Tag: chemical changes

Questions Related to chemical changes

Electrolysis rules of Faraday's states that mass depends on electrodes is proportional to:-

  1. $m \propto I^2$

  2. $m \propto Q$

  3. $m \propto Q^2$

  4. None of these


Correct Option: B

The process of electrolysis is used in:

  1. extraction of metals

  2. electroplating

  3. refining of metals

  4. all of the above


Correct Option: D
Explanation:

Answer is D.
Electrolysis has wide applications in industries. Some of the important applications are, as follows, 
(i) Production of hydrogen by electrolysis of water. 
(ii) Manufacture of heavy water. 
(iii) The metals like K, Mg, Al, etc., are obtained by electrolysis of fused electrolytes. 
(iv) Non-metals like hydrogen, fluorine, chlorine are obtained by electrolysis. 
(v) In this method pure metal is deposited at cathode from a solution containing the metal ions, etc. 
(vi) Compounds like NaOH, KOH, white lead, etc. are synthesised by electrosynthesis method.
(vii) Electroplating: The process of coating an inferior metal with a superior metal by electrolysis is known as electroplating. 
Hence, the options A, B and C are correct.

Electromeric effect involves the completer transfer of 

  1. $\sigma $ electorn

  2. $\pi $ electron

  3. proton

  4. Both $\sigma $ and $\pi $ electrons


Correct Option: B

Refining an impure metal to give a pure metal is known as:

  1. electro-refining

  2. electrode refining

  3. electroplating

  4. none of the above


Correct Option: A
Explanation:

Refining an impure metal to give a pure metal is known as electrorefining.

The most durable metal plating on iron to protect against corrosion is:

  1. tin plating

  2. zinc plating

  3. copper plating

  4. nickel plating


Correct Option: B
Explanation:

Metal plating protects the iron against corrosion by forming a sacrificial layer. This metal layer itself gets oxidize instead of iron and thus protects the iron itself. 


For a metal to be used as plating, its oxidation should be preferred over iron and thus should have lower reduction potential than iron. 

Among the given options, zinc has a lower reduction potential than iron and thus can be used for plating. The process is called galvanization.

Hence, option B is correct. 

Extracting metals in the pure form from its compound is known as:

  1. electrorefining

  2. electrochemicals

  3. extraction of metals

  4. electroplating


Correct Option: C
Explanation:

Extracting metals in the pure form from its compound is known as extraction of metals.

Electroplating is:

  1. the deposition of superior metal on the surface of inferior metal

  2. the deposition of inferior metal on the surface of inferior metal

  3. both A and B

  4. none of the above


Correct Option: A
Explanation:

The deposition of superior metal on the surface of inferior metal is known as electroplating.

In the electrolysis of molten $Al _2O _3$ with inert electrodes:

  1. Al is oxidized at anode to $Al^{3+}$

  2. $O _2$ gas is produced at anode

  3. $O^{2-}$ is reduced at cathode

  4. $O$ is oxidized at anode


Correct Option: B
Explanation:

The electrodes are inert so that they do not involve the electrode reactions but transfer electrons through them. Since the silvery metal i.e. $Al$ is produced at cathode.

At cathode: $Al^{3+}+3e^- \longrightarrow Al$
At anode, $O _2$ bubbles off.
At anode; $2O^{2-}\longrightarrow O _2+4e^-$
The cell reaction: $2Al _2O _3\longrightarrow 4Al+3O _2$

Cost of electricity for the production of $X$ litres of $H _2$ at $NTP$ at the cathode is Rs $X$, then cost of electricity for the production $X$ litres of $O _2$ gas at $NTP$ at the anode will be:


[Assume $1$ mole of electrons as one unit of electricity]

  1. $2X$

  2. $4X$

  3. $16X$

  4. $32X$


Correct Option: A
Explanation:

Electrolysis of water gives:


${ H } _{ 2 }O\rightarrow { H } _{ 2 }+\tfrac { 1 }{ 2 } { O } _{ 2 }$


On electrolysis of water, hydrogen, and oxygen formed in the ratio $2:1$.
Since $X$ litres of ${ H } _{ 2 }$ is formed. Amount of ${ O } _{ 2 }$ formed will be $\tfrac { X }{ 2 } $. Since cost of production of electricity from $\tfrac { X }{ 2 } $ litres of ${ O } _2$ = Rs $X$

So, cost of production of electricity from $X$ litres of ${ O } _2$ = Rs $2X$.

So, the correct answer is option $A$.

A solution of $CuSO _4$ is electrolysed for $7$ minutes with a current of $0.6A$. The amount of electricity passed is equal to:

  1. $4.2C$

  2. $2.6\times 10^{-3}F$

  3. $126C$

  4. $36C$


Correct Option: B
Explanation:

Quantity of electricity passed $\displaystyle Q(C) = I(A) \times t(s)$
$\displaystyle Q(C)=0.6 \ A \times 7 \ min \times 60 \ s/min$
$\displaystyle Q(C)=252 \ C$
Number of faraday passed $\displaystyle = \dfrac {252 \ C}{96500 \ C/F}=2.6 \times 10^{-3} \ F$