Tag: metallic and electrolytic conduction

Questions Related to metallic and electrolytic conduction

Electrical conductivity of $Se$ is negligible in dark but increases to exposure to light. Because of this property it is used in:

  1. photoelectric cells

  2. semiconductors

  3. high voltage batteries

  4. lasers


Correct Option: A
Explanation:

In photoelectric cells, light energy is converted into electrical energy .Thus $Se$ is used in photocells because of its behaviour to become a conductor in presence of light.

Pure water is a good conductor of electricity.

  1. True

  2. False


Correct Option: B
Explanation:

For pure water ${ K } _{ a }=1.8\times { 10 }^{ -16 }$ at $25℃$

Since dissociation constant is very very low.
So pure water doesn't conduct electricity. Because free ions are not present in sufficient amount for conduction.
${ H } _{ 2 }{ O\rightleftharpoons  }{ H }^{ + }+{ OH }^{ - }$ (weak dissociation constant)

Molten sodium chloride conducts electricity due to presence of:

  1. free electrons

  2. free ions

  3. free molecules

  4. free atoms


Correct Option: B
Explanation:
Molten sodium chloride conducts electricity due to the presence of free ions.
In order to conduct electricity a substance must have charge particles, such as electrons and ions, that are free to move freely through it. In the solid state, ionic compounds such as sodium chloride have their ions fixed in position and therefore these ions cannot move so solid ionic compounds cannot conduct electricity. However in the molten state, ions in ionic compounds are free to flow and therefore molten sodium chloride can conduct electricity.

Metallic conduction depends upon which of the following?

  1. The nature of the metal

  2. Number of valence electrons per atom

  3. Density of metal

  4. Temperature


Correct Option: A
Explanation:

For metals the thermal conductivity is mainly a function of the motion of free electrons.As the temperature increases the molecular vibrations increases so they obstruct the free flow of electrons thus reducing conductivity.

How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere?

  1. 26.805 h

  2. 153.61 h

  3. 107.22 h

  4. None of these


Correct Option: D
Explanation:
Solution:- (D) none of these
$2 {H} _{2}O \longrightarrow 2 {H} _{2} + {O} _{2}$
From the above reaction-
$1$ mole of ${H} _{2}O$ exchanges $2$ moles of electrons, then $2$ moles of ${H} _{2}O$ will exchange $4$ moles of electrons.
From Faraday's law of electrolysis,
$q = nF$
$\Rightarrow i \times t = nF \; \left( \because q = i \times t \right)$
$\Rightarrow t = \cfrac{nF}{i} = \cfrac{4 \times 96500}{2} = 193000 \; s = 53.61 \; hr$
Hence the time required is $53.61$ hours.

A dilute solution of $H _2SO _4$ was electrolyzed by passing a current of 2 amp. The time required for formation of 0.5 mole of oxygen is:

  1. 26.8 hours

  2. 13.4 hours

  3. 6.7 hours

  4. 28.6 hours


Correct Option: A
Explanation:
Let current of $2$ amp is passed through the solutions for $t$ seconds
$\therefore$   Charge passed $=2\times t$

$\therefore$   moles of electrons passed $=\dfrac { 2t }{ 96500 } $
At anode :

${ 2OH }^{ \left( - \right)  }\rightarrow 1/2{ O } _{ 2 }+{ H } _{ 2 }O+{ 2e }^{ - }$
$\therefore$   moles of ${ O } _{ 2 }$ released at anode $=\dfrac { 2t }{ 96500 } \times \dfrac { 1 }{ 4 } $

$\therefore$   $\dfrac { 2t }{ 96500 } \times \dfrac { 1 }{ 4 } =0.5$
$\Rightarrow t=96500$ secs $=26.8$ hours

$\therefore$   The time required for formation of $0.5$ mole of ${ O } _{ 2 }$ is $26.8$ hours.

Hence, the correct option is A.

A current being passed for two hour through a solution of an acid liberating 11.2 litre of oxygen at NTP at anode. What will be the amount of copper deposited at the cathode by the same current when passed through a solution of copper sulphate for the same time?

  1. 16 g

  2. 63 g

  3. 31.5 g

  4. 8 g


Correct Option: B
Explanation:

Solution:- (B) $63 \; g$

At STP,
$1$ mole of oxygen gas $= 32 \; g = 22.4 \; L$
Therefore,
$11.2 \; L$ of ${O} _{2} = \cfrac{32}{22.4} \times 11.2 = 16 \; g$
Therefore,
$\cfrac{{M} _{Cu}}{{M} _{{O} _{2}}} = \cfrac{{E} _{Cu}}{{E} _{O}}$
$\Rightarrow \cfrac{{M} _{Cu}}{16} = \cfrac{\left( \cfrac{63}{2} \right)}{\left( \cfrac{16}{2} \right)}$
$\Rightarrow {M} _{Cu} = 63 \; g$
Hence the amount of copper deposited at the cathode is $63 \; g$.

A current of 9.65 Ampere flowing for 10 minutes deposits 3 g of metal which is monovalent, the atomic mass of metal is a:

  1. 10

  2. 50

  3. 30

  4. 96.5


Correct Option: B
Explanation:

Solution:- (B) $50$

Weight of metal deposited $= 3 \; g$
Quantity of electricity passed $\left( q \right)$-
$q = I \times t$
Given:-
$I = 9.65 A$
$t = 10 \; min = 10 \times 60 = 600 \; sec$
$\therefore q = 9.65 \times 600 = 5790 \; C$
Now, as we know that,
Eq. wt. of metal $= \cfrac{\text{Weight of metal}}{q} \times F$
$\Rightarrow$ Eq. wt. of metal $= \cfrac{3}{5790} \times 96500 = 50 \; g$
As the metal is monovalent, atomic weight of metal will be equal to its equivalent weight.
Therefore,
Atomic weight of metal $= 50 \; g$

Which of the following ($1\ M$) conducts more electricity?

  1. Sulphuric acid

  2. Boric acid

  3. Nitric acid

  4. Phosphorous acid


Correct Option: A

In metallic conductor the current is conducted by flow of:

  1. ions

  2. atoms

  3. electrons

  4. molecules


Correct Option: C
Explanation:

In metallic conductor the current is conducted by flow of free electrons.Ions contain the free electron therefore they conducts the electricity.