Tag: chemical changes

Questions Related to chemical changes

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

  1. anode

  2. cathode

  3. electrode

  4. electrolyte


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.

  1. True

  2. False


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:

  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


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:

  1. anode

  2. cathode

  3. both anode and cathode

  4. none of the above


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:

  1. $100$%

  2. $95.185$%

  3. $80$%

  4. $82.951$%


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 $Cr _2O _7^{2-}$?

  1. $6 \times 10^6C$

  2. $5.79\times 10^5C$

  3. $5.25 \times 10^5C$

  4. None of these


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. 

  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


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.

  1. $0.20$

  2. $0.30$

  3. $0.40$

  4. $0.50$


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. $

A solution of ${ AgNO } _{ 3 }$ is a good electrolyte but is not used for electroplating an article with silver because :

  1. rapid deposition of silver takes place resulting in uneven coating

  2. slow deposition of silver takes place resulting in uneven coating

  3. poor conduction of silver

  4. none of the above


Correct Option: A
Explanation:

A solution of $AgNO _3$ is a good electrolyte but is not used for electroplating an article with silver because it undergoes very fast redox process and electroplating done through this process is not uniform.