Tag: applications of electrolysis

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Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

The main applications of electrolysis are:

  1. electroplating

  2. electrorefining

  3. extraction of metals

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The main application of electrolysis are:


Electroplating: Electroplating is a process of depositing a layer of any desired metal on another material by means of electricity.

Electrorefining: Electrorefining refers to the process of using electrolysis to increase the purity of a metal extracted from its ore.

Extraction of metals: Extractions are a way to separate the desired substance when it is mixed with others. 

Hence, option D is correct.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis
In an electroplating experiment with a $Cu^{2+}$ solution, $10.0$ amp is applied for $965\ sec$. How many moles of $Cu$ will be plated?
  1. $0.05\ \text{moles}$
  2. $0.1\ \text{moles}$
  3. $0.001\ \text{moles}$
  4. $0.005\ \text{moles}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$w = {Z\times I\times t}$

$w = \dfrac{E\times I\times t}{96500}$

$w = \dfrac{Molecular\ weight \times I\times t}{2\times 96500}$                                                 [$\therefore E= M/n$]

n=2

$n = \dfrac{w}{M} = \dfrac{I\times t}{2\times 96500}$

$ n=$ $\dfrac{10\times 965}{2\times 96500} = 0.05\ moles$

Hence, option A is correct.
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

In the electrolysis of $CuCl _{2}$ solution, the mass of cathode increased by $6.4\ g$. What occurred at copper anode?

  1. $0.224$ litre of $Cl _{2}$ was liberated
  2. $1.12$ litre of oxygen was liberated
  3. $0.05\ mole\ Cu^{2+}$ passed into the solution
  4. $0.1\ mole\ Cu^{2+}$ passed into the solution
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In the electrolysis process,

reduction takes place at cathode and oxidation takes place at anode
so, if we increase the mass on cathode then same number of moles get oxidised at anode and go into the solution,
so, moles of Cu passed into the solution = $\dfrac{6.4}{63.5} = 0.1 mole$

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Which is correct about silver plating?

  1. Anode - pure $Ag$

  2. Cathode - object to be electroplated
  3. Electrolyte - $Na[Ag(CN) _{2}]$

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
In silver plating, the object to be plated (e.g., a spoon) is made from the cathode of an electrolytic cell. 

The anode is a bar of silver metal, and the electrolyte (the liquid in between the electrodes) is a solution of silver cyanide, $AgCN$, in water.

When a direct current is passed through the cell, positive silver ions ($Ag^+$) from the silver cyanide migrate to the negative anode (the spoon), where they are neutralized by electrons and stick to the spoon as silver metal.

Hence, option D is correct.
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

When water is electrolysed, hydrogen and oxygen gases are produced. If $1.008\ g$ of $H _{2}$ is liberated at cathode, what mass of $O _{2}$ is formed at the anode?

  1. $32\ g$
  2. $16\ g$
  3. $8\ g$
  4. $4\ g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\dfrac {W _{1}}{W _{2}} = \dfrac {E _{1}}{E _{2}}$
$\dfrac {1.008}{W _{2}} = \dfrac {1.008}{8}$
$\therefore W _{2} = 8\ g$
where, $E _{1}$ and $E _{2}$ are equivalent masses of hydrogen and oxygen respectively.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Zn metal reduces ${SO _{3}}^{2-}$ ions into $H _{2}S$in presence of concentrated $H _{2}SO _{4}$ What weight of Zn is required for
reduction of 6.3 g $Na _{2}SO _{3}$ in presence of concentrated acid.

  1. 9.75 g

  2. 13 g

  3. 130 g

  4. 23 g
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

eq. of $Zn=$ eq.of ${SO _{3}}^{2-}$
$\frac{w}{65}\times 2 =\frac{6.3}{126}\times 6\Rightarrow w=\frac{6.3}{126}\times \frac{6\times 65}{2}=9.75 g$


Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

$Zn\left( s \right) \left|\ Zn{ { \left( CN \right)  } } _{ 4 }^{ 2- }\ \left( 0.5\ M \right) ,{ CN }^{ - }\left( 0.01 \right)  \right| \left|\ Cu{ \left( { NH } _{ 3 } \right)  } _{ 4 }^{ 2+ }\ \left( 0.5\ M \right) ,{ NH } _{ 3 }\left( 1\ M \right)  \right|\ Cu\left( s \right) $
Given: ${ K } _{ f }$ of $Zn{ { \left( CN \right)  } } _{ 4 }^{ -2\  }=\ { 10 }^{ 16 }$, $\quad \quad \quad$ ${ K } _{ f }$ of $Cu{ \left( { NH } _{ 3 } \right)  } _{ 4 }^{ 2+ }\ =\ { 10 }^{ 12 }$
$\displaystyle \quad \quad \ \ { E } _{ Zn|{ Zn }^{ -2 } }\ =\ 0.76V\ ;\ { E } _{ { Cu }^{ +2 }|Cu }\ =\ 0.34V\ ,\ \dfrac { 2.303RT }{ F } =0.06$
The emf of above cell is:

  1. $1.22\ V$
  2. $1.10\ V$
  3. $0.98\ V$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer