Chemistry

Electrochemistry and Solutions

205 Questions

Work through advanced chemistry questions focusing on electrochemistry and properties of solutions. Key themes include electrolytic conductivity, colloidal solutions, and solubility products. These concepts are frequently tested in JEE advanced, NEET, and university level chemistry competitive examinations.

Colloidal solutionsMolar conductivitySolubility productElectrolytic conductivityHydrolysis of saltsEquivalent conductivity

Electrochemistry and Solutions Questions

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

An experiment showed that a lead chloride solution is formed when 6.21 g of lead combines with 4.26 g of chlorine. What is the empirical formula of this chloride? 

[Pb = 207; Cl = 35.5]

  1. $PbCl _3$
  2. $PbCl _2$
  3. $PbCl _4$
  4. $PbCl$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
  Mass        Atomic weight    Relative no. of moles    Simplest ratio
Lead           6.21 g        207     6.21/207 = 0.03    0.03/0.03 = 1
Chlorine 4.26 g        35.5     4.26/35.5 = 0.12    0.12/0.03 = 4


Hence, empirical formula is $PbCl _4$

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Common salt obtained from Clifton beach contained $60.75\%$ chlorine while $6.40$ g of a sample of common salt from Khewra mine contained $3.888$ g of chlorine. State the law illustrated by these chemical combinations.

  1. Law of reciprocal proportion

  2. Law of multiple proportion

  3. Law of constant composition

  4. None of the above

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

First case :


Common salt from Clifton beach contains $=$ $60.75\%$ $Cl _2$

$100$ g of salt $=60.75$ g of $Cl _2$

$1$ g of salt $=\dfrac {60.75}{100}=0.6075$ g of $Cl _2$

Second case :


$6.40$ g of $NaCl$ from Khewra mine $=3.888$ g of $Cl _2$

$1$ g of $NaCl$ from Khewra mine $=\dfrac {3.888}{6.40}=0.6075$ g of $Cl _2$

Thus, the weight of $Cl _2$ in $1$ g of salt in both the cases is same. Hence, the law of constant composition is verified.


Hence the correct option is C.

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Potassium combines with two isotopes of chlorine $(^{35} Cl\,\, and\,\, ^{37}Cl)$ respectively to form two samples of $KCl$ Their formation follows the law of:

  1. constant proportions

  2. multiple proportions

  3. reciprocal proportions

  4. none of these

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
According to the Law of Definite Proportions, a chemical compound will always have exactly the same proportion of elements by mass.
This means that the elements that make up a compound will always have the same per cent composition by mass, regardless of the actual mass of the sample.
In this case, potassium chloride has a molar mass of 74.551 g/mol. The two elements that form potassium chloride are potassium, which has a molar mass of 39.0983 g/mol, and chlorine, which has a molar mass of 35.4527 g/mol.
This tells you that every mole of potassium chloride weighs 74.551 g, out of which 39.0983 g is potassium and 35.4527 g is chlorine.

Therefore, chlorine and potassium will always be in a ratio by mass of
$\dfrac{35.4527g}{39.0983g}$=0.90681
so the constant proportion is the right answer 

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

The % loss in mass after heating a pure sample of potassium chlorate (Mol. mass = 122.5) will be:

  1. 12.25

  2. 24.50

  3. 39.17

  4. 49.0

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2KClO _3\rightarrow 2KCl+3O _2$
$2\times 122.5$ grams shows a wieght loss of  $3\times32 grams$
So 245 grams of $KClO _3$ is 100%
96 grams of $O _2$ is X%
$x=\frac{100\times 96}{245}=39.17$.
Hence option C is correct.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

An aqueous solution of volume 500 ml, when the reaction : $ 2Ag^+(aq) +Cu(s) \leftrightharpoons Cu^{2+}(aq) +2Ag(s) $ reached equilibrium, the concentration of $ Cu^{2+} $ ions was xM. to this solution, 500 ml of water is added.at new equilibrium , the concentration of $ Cu^{2+} $ ions would be 

  1. 2x M

  2. x M

  3. between x and 0.5 x M

  4. less than 0.5 x M

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Adding water increases the volume, which shifts the equilibrium. According to Le Chatelier's principle, the system will try to increase the number of ions. However, the dilution effect (concentration = moles/volume) dominates, leading to a decrease in concentration.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

$NaCl$ is doped with $2\times { 10 }^{ -3 }$ mole % $Sr{Cl} _{2}$, the concentration of cation vacancies is

