Tag: extraction of metals by electrolysis

Questions Related to 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?

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

  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

Show answer
Correct Option: D
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 = $\dfrac{6.4}{63.5} = 0.1 mole$

Which is correct about silver plating?

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

  1. Anode - pure $Ag$

  2. Cathode - object to be electroplated

  3. Electrolyte - $Na[Ag(CN) _{2}]$

  4. All of the above

Show answer
Correct Option: D
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.

What is the number of moles of oxygen gas evolved by electrolysis of $180\ g$ of water?

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

  1. $2.5$

  2. $5.0$

  3. $7.5$

  4. $10.0$

Show answer
Correct Option: B
Explanation:

$2{ H } _{ 2 }O(l)\rightarrow2{ H } _{ 2 }(g)+{ O } _{ 2 }(g)$

moles of water=$\frac { 180 }{ 18 } =10$
1 mole of oxygen for 2 moles of water.
So, 5 moles of oxygen for 10 moles of water.

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?

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

  1. $32\ g$

  2. $16\ g$

  3. $8\ g$

  4. $4\ g$

Show answer
Correct Option: C
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.

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.

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

  1. 9.75 g

  2. 13 g

  3. 130 g

  4. 23 g

Show answer
Correct Option: A
Explanation:

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


$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:

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

  1. $1.22\ V$

  2. $1.10\ V$

  3. $0.98\ V$

  4. $None\ of\ these$

Show answer
Correct Option: A

Calculate the mass of Ag deposited at cathode when a current of 2A was passed through a solution of $Ag{ NO } _{ 3 }$ for 15 min.
(Given : Molar mass of $Ag = 108\ g\ { mol }^{ -\ 1 }$ $\ 1F=96500\ C\ { mol }^{ -1 }$).

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

  1. $3.015\ g$

  2. $2.015\ g$

  3. $4.2\ g$

  4. $3.1\ g$

Show answer
Correct Option: B
Explanation:

Given:

Molar Mass of Ag = 108 g/mol

$1F = 96500\ C mol^{−1}$

Reaction at cathode = $Ag + e^-  \rightarrow   Ag(s)$ 

$w = Zlt$

Where, w = Mass deposited at cathode

Z = electrochemical constant

I = current

t = time

Now I = 2amp

$t = 15\ min = 15\times 60 = 900\ seconds$

Z = Eq. wt of substance $/ 96500 = 108/96500$ 

So,

$w = \dfrac{108}{96500} \times 900 \times 2 $

= $2.015g$

The electrochemical equivalent of silver is $0.0011180g$. When an electric current of $0.5$ ampere is passed through an aqueous silver nitrate solution for $200sec$, the amount of silver deposited is:

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

  1. $1.1180g$

  2. $0.11180g$

  3. $5.590g$

  4. $0.5590g$

Show answer
Correct Option: B
Explanation:

According to the data given when $1$ e is passed the amount of silver,

deposited is $0.0011180\,g$
$\therefore (0.5\times 200)c$ is passed
then $(0.0011180\times 100)g$
of silver gets deposited 
$=0.11180\,g$

Brine solution on electrolysis will not give__________.

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

  1. $NaOH$

  2. ${Cl} _{2}$

  3. ${H} _{2}$

  4. ${O} _{2}$

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Correct Option: D

$H _2(g)$ and $O _2(g)$ , can be produced by the electrolysis of water. What total volume (in $L$) of $O _2$ and $H _2$ are produced at $STP$ when a current of $30$ A is passed through a $K _2SO _4\, (aq)$  solution for 193 minutes?

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

  1. 20.16

  2. 40.32

  3. 60.48

  4. 80.64

Show answer
Correct Option: D