Tag: extraction of metals by electrolysis

Questions Related to extraction of metals by electrolysis

During the mid ninteenth century, Daguerro-type portraits were very popular. A portrait image of a person was made on silver plated copper by developing the image with ?

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

  1. Silver bromide

  2. Mercury vapour

  3. 'Hypo' (sodium hypochlorite)

  4. An iodine solution

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Correct Option: B
Explanation:
The latent image was developed to visibility by several minutes of exposure to the forms given off by heated mercury in purpose made developing box.
So development process of Daguerro-type portraits is done with Mercury Vapour.

Silver can be spread as a thin sheet on another metal by electroplating. The film of silver sticks strongly to the metal. Which of the following metals cannot be properly plated with silver?

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

  1. Copper

  2. Iron

  3. Nickel

  4. Brass

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Correct Option: B
Explanation:

Iron cannot be property plated with silver it becomes peel if plated with silver are to peeling of silver it cannot properly plated iron.

In electroplating of copper by silver, the silver acts as an :

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

  1. anode

  2. cathode

  3. electrode

  4. electrolyte

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Correct Option: A
Explanation:

Silver plating on Steel spoon: A silver rod is made as anode and steel spoon as cathode. Potassiumargentocyanide $(K[Ag(CN) _2])$ is used as electrolyte.
$AgNO _3$ is not used as electrolyte because in that case the silver coating will be non-uniform as the electrolysis takes place rapidly.
In electroplating of copper by silver, silver rod is taken as anode.

Electrolysis of brine produces Na at cathode.

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

  1. True

  2. False

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Correct Option: B
Explanation:

Electrolysis of brine or NaCl produces hydrogen at cathode and chlorine at anode while the electrolyte NaOH sodium hydroxide remains in the solution.

 In the silver plating of copper, $K[Ag(CN) _2]$ is used instead of $AgNO _3$. The reason is:

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

  1. a thin layer of Ag is formed on Cu

  2. more voltage requirement

  3. $Ag^+$ ions are completely removed from solution

  4. less availability of $Ag^+$ ions, as Cu cannot displace Ag from $[Ag(CN) _2]^-$ ion

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Correct Option: D
Explanation:

Option D is correct 

less avaibality of $Ag+$ions as, $Cu$ cannot displace $Ag$ from $[Ag(CN) _2]$
$Cu$ can replace $Ag$ in $AgNO _3$. But In coordination compound $Ag$ can't be replaced.

In $Ag$ - $CuSO _{4}$ cell, silver electrode will serve as:

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

  1. anode

  2. cathode

  3. both anode and cathode

  4. none of the above

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Correct Option: B
Explanation:
In $Ag - CuSO _4$ cell, silver electrode will serve as cathode, as reduction occurs here and silver metal is deposited. Copper being high in the reactivity series than silver, undergoes oxidation, and silver ions undergo reduction.

Electrolysis of a solution of $Mn{ SO } _{ 4 }$ in aqueous sulphuric acid is a method for the preparation of $MnO _ 2$. Passing a current of $27A$ for $24$ hours gives $1kg$ of $MnO _2$. The current efficiency in this process is:

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

  1. $100$%

  2. $95.185$%

  3. $80$%

  4. $82.951$%

Show answer
Correct Option: B
Explanation:

The reaction taking place on the given reaction are,

$Mn^{2+}+2H _2O\rightarrow MnO _2+2H^{+}+H _2$

Amount of current is given by,
$m= Z \times I \times t$

$I=m\dfrac{F\times x}{t\times M}$

    $=1000g\times{\dfrac{96500\times 2}{24\times 60 \times 60 \times 86.9}}$

    $\implies I=25.70A$

Now current efficiency$=\dfrac{25.7}{27} \times 100=95.185$ %

Hence,option B is correct answer.

Consider the reaction : $Cr _2O _7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+}+7H _2O$


What is the quantity of electricity in coulombs needed to reduced $1\ mol$ of $Cr _2O _7^{2-}$?

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

  1. $6 \times 10^6C$

  2. $5.79\times 10^5C$

  3. $5.25 \times 10^5C$

  4. None of these

Show answer
Correct Option: B
Explanation:
$Cr _2O _7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+}+7H _2O$

The valency factor for Chromium is $6$.

Applying faraday's first law-
$ \dfrac{Q} {F} =mole \times V.f.$

$  Q            =mole \times V.f. \times F$ 

$  Q            =1 \times 6 \times 96500$

$  Q            =5.79\times 10^5\ C$

Calculate the amounts of Na and chlorine gas produced during the electrolysis of fused $NaCl$ by the passage of 1 ampere current for 25 minutes. 

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

  1. 0.3565 gm and 0.55 gm

  2. 0.3565 gm and 0.66 gm

  3. 0.55 gm and 0.3565

  4. None of above

Show answer
Correct Option: A
Explanation:
$i= 1 \ amp$
$t= 25 \ min = 25 \times 60 \ sec$
Molar mass of $Na= 23 \ g \ mol^{-1}$
Molar mass of $Cl _2= 71 \ g \ mol^{-1}$
$Cl^-= \cfrac 12 {(12)}=35.5 \ g$
Formula used, $m= \cfrac {Molar \ mass}{n \times F} \times i \times t$
For $Na$, $n=1$
$m= \cfrac {23}{1 \times 96500} \times 1 \times 25 \times 60$
$=0.3575 \ g$

For $Cl$, $n=1$
$m= \cfrac {35.5}{1 \times 96500} \times 1 \times 25 \times 60$
$=0.5518 \ g$

$\therefore$ The amounts of $Na$ and $Cl$ gas produced are $0.3575 \ g$ and $0.5518 \ g$ respectively.

A current of  $13.4\mathrm { A }$  is passed through  $1.0\mathrm { L }$  of  $1.0\mathrm { M }$  $HCl$  solution by using  $Pt$  electrodes for  $1.0\  \mathrm { hr }.$  The  $pH$  of the solution after the experiment is over at  $298 K$  will be about.

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

  1. $0.20$

  2. $0.30$

  3. $0.40$

  4. $0.50$

Show answer
Correct Option: B
Explanation:
$ (B) 0.30 $ 
Change in coulombs $ = 13.4\times 1\times 60\times 60 = 48240 $

Moles of electric charge $ = \dfrac{48240}{96500} = 0.5 F $

Equivalent of HCl decomposed on electrolysis $ = 0.5 $

Mili equivalents of HCl along electrolysis $ = 500 $

Molarity of HCl after electrolysis $ = \dfrac{500}{1000} = 0.5 M$

$ [H ^{+}] = 5\times 10^{-1} $

$ p{H} = -\log [H^{+}] $

$ p{H} = -\log 5\times 10^{-1} = 0.30. $