  1. $6.02\times { 10 }^{ 18 }{ mol }^{ -1 }$
  2. $1.204\times { 10 }^{ 19 }{ mol }^{ -1 }$
  3. $3.01\times { 10 }^{ 18 }{ mol }^{ -1 }$
  4. $1.204\times { 10 }^{ 21 }{ mol }^{ -1 }\quad $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Each Sr2+ ion replaces two Na+ ions in the lattice, creating one cation vacancy to maintain charge neutrality. 2 * 10^-3 mole % means 2 * 10^-5 moles of SrCl2 per mole of NaCl. This corresponds to 2 * 10^-5 * 6.022 * 10^23 vacancies per mole, which is 1.204 * 10^19.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

A big irregular shaped vessel contained water, conductivity of which was $ 2.56 \times 10^{-3}\, S^{-1} \, m^{-1}.$ 585 g of NaCI was then added to the water and conductivity after the addition of Nacl, was found to be $ 3.06 \times 10^{-3}\,S^{-1}\, m^{-1}. $ The molar conductivity of Nacl at this concentration is $ 1.5 \times 10^{-2}\, S^{-1}\, mol^{-1}.$The capacity of vessel if it is fulfilled with water, is

  1. $ 3 \times 10^{4} 1$
  2. $ 30 \,1$
  3. $ 3 \times 10^{8} 1$
  4. $ 3 \times 10^{5} 1 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

The electrolysis of which electrolyte gives the same products in the fused state as well as in the aqueous solution state?

  1. NaCl

  2. KCl

  3. AuCl$ _{3}$
  4. BaCl$ _{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

All have $Cl^-$ ions, $\therefore  Cl _2$, is released at anode
The element with reduction potential > red. potential of $H^+$ will give same results for aqueous and molter states.
From the given options only $SRP  of  Au^+  >  SRP  of  H^+$
Hence, for other salts in their aqueous solutions $H^+$  gets reduced

Hence, option C is correct.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis
Aqueous solution of nickel sulphate contains $Ni^{2+}$ and ${SO _{4}}^{2-}$ ions. What will be the product at the nickel anode?
  1. $Ni^{2+}$
  2. ${SO _{4}}^{2-}$
  3. $Ni$
  4. $H _2SO _4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At the anode, oxidation occurs. Nickel metal (the anode) loses electrons to form Ni2+ ions, which dissolve into the solution.

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

$ A \,Tl^{+} |Tl$ couple was prepared by saturating $ 0.10 M-KBr $ with TlBr and allowing $ Tl^{+}$ ions form the insoluble bromide to equilibrate. This couple was observed to have a potential $ -0.444 V $ with respect to $ PB^{2+} | Pb $ couple in which $ Pb^{2+}$ was 0.10 M. What is the $K _{sp} $ of $ TlBr.$ [Given :$ E _{Pb^{2+}|Pb}^{0} = -0.126 V, E _{Tl^{+}|Tl}^{0} = -0.336 V,$
$ log 2.5 = 0.4, 2.303 RT/F = 0.06]$

  1. $ 4.0 \times 10^{-6}$
  2. $ 2.5 \times 10^{-4}$
  3. $ 4.0 \times 10^{-5} $
  4. $ 6.3 \times 10^{-3} $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Use the Nernst equation for the cell potential: E_cell = E_cell_0 - (0.059/n) * log(Q). The cell reaction involves Tl+ and Pb2+. Use the given potential and concentrations to solve for [Tl+] at equilibrium, then use Ksp = [Tl+][Br-].

Multiple choice chemistry chemical changes electrochemical reactions electrolysis and its applications electrolytic cells and electrolysis

Which of the following methods can be used to separate hydrogen and oxygen in water?

  1. Boiling

  2. Electrolysis

  3. Distillation

  4. None of the above

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Electrolysis of ${ H } _{ 2 }O$ helps in separation of ${ H } _{ 2 }O$ to ${ H } _{ 2 }$ & ${ O } _{ 2 }$

${ 2H } _{ 2 }O\longrightarrow { 2H } _{ 2 }+{ O } _{ 2 }$

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

Charge required for liberating $710 g$ of $Cl _{2}(g)$ by electrolyzing a concentrated solution of $NaCl$ will be:

  1. $1.93$ x $10^{5}$ $C$
  2. $1.93$ x $10^{6}$ $C$
  3. $9.65$ x $10^{6}$ $C$
  4. $9.65$ x $10^{5}$ $C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The reaction taking place at anode is given by:

$2Cl^- \rightarrow Cl _2+2e^-$
Thus, $2$ moles of $e^-$ are required to liberate $1$ mole of $Cl _2$
Moles of $Cl _2=\dfrac{710}{71}=10$
Hence moles of $e^-$ required $=2\times10=20$
Hence $Q=20F$
$\Rightarrow Q=20\times 96500$
$\Rightarrow Q=1.93\times10^6 C$