Tag: extraction of metals by electrolysis

Quizzes (0) Questions (130)

Questions Related to extraction of metals by electrolysis

Multiple choice chemistry production of metals extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

Identity the principle behind the concentration of bauxite ore.

  1. Magnetic property of impurities in white bauxite
  2. Low specific gravity of $Al _{2}O _{3}$
  3. Amphoteric nature of $Al _{2}O _{3}$
  4. High melting point of $Al _{2}O _{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Al _2O _3 + 2NaOH + 3H _2O \rightarrow 2Na[Al(OH) _4]$
$Al _2O _3 + 6HCl \rightarrow 2AlCl _3 + 3H _2O$

Hence, in both the ways, Al can be concentrated, which owes to its amphoteric nature. 

Multiple choice chemistry production of metals extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

Use the relationship $\Delta \,G^{\circ}\,=\,- nFE^{\circ} _{cell}$ to estimate the minimum voltage required to electrolyse $Al _2O _3$ in the Hall-Heroult process.

$\Delta G^{\circ} _{f}(Al _2O _3)\,=\,-1520\,kJ\,mol^{-1}$

$\Delta G^{\circ} _{f}(CO _2)\,=\,-394\,kJ\,mol^{-1}$

The oxidation of the graphite anode to $CO _2$ permits the electrolysis to occur at a lower voltage than if the electrolysis reactions were :

$Al _2O _3\,\rightarrow\,2Al\,+\,3O _2$.

What is the approximate value of low voltage?

  1. 2 V

  2. 3 V

  3. 4V

  4. 5V
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Net reaction in Hall-Heroult process is

$3C\,+\,2Al _2O _3\,\rightarrow\,4Al\,+\,3CO _2$

Or $4Al^{3+}\,+\,12e^{-}\,\rightarrow \,4Al,$

Number of electrons (n) = 12

$\Delta\,G^{\circ }\,=\,3 \Delta\, G^{\circ } _f (CO _2)\,-\,2 \Delta\,G^{\circ } _f(Al _2O _3)$

$=-3\,\times\,394\,-\,2(-1520)$

$= 1858\ kJ$


$\Delta G^{\circ}\,=\, -nFE^{\circ} _{cell}$

$-E^{\circ} _{cell}\,=\, \displaystyle \frac {\Delta G^{\circ}}{nF}\, =\, \displaystyle\frac {1858\,\times\, 1000}{12\, \times\, 96500}$

$= 1.60\ V$

Thus, Hall-Heroult process takes place at lower voltage.

Multiple choice chemistry production of metals extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

In electrtolysis of $Al _2O _3$ by Hall-Heroult process:

  1. Cryolite $Na _3[AIF _6]$ lowers the melting point of $Al _2O _3$ and increases its electrical conductivity.
  2. $Al$ is obtained at cathode and probably $CO _2$ at anode

  3. both (a) and (b) are correct

  4. none of the above is correct
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The electrolysis of alumina by Hall and Heroult's process is carried by using a fused mixture of alumina and cryolite  along with minor quantities of aluminum fluoride and fluorspar. The addition of cryolite and fluorspar increases the electrical conductivity of alumina and  lowers the fusion temperature. Thus, the option A is correct. Also at anode, alumina reacts with fluorine to give oxygen. The liberated oxygen reacts with carbon to form $CO$ and $CO _2$. These gases are liberated at anode.

Multiple choice chemistry production of metals extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

Which one of the following statements is incorrect?

  1. In the aluminothermite process, aluminium acts as reducing agent.

  2. Lead is extracted from its chief ore by both carbon reduction and self-reduction.

  3. Zinc is extracted from its chief ore by carbon reduction.

  4. Extraction of gold involves the leaching of ore with cyanide solution followed by reduction with zinc.

  5. In Hall-Heroult process, the electrolyte used is a molten mixture of alumina, sodium hydroxide and cryolite.
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation
(1) $Cr _{2}O _{3}+2Al\rightarrow Al _2O _3+2Cr^{3+}$
(2) $2PbS+2PbO\rightarrow 2PbO+2SO _{2}$
$PbS+2PbO\rightarrow 3Pb+SO _{2}$ (self reduction)
$PbO+C\rightarrow Pb+CO$ (carbon reduction)
(3) $ZnO+C\rightarrow Zn+CO$ (carbon reduction)
(4) $4Au(S)+8CN(aq)+2H _{2}O(aq)+O _{2}(g)\rightarrow 4[Au(CN) _{2}]^{-}(aq)+4OH^{-}(aq)$ - leaching
$2[Au(CN] _{2}]^{-}(aq)+Zn(S)\rightarrow 2Au(S)+[Zn(CN) _{4}]^{2-}(aq)$ - reduction
(5) It is molten mixture of alumina and cryolite
Multiple choice chemistry industrial inorganic chemistry extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

The following question is relevant to the extraction of Aluminium:
Along with cryolite and alumina, another substance is added to the electrolyte mixture. the substance is ___________.

  1. Fluorspar

  2. Mica

  3. NaCl

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Fluorspar is the substance act as a solvent and it increases conductivity. (Pure alumina melts at about $2000^oC$ and is a bad conductor of electricity. If fused cryolite and fluorspar is added, the mixture melts at $900^oC$ and  $Al _2O _3$ becomes a good conductor of electricity.)

Multiple choice chemistry industrial inorganic chemistry extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

A compound added to lower the fusion temperature of the electrolytic bath in the extraction of Al is ___________.

  1. cryolite and fluorspar

  2. Bauxite

  3. Iron and fluorspar

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Pure alumina melts at about $2000^oC$ and is a bad conductor of electricity. If fused cryolite and fluorspar is added, the mixture melts at $900^0C$ and $Al _2O _3$ becomes a good conductor of electricity.

Multiple choice chemistry chemical effects of electric current applications of electrolysis electroplating extraction of metals by electrolysis

If $Zn/Zn^{+2}$ electrode is diluted 100 times then the change in electromotive force will be :-

  1. increases of 59 mV

  2. decreases of 59 mV

  3. increases of 29.5 mV

  4. decreases of 29.5 mV
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the Nernst equation: E = E_cell - (0.059/n) * log(1/[Zn+2]). Diluting 100 times increases the concentration of the denominator in the log term, making the log term more negative, which increases the overall potential by (0.059/2) * log(100) = 0.059 V = 59 mV.

Multiple choice chemistry chemical effects of electric current applications of electrolysis electroplating extraction of metals by electrolysis

The process of depositing thin layer of a metal over an another metal with a help of electric current is called :

  1. electrorefining

  2. electroplating

  3. thermoplating

  4. none of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Electroplating is the process of plating one metal onto another, for decorative purposes or to prevent corrosion of a metal. 

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

Which pair of electrolytes could not be distinguished by the products of electrolysis using inert electrodes.

  1. $\text{1M CuSO} _4$ solution, $\text{1M CuCl} _2$ solution
  2. $\text{1M KCl}$ solution, $\text{1M Kl}$ solution
  3. $\text{1M AgNO} _3$ solution, $\text{1M Cu(NO} _3) _2$ solution
  4. $\text{1M KCl}$ solution, $\text{1M NaCl}$ solution
  5. $\text{1M CuBr} _2$ solution, $\text{1M CuSO} _4$ solution
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Electrolysis of 1M KCl and 1M NaCl using inert electrodes both yield potassium or sodium ions remaining in solution while hydrogen gas is evolved at the cathode and chlorine gas at the anode, making their electrolysis products identical and indistinguishable.