मराठी

Calculate the λ0m for Cl- ion from the data given below: ∧0m MgCl2 = 258.6 Scm2 mol-1 and λ0m Mg2+ = 106 Scm2 mol-1

Advertisements
Advertisements

प्रश्न

Calculate the λ0m for Cl- ion from the data given below:

0m MgCl2 = 258.6 Scm2 mol-1 and λ0m Mg2+ = 106 Scm2 mol-1

संख्यात्मक
Advertisements

उत्तर

0m (MgCl2) = λ0m (Mg2+) + 2 λ0m (Cl-)

258.6 = 106 + 2 λ0m (Cl-)

λ0m (Cl-) = 76.3 Scm2 mol-1

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2023-2024 (March) Board Sample Paper

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

A current strength of 3.86 A was passed through molten Calcium oxide for 41minutes and 40 seconds. The mass of Calcium in grams deposited at the cathode is (atomic mass of Ca is 40g/mol and 1F = 96500 C).


A gas X at 1 atm is bubbled through a solution containing a mixture of 1 MY and 1 MZ at 25°C. If the reduction potential of Z > Y > X, then ______.


Describe the construction of Daniel cell. Write the cell reaction.


Use the data given in below find out which option the order of reducing power is correct.

`"E"_("Cr"_2"O"_7^(2-)//"Cr"^(3+))^⊖`= 1.33 V `"E"_("Cl"_2//"Cl"^-)^⊖` = 1.36 V

`"E"_("MnO"_4^-//"Mn"^(2+))^⊖` = 1.51 V `"E"_("Cr"^(3+)//"Cr")^⊖` = - 0.74 V


For the given cell, \[\ce{Mg | Mg^{2+} || Cu^{2+} | Cu}\]

(i) \[\ce{Mg}\] is cathode

(ii) \[\ce{Cu}\] is cathode

(iii) The cell reaction is \[\ce{Mg^+ Cu^{2+} -> Mg^{2+} + Cu}\]

(iv) \[\ce{Cu}\] is the oxidising agent


The electrochemical cell stops working after some time because


A current of 2.0 ampere passed for 5 hour through a molten salt deposits 22 g of the metal (Atomic mass = 177). The oxidation state of the metal in the metal salt is


Standard reduction electrode potentials of three metals A, B and C are respectively + 0.5 V, – 3.0 V. and – 1.2 V. The reducing powers of there metal is.


Read the passage given below and answer the questions that follow:

Oxidation-reduction reactions are commonly known as redox reactions. They involve transfer of electrons from one species to another. In a spontaneous reaction, energy is released which can be used to do useful work. The reaction is split into two half-reactions. Two different containers are used and a wire is used to drive the electrons from one side to the other and a Voltaic/Galvanic cell is created. It is an electrochemical cell that uses spontaneous redox reactions to generate electricity. A salt bridge also connects to the half-cells. The reading of the voltmeter gives the cell voltage or cell potential or electromotive force. If \[\ce{E^0_{cell}}\] is positive the reaction is spontaneous and if it is negative the reaction is non-spontaneous and is referred to as electrolytic cell. Electrolysis refers to the decomposition of a substance by an electric current. One mole of electric charge when passed through a cell will discharge half a mole of a divalent metal ion such as Cu2+. This was first formulated by Faraday in the form of laws of electrolysis.
The conductance of material is the property of materials due to which a material allows the flow of ions through itself and thus conducts electricity. Conductivity is represented by k and it depends upon nature and concentration of electrolyte, temperature, etc. A more common term molar conductivity of a solution at a given concentration is conductance of the volume of solution containing one mole of electrolyte kept between two electrodes with the unit area of cross-section and distance of unit length. Limiting molar conductivity of weak electrolytes cannot be obtained graphically.

  1. Is silver plate the anode or cathode?  (1)
  2. What will happen if the salt bridge is removed?  (1)
  3. When does electrochemical cell behaves like an electrolytic cell?  (1)
  4. (i) What will happen to the concentration of Zn2+ and Ag+ when Ecell = 0.   (1)
    (ii) Why does conductivity of a solution decreases with dilution?  (1)
    OR
    The molar conductivity of a 1.5 M solution of an electrolyte is found to be 138.9 S cm2mol-1. Calculate the conductivity of this solution.  (2)

Galvanic cell is a device in which ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×