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Only One Option Correct Type
This section contains 29 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE is correct.
Two Faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (atomic mass of Cu = 63.5 amu)
(JEE Main 2015)
Cu formed by two Faraday =
1 mole = 63.5 g
Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 Sm-1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω The molar conductivity of 0.5 Msolution of the electrolyte in S mol-1 is
(JEE Main 2014)
The resistivity of the solution is related to specific conductance by
The specific conductance of the solution is given by
Now, the molar conductivity of the solution is given by
Hence, the molar conductivity of 0.5 M solution of the electrolyte is 5 x 10 -4 S m2 mol-1.
The equivalent conductance of NaCI at concentration C and at infinite dilution are λc and λ∞ respectively. The correct relationship between λc and λ∞ is given as
(JEE Main 2014)
Debye-Huckel Onsager equation can be written as
λc = Molar conductivity of the solution at certain concentration
λ∞ = Limiting molar conductivity
C = Concentration
B = Constant that depends on temperature, charges on the ions and dielectric constant as well as viscosity of the solution
Given below are the half-cell reactions
(JEE Main 2014)
Since, the value of E° is -ve, therefore the reaction is non-spontaneous.
Based on the data given above, strongest oxidising agent will be
Higher the SRP (Standard Reduction Potential), stronger is the oxidising agent. Among the given, electrode potentials, is highest. Hence, is the strongest oxidising agent.
The standard reduction potentials for Zn2+ / Zn, Ni2+ /Ni and Fe2+ / Fe are - 0.76, - 0.23 and - 0.44 V, respectively. The reaction X + y2+→ X2+ + Y will be spontaneous when
A cell reaction is spontaneous, if ΔG° < 0.
The reduction potential of hydrogen half-cell will be negative if
For hydrogen electrode,
Here, n = number of electrons involved in the reaction
Hence, when p(H2) is 2 atm and [H+] is 1.0 M, Ered is negative.
Resistance of 0.2 M solution of an electrolyte is 50 Ω The specific conductance of the solution is 1.3 Sm-1. If resistance of the 0.4 M solution of the same electrolyte is 260 Ω, its molar conductivity is
Molar conductivity of the solution is given by
The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows
The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least
The value of standard electrode potenital for the charge,
It is given that
The potential for the cell
Cr | Cr3+ (0.1 M) || Fe2+ (0.01 M) | Fe is
The equivalent conductances of two strong electrolytes at infinite dilution in H2O (where ions move freely through a solution) at 25°C a re given below
What additional information /quantity one needs to calculateof an aqueous solution of acetic acid?
Molar conductivity of aqueous CH3COOH solution can be calculated from Kohlraush’s law as
Hence, for determining the molar conductivity of CH3COOH solution, is required.
The cell, Zn | Zn2+ (1 M) || Cu2+ (1 M) | Cu
(E°cell = 1.10 V), was allowed to be completely discharged at 298 K. The relative concentration
Cell is completely discharged. It means equilibrium gets established, Ecell = 0
Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω. The conductivity of this solution is 1.29 Sm-1. Resistance of the same cell when filled with 0.2 M of the same solution is 520 Ω. The molar conductivity of 0.2 M solution of the electrolyte will be
For 0.1 M solution,
For 0.2 M solution
Given, ft = 520 Ω, C = 0.2 M,
μ (molar conductivity) = ?
k can be calculated as now cell constant is known]
The molar conductivities and at infinite dilution in water at 25°C are 91.0 and 426.2S cm2/mol, respectively. To calculate a additional value required is
According to Kohlrausch's law
Thus, on adding Eqs. (ii) and (ill),
subtracted, we can obtained the value of . Thus, additional value required is λ°NaCl.
Given the data at 25°C,
What is the value of log Ksp for Agl?
Calculate using appropriate molar conductances of the electrolytes listed above at infinite dilution in H2O at 25°C
Aluminium oxide may be electrolysed at 1000°C to furnish aluminium metal (atomic mass = 27 u; 1 Faraday = 96500 C). The cathode reaction is
To prepare 5.12 kg of aluminium metal by this method would require
where, w = amount of metal = 5.12 kg = 5.12 x 103 g
z = electrochemical equivalent
The limiting molar conductivities for NaCI, KBr and KCI are 126, 152 and 150S cm2 mol-1, respectively . The for NaBr is
By Kohlrausch’s law.
The E°m3+/m2+ values for Cr, Mn, Fe and Co are
- 0.41, + 1.57, + 0.77 and + 1.97 V,
respectively . For which one of these metals, the change in oxidation state from + 2 to + 3 is easiest?
More negative value of E°red indicates better reducing agent. Thus, easily oxidised. Hence, oxidation of Cr2+ to Cr3+ is the easiest.
Consider the following Eo values :
Under standard conditions, the potential for the reaction,
In a cell that utilises the reaction,
Zn (s)+ 2H + (aq) → Zn2+ (aq),+ H2(g)
addition of H2SO4 to cathode compartment will
If H2SO4 is added to cathodic compartment, (towards reactant side), then Q decrease (due to increase in H+).
Hence, equilibrium is displaced towards right and Ecell increases.
The standard emf of a cell, involving one electron change is found to be 0.591 V at 25° C. The equilibrium constant of the reaction is (F= 96500 C mol-1)
Relation between Keq and E°cell is
When during electrolysis of a solution of AgNO3, 9650 C of charge pass through the electroplating bath, the mass of silver deposited on the cathode will be
Standard reduction electrode potentials of three metals A, B and C are + 0.5 V, - 3.0 V and - 1.2 V, respectively. The reducing power of these metals are
More negative value of E° means higher reducing power.
For a cell reaction involving a two electron change, the standard emf of the cell is found to be 0.295 V at 25°C. The equilibrium constant of the reaction at 25°C will be
The standard emf of the cell,
Conductivity (Seimen’s S) is directly proportional to area of the vessel and the concentration of the solution in it and is inversely proportional to the length of the vessel, then constant of proportionality is expressed in
At LHS (oxidation)
Note E° values remain constant when half-cell equation is multiplied/divided.
For the following cell with hydrogen electrodes at two different pressures p1 and p2
emf is given by