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

Which of the following is the preferred electrolyte for silver plating?

  1. $AgCl+HCl$
  2. $Ag{NO} _{3}+H{NO} _{3}$
  3. ${Ag} _{2}{SO} _{4}$
  4. $AgCN+NaCN$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Silver plating is an electroplating process utilizing an electrolyte containing silver cyanide solution and some free cyanide ions and operating at pH value not less than $8$. Free cyanide prevents precipitation of silver cyanide salt from the solution, provides electrical conductivity of the electrolyte and helps in the dissolution of silver anodes.

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

The main applications of electrolysis are:

  1. electroplating

  2. electrorefining

  3. extraction of metals

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The main application of electrolysis are:


Electroplating: Electroplating is a process of depositing a layer of any desired metal on another material by means of electricity.

Electrorefining: Electrorefining refers to the process of using electrolysis to increase the purity of a metal extracted from its ore.

Extraction of metals: Extractions are a way to separate the desired substance when it is mixed with others. 

Hence, option D is correct.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis
In an electroplating experiment with a $Cu^{2+}$ solution, $10.0$ amp is applied for $965\ sec$. How many moles of $Cu$ will be plated?
  1. $0.05\ \text{moles}$
  2. $0.1\ \text{moles}$
  3. $0.001\ \text{moles}$
  4. $0.005\ \text{moles}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$w = {Z\times I\times t}$

$w = \dfrac{E\times I\times t}{96500}$

$w = \dfrac{Molecular\ weight \times I\times t}{2\times 96500}$                                                 [$\therefore E= M/n$]

n=2

$n = \dfrac{w}{M} = \dfrac{I\times t}{2\times 96500}$

$ n=$ $\dfrac{10\times 965}{2\times 96500} = 0.05\ moles$

Hence, option A is correct.
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 = $\dfrac{6.4}{63.5} = 0.1 mole$