Tag: extraction of metals by electrolysis

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

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 $Ag = 108\ g\ { mol }^{ -\ 1 }$ $\ 1F=96500\ C\ { mol }^{ -1 }$).

  1. $3.015\ g$
  2. $2.015\ g$
  3. $4.2\ g$
  4. $3.1\ g$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $

= $2.015g$

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

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:

  1. $1.1180g$
  2. $0.11180g$
  3. $5.590g$
  4. $0.5590g$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 
$=0.11180\,g$

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

$H _2(g)$ 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?

  1. 20.16

  2. 40.32

  3. 60.48

  4. 80.64
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total charge passed is Q = I * t = 30 A * (193 * 60 s) = 347400 C. The moles of electrons transferred are n(e-) = Q / F = 347400 / 96500 = 3.6 mol. From the electrolysis of water, 2 moles of electrons produce 1 mole of gas total (0.5 mol O2 and 1 mol H2 per 2 moles of electrons, meaning 3 moles of total gas per 4 moles of electrons; specifically, 4 e- give 1 mole O2 (22.4 L) and 2 moles H2 (44.8 L)). Scaling by 3.6 mol of electrons gives (3.6 / 4) * 3 * 22.4 = 60.48 L total gas at STP.