Q1: Plotting against for aqueous solutions of a monobasic weak acid (HX) resulted in a straight line with y-axis intercept of P and slope of S. The ratio P/S is [JEE Advanced 2023 Paper 1]
[ = molar conductivity
= limiting molar conductivity
c = molar concentration
Ka = dissociation constant of HX]
(a)
(b)
(c)
(d)
Ans: (a)
Q1: Consider the strong electrolytes ZmXn , Um YP and Vm Xn . Limiting molar conductivity (Λ0) of Um Yp and Vm Xn are 250 and 440 S cm2 mol−1 , respectively. The value of (m + n + p) is
Given:
λ0 is the limiting molar conductivity of ions
The plot of molar conductivity (Λ) of Zm Xn vs c1/2 is given below. [JEE Advanced 2022 Paper 2]
Ans: 7
Given,
It is also given that
In the question, a graph of ( Λ ) of ZmXn Vs C1/2 is given,
For electrolyte ZmXn and from given curve
so
Solving eqn (2) and eqn (3),
⇒ n = 3
Substituting the value of n in eqn (2), we get,
Now, substituting the value of m is eqn (1), we get
Hence, the required value of m + n + p is 7.
Q2: The reduction potential l (E0 , in V) of MnO4− (aq) / Mn (s) is __________. [JEE Advanced 2022 Paper 1]
[Given:
Ans: 0.74 to 0.80
Adding (1), (2) and (3),
Q1: At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 102 S cm2 mol−1. At 298 K, for an aqueous solution of the acid the degree of dissociation is α and the molar conductivity is y × 102 S cm2 mol−1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y × 102 S cm2 mol−1. [JEE Advanced 2021 Paper 2]
The value of α is __________.
Ans: 0.22
Degree of dissociation = α
Limiting molar conductivity,
Molar conductivity,
Molar conductivity of dilution,
Concentration before dilution = C
Concentration after dilution = C/20
Dissociation constant,
Putting Eq. (i),
Dissociation constant before dillution,
Dissociation constant after dilution,
Comparing Eqs. (ii) and (iii),
Putting in Eq. (i),
The value of α is 0.22.
Q2: At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 × 102 S cm2 mol−1. At 298 K, for an aqueous solution of the acid the degree of dissociation is α and the molar conductivity is y × 102 S cm2 mol−1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y × 102 S cm2 mol−1. [JEE Advanced 2021 Paper 2]
The value of y is __________.
Ans: 0.86
Degree of dissociation = α
Limiting molar conductivity,
Molar conductivity,
Molar conductivity of dilution,
Concentration before dilution = C
Concentration after dilution = C/20
Dissociation constant,
Putting Eq. (i),
Dissociation constant before dillution,
Dissociation constant after dilution,
Comparing Eqs. (ii) and (iii),
Q3: Some standard electrode potentials at 298 K are given below :
Pb2+ /Pb − 0.13 V
Ni2+ /Ni − 0.24 V
Cd2+ /Cd − 0.40 V
Fe2+ /Fe − 0.44 V
To a solution containing 0.001 M of X2+ and 0.1 M of Y2+, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is(are)
(Given : Gas constant, R = 8.314 JK− mol−1, Faraday constant, F = 96500 C mol−1) [JEE Advanced 2021 Paper 2]
(a) Cd and Ni
(b) Cd and Fe
(c) Ni and Pb
(d) Ni and Fe
Ans: (a, b, c)
Given,
Nernst equation,
Ecell should be positive for a reaction to be spontaneous.
Reaction is spontaneous.
Reaction is spontaneous.
Reaction is spontaneous.
Reaction is non-spontaneous.
Therefore, the correct combinations of X and Y are (a), (b) and (c).
Q1: Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is
The work derived from the cell on the consumption of 1.0 × 10−3 mole of H2(g) is used to compress 1.00 mole of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in K) of the ideal gas?
The standard reduction potentials for the two half-cells are given below : [JEE Advanced 2020 Paper 1]
Ans: 13.32
Vessel is insulated, thus q = 0
For the given reaction :
= − 2 × 96500 × 1.23 J/mol
Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 × 10−3 mol of H2(g)
= 2 × 96500 × 1.23 × 0.7 × 1 × 10−3
= 166.17 J
This work done = change in internal energy (for monoatomic gas, Cv'm = 3R/2),
Q1: Molar conductivityof aqueous solution of sodium stearate, which behaves as a strong electrolyte, is recorded at varying concentrations (C) of sodium stearate. Which one of the following plots provides the correct representation of micelle formation in the solution?
(critical micelle concentration (CMC) is marked with an arrow in the figures) [JEE Advanced 2019 Paper 1]
(a)
(b)
(c)
(d)
Ans: (c)
At normal or low concentration, sodium stearate (CH3(CH2)16COO-Na+] behaves as strong electrolyte and for strong electrolyte, molar conductance decreases with increase in concentration. Above particular concentration, sodium stearate forms aggregates known as micelles. The concentration is called as CMC. Since, number of ions decreases and hence also decreases.
Hence, option (c) is correct.
Q1: Consider an electrochemical cell:
The value of ΔH∘ for the cell reaction is twice that of ΔG∘ at 300 K . If the emf of the cell is zero, the ΔS∘ (in JK−1 mol−1) of the cell reaction per mole of B formed at 300 K is ___________.
(Given: ln (2) = 0.7 , R (universal gas constant) = 8.3 JK−1 mol−1. H , S and G are enthalpy, entropy and Gibbs energy, respectively.) [JEE Advanced 2018 Paper 2]
Ans: -11.62
At 300 K, following electrochemical cell operates:
The reactions at:
(i) Anode:
(ii) Cathode :
Multiplying equation (i) by 2,
The net electrochemical cell is written as
Given : Enthalpy change ( ΔH∘) for cell reaction = 2 × Gibbs free energy than for cell reaction
According to the Nernst equation :
For a spontaneous reaction,
The change in entropy (ΔS∘) per mol of B is
Q2: For the electrochemical cell,
the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+ is changed to x M , the cell potential changes to 2.67 V at 300 K. The value of x is ___________.
(given, F/R = 11500 KV-1, where F is the Faraday constant and R is the gas constant, In (10 = 2.30) [JEE Advanced 2018 Paper 1]
Ans: 10
Equation of cell reaction according to the cell notation given, is
Given, E0cell = 2.70 V, T = 300 K with [Mg2+(aq)] = 1 M and [Cu2+(aq)] = 1 M and n = 2
Further, Ecell = 2.67 V with [Cu2+(aq)] = 1 M and [Mg2+(aq)] = xM and F/R = 11500 KV−1 where F = Faraday constant, R = gas constant
From the formula,
After putting the given values
or x = 10 (as given ln (10) = 2.30)
352 videos|596 docs|309 tests
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1. What is the Nernst equation and how is it used in electrochemistry? |
2. How does the concentration of electrolytes affect the conductivity of a solution in electrochemistry? |
3. What is the significance of the standard hydrogen electrode (SHE) in electrochemistry? |
4. How does the concept of overpotential affect the efficiency of electrochemical reactions? |
5. What is the role of the salt bridge in an electrochemical cell and why is it necessary? |
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