Tag: extraction of metals by electrolysis

Quizzes (0) Questions (130)

Questions Related to extraction of metals by electrolysis

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

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. $32X$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $A$.
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

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. $36C$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\displaystyle = \dfrac {252 \ C}{96500 \ C/F}=2.6 \times 10^{-3} \ F$

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

State True or False.
Electrotyping is an application of electrolysis.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Electrotyping is a chemical method for forming metal parts that exactly reproduce a model.
Electrotyping is related to electroplating, which permanently adds a thin metallic overlayer to a metallic object instead of creating a freestanding metal part. So it is basically an application of electrolysis.

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

During electrolysis of an aqueous solution of a salt, pH in the space near one of the electrodes is increased, which of the following salt solution was electrolysed?

  1. $KCl$
  2. ${ CuCl} _{ 2 }$
  3. ${ Cu(NO } _{ 3 }{ ) } _{ 2 }$
  4. ${ CuSO } _{ 4 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
As the cation $\left( {K}^{+} \right)$ in $KCl$ has lower electrode potential than $H$, hydrogen is liberated at cathode.

There is an accumulation of ${H}^{+}$ at one electrode, resulting in an increase in pH.

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

A 5-ampere current is passed through a solution of zinc sulphate for $40 $ minutes. The amount of zinc deposited at the cathode is:

  1. $0.4065 g$
  2. $65.04 g$
  3. $40.65 g$
  4. $4.065 g$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\because$ $W=z.i.t$=$\cfrac{E}{F}\times{i}.{t}$$=\left(\cfrac{65.38\times5\times40\times60}{2\times96500}\right)g$.


                                $=4.065 g.$

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

The same amount of electricity was passed through two separate electrolytic cells containing solutions of nickel nitrate $\left[ Ni{ \left( { NO } _{ 3 } \right)  } _{ 2 } \right]$ land chromium nitrate $\left[Cr{ \left( { NO } _{ 3 } \right)  } _{ 3 } \right]$ respectively. If $0.3g$ of nickel was deposited in the first cell, the common of chromium deposited is :
$(at. Wt. Of Ni=59, at. Wt. Of Cr=52)$

  1. $0.1g$
  2. $0.17g$
  3. $0.3g$
  4. $0.6g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Faraday's laws of electrolysis, the mass deposited is proportional to the equivalent weight. Equivalent weight of Ni is 59/2 = 29.5, and for Cr is 52/3 = 17.33. Mass of Cr = (Mass of Ni * Eq. Wt. Cr) / Eq. Wt. Ni = (0.3 * 17.33) / 29.5 = 0.176g.

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

A certain quantity of electricity when passed through solution of ${ AgNO } _{ 3 }$, ${ ZnSO } _{ 4 }$, ${ CrI } _{ 3 }$. If X moles of Cr are deposited at its cathode, how many moles of Ag and Zn are deposited at their respective cathodes.

  1. X, X

  2. 3X, 2X

  3. 3X, 1.5X

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

$Ag^++e^- \longrightarrow Ag$

$Zn^{2+}+2e^- \longrightarrow Zn$
$Cr^{3+}+3e^-\longrightarrow Cr$
Given that $x$ moles of $Cr$ is deposited at it's cathode means in case of $Ag$ it will be $3X$  and for $Zn$ it will be $\cfrac {3X}{2}$ moles deposits at cathode.

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

How long (approximate) should water be electrolysed by passing through $100$ amperes current so that the oxygen realised can completely burn $27.66\ g$ of diborane?


(Atomic weight of $B=10.8\ u$ )

  1. $0.8$ hours
  2. $3.2$ hours
  3. $1.6$ hours
  4. $6.4$ hours
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The combustion of diborane (B2H6 + 3O2 -> B2O3 + 3H2O) requires 3 moles of O2 per mole of B2H6. 27.66g of B2H6 is 1 mole (MW=27.66). Thus, 3 moles of O2 are needed. Electrolysis of water (2H2O -> 2H2 + O2) requires 4 moles of electrons per mole of O2. Total charge Q = 3 * 4 * 96500 C. Time = Q / I = 1158000 / 100 = 11580 seconds, which is approximately 3.2 hours.

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

The mass of carbon anode consumed (giving only carbondioxide) in the production of 270 kg of Aluminium metal from bauxite by the Hall process is :

  1. 270 kg

  2. 540 kg

  3. 90 kg

  4. 180 kg
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In the Hall-Heroult process, aluminum is produced along with the consumption of carbon anodes which react with oxygen to form carbon dioxide according to the reaction 2Al2O3 + 3C -> 4Al + 3CO2. Stoichiometric calculations for 270 kg of Al give 90 kg of carbon consumed.