain this answer?
During electrolysis, the reaction that takes place at anode is:
a)
Reduction
b)
Neutralization
c)
Hydrolysis
d)
Oxidation
|
|
Rajat Patel answered |
Oxidation takes place at the right electrode, so the right one is the anode. While in electrolytic cell reduction takes place at the right electrode, so right one is the cathode. Oxidation takes place at the left electrode, so the left one is the anode.
The ionic conductivity of the cation and anion of the uni-univalent salt is 140 and 80 ohm-1 cm2 mol-1 respectively. The molar conductivity of the salt is:- a)280 ohm-1 cm2 mol-1
- b)60 moles
- c)160 ohm-1 cm2 mol-1
- d)220 ohm-1 cm2 mol-1
Correct answer is option 'D'. Can you explain this answer?
The ionic conductivity of the cation and anion of the uni-univalent salt is 140 and 80 ohm-1 cm2 mol-1 respectively. The molar conductivity of the salt is:
a)
280 ohm-1 cm2 mol-1
b)
60 moles
c)
160 ohm-1 cm2 mol-1
d)
220 ohm-1 cm2 mol-1
|
Hansa Sharma answered |
The correct answer is option D
The molar conductivity of uni-univalent electrolyte like AB = sum of their ionic conductivities = 140 + 80 = 220 ohm-1 cm2 mol-1
The molar conductivity of uni-univalent electrolyte like AB = sum of their ionic conductivities = 140 + 80 = 220 ohm-1 cm2 mol-1
At equilibrium:- a)Cell potential’ E cell‘ becomes zero
- b)Equilibrium constant becomes equal to electrode potential
- c)Equilibrium constant becomes zero
- d)Cell potential ‘E cell‘ becomes unity
Correct answer is option 'A'. Can you explain this answer?
At equilibrium:
a)
Cell potential’ E cell‘ becomes zero
b)
Equilibrium constant becomes equal to electrode potential
c)
Equilibrium constant becomes zero
d)
Cell potential ‘E cell‘ becomes unity
|
Ayush Joshi answered |
E cell is 0 in equilibrium that is E cathode becomes equal to E anode ………. ... So E cell is zero at equilibrium that is when the E(cathode) becomes equal to E(anode). E deg cell is zero in the concentration cell when both the electrodes are of the same metal.
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Chemistry for BMAT (Section 2)
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