Tag: electrochemistry

Questions Related to electrochemistry

Metallic conduction depends upon which of the following?

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. The nature of the metal

  2. Number of valence electrons per atom

  3. Density of metal

  4. Temperature

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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?

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. 26.805 h

  2. 153.61 h

  3. 107.22 h

  4. None of these

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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:

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. 26.8 hours

  2. 13.4 hours

  3. 6.7 hours

  4. 28.6 hours

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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?

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. 16 g

  2. 63 g

  3. 31.5 g

  4. 8 g

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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:

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. 10

  2. 50

  3. 30

  4. 96.5

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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?

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. Sulphuric acid

  2. Boric acid

  3. Nitric acid

  4. Phosphorous acid

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Correct Option: A

In metallic conductor the current is conducted by flow of:

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. ions

  2. atoms

  3. electrons

  4. molecules

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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.

The charge required to deposit 40.5 g of Al(atomic mass=27.0 g) from the fused ${ Al } _{ 2 }{ \left( { SO } _{ 4 } \right) } _{ 3 }$ is :

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. $4.34\times { 10 }^{ 5 }C$

  2. $43.4\times { 10 }^{ 5 }C$

  3. $1.44\times { 10 }^{ 5 }C$

  4. None of these

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Correct Option: A

Electrolytic conductivity of $0.3 M$ solution of $KCI$ at $298 K $is $3.72$ x ${10}^{-2}Scm^{-1}$.Calculate its molar conductivity ($S cm^{2} mol ^{-1}$) 

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. $124$

  2. $30.56$

  3. $192$

  4. $185$

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Correct Option: A
Explanation:

Given,

$k=3.72\times 10^{-2}\ Scm^{-1}$
$c=0.3M$

Molar conductivity($m$) is given by,

$m=\dfrac{k\times 1000}{c}$

$\Rightarrow m=\dfrac{3.72\times 10^{-2}\times 1000}{0.3}$

$\Rightarrow m=124\ S\ cm^2mol^{-1}$

Resistance of decimolar solution is 50 ohm. If electrodes of surface area 0.0004 $m^2$ each are placed at a distance of 0.02 m then conductivity of solution is : 

metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

  1. $1 \,s\,cm^{-1}$

  2. $0.01 \,s\,cm^{-1}$

  3. $0.001 \,s\,cm^{-1}$

  4. $10 \,s\,cm^{-1}$

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Correct Option: C