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Assertion: pure iron when heated in dry air is converted with a layer of rust. Reason: Rust has the compositionFe3O4. - Chemistry

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प्रश्न

Assertion: pure iron when heated in dry air is converted with a layer of rust.

Reason: Rust has the compositionFe3O4.

पर्याय

  • if both assertion and reason are true and reason is the correct explanation of assertion.

  • if both assertion and reason are true but reason is not the correct explanation of assertion.

  • assertion is true but reason is false.

  • both assertion and reason are false.

MCQ

उत्तर

both assertion and reason are false.

Explanation:

Both are false

  1. Dry air has no reaction with iron
  2. Rust has the composition \[\ce{Fe2O3.{x} H2O}\]
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पाठ 9: Electro Chemistry - Evaluation [पृष्ठ ६४]

APPEARS IN

सामाचीर कलवी Chemistry - Volume 1 and 2 [English] Class 12 TN Board
पाठ 9 Electro Chemistry
Evaluation | Q 15. | पृष्ठ ६४

संबंधित प्रश्‍न

At 25°C, the emf of the following electrochemical cell.

\[\ce{Ag_{(s)} | Ag^+ (0.01 M) | | Zn^{2+} {(0.1 M)} | Zn_{(s)}}\] will be:

(Given \[\ce{E^0_{cell}}\] = −1.562 V)


Define cathode


Is it possible to store copper sulphate in an iron vessel for a long time?

Given: \[\ce{E^0_{{Cu^{2+}|{Cu}}}}\] = 0.34 V and \[\ce{E^0_{{Fe^{2+}|{Fe}}}}\] = −0.44 V


A copper electrode is dipped in 0.1 M copper sulphate solution at 25°C. Calculate the electrode potential of copper.
[Given: \[\ce{E^0_{{Cu^{2+}|Cu}}}\] = 0.34 V]


Use the data given in below find out the most stable oxidised species.

`E^0 (Cr_2O_1^(2-))/(Cr_(3+))` = 1.33 V   `E^0 (Cl_2)/(Cl^-)` = 1.36 V

`E^0 (MnO_4^-)/(MN^(2+))` = 1.51 V   `E^0 (Cr^(3+))/(Cr)` = – 0.74 V


The quantity of charge required to obtain one mole of aluminium from Al2O3 is ______.


Match the terms given in Column I with the items given in Column II.

Column I Column II
(i) Λm (a) intensive property
(ii) ECell (b) depends on number of ions/volume
(iii) K (c) extensive property
(iv) ∆rGCell (d) increases with dilution

Cell reaction is spontaneous when


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)

The cell constant of a conductivity cell is 0.146 cm-1. What is the conductivity of 0.01 M solution of an electrolyte at 298 K, if the resistance of the cell is 1000 ohm?


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