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 chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

$\displaystyle { BaCl } _{ 2 }$ dissociates in water to give one $\displaystyle { Ba }^{ 2+ }$ ion and two $\displaystyle { Cl }^{ - }$ ions. If concentrated $\displaystyle HCl$ is added to this solution :

  1. $\displaystyle \left[ { Ba }^{ 2+ } \right] $ increases
  2. $\displaystyle \left[ { Ba }^{ 2+ } \right] $ remains constant
  3. $\displaystyle \left[ { OH }^{ - } \right] $ increases
  4. The number of moles of undissociated $\displaystyle { BaCl } _{ 2 }$ increases
  5. $\displaystyle \left[ { H }^{ + } \right] $ decreases
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The common ion effect describes the changes that occur with the introduction of ions to a solution containing that same ion.
The role that the common ion effect in solutions is mostly visible in the decrease of solubility of solids. Through the addition of common ions, the solubility of a compound generally decreases due to a shift in equilibrium.
$BaCl _2\rightarrow Ba^{2+} + 2Cl^{-}$
$HCl \rightarrow H^{+} + Cl^{-}$
As $Cl^{-} $ is the common ion so, the $ Ba^{2+}$and $ 2Cl^{-}$ combines ( associate ) to give the undissociated $BaCl _2$. 
Hence , there is increase in concentration of undissociated $BaCl _2$.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

The following reaction occurs in a beaker: $\displaystyle { Ag }^{ + }\left( aq \right) +{ Cl }^{ - }\left( aq \right) \rightarrow AgCl\left( s \right) $. If a solution of sodium chloride were added to this beaker,

  1. The solubility of the sodium chloride would decrease

  2. The reaction would shift to the left

  3. The concentration of silver ions in solution would increase

  4. The solubility of the silver chloride would decrease

  5. The equilibrium would not shift at all
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Ag^+ + Cl^- \rightarrow AgCl$
$NaCl \rightarrow Na^+ + Cl^-$
As $Cl^- $ is the common ion in both the solution ( i.e. $NaCl $ and $AgCl$ ) , hence $AgCl$ being weaker electrolyte is precipitated or solubility of the silver chloride would decrease.
Common ion Effect : The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions. Some of the salt will be precipitated until the ion product is equal to the solubility product. In short, the common ion effect is the suppression of the degree of dissociation of a weak electrolyte containing a common ion

Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

In KI solution, $ I _2 $ readily dissolves and forms :

  1. $ I^{-}$
  2. $ KI _2 $
  3. $ KI _2^- $
  4. $ KI _3 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In KI solution, $I _2$ readily dissolves and forms $KI _3$ which contains polyhalide ion $I _3^-$


$KI + I _2  \rightarrow KI _3$

Due to this, iodine is partly soluble in water but highly soluble in the $KI$ solution.

So, the correct option is $D$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Conductivity ($\kappa $) of $0.01M$ $NaCl$ solution is $0.000145S$ ${cm}^{-1}$. What happens to the conductivity if extra $100mL$ of ${H} _{2}O$ be added to the above solution?

  1. Decreases

  2. Increases

  3. Remains unchanged

  4. First increase and then decrease
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Conductivity (kappa) is defined as the conductance of a solution contained between two electrodes of unit area and unit distance apart. Adding water (dilution) decreases the number of ions per unit volume, thus decreasing the conductivity.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Electrolytic conductivity of $0.3 M$ solution of $KCI$ at $298 K $is $3.72$ x ${10}^{-2}Scm^{-1}$.Calculate its molar conductivity ($S cm^{2} mol ^{-1}$) 

  1. $124$
  2. $30.56$
  3. $192$
  4. $185$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,

$k=3.72\times 10^{-2}\ Scm^{-1}$
$c=0.3M$

Molar conductivity($m$) is given by,

$m=\dfrac{k\times 1000}{c}$

$\Rightarrow m=\dfrac{3.72\times 10^{-2}\times 1000}{0.3}$

$\Rightarrow m=124\ S\ cm^2mol^{-1}$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Which among the following solutions is NOT used in determination of the cell constant?

  1. ${ 10 }^{ -2 }MKCl$
  2. ${ 10 }^{ -1 }MKCl$
  3. $1 M KCl$
  4. Saturated $KCl$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Standard KCl solutions (like 0.1M or 0.01M) are used for cell constant determination because their conductivity is accurately known. Saturated KCl is not used because its concentration and conductivity can be unstable or difficult to reproduce precisely.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Ionic conductance of ${ H }^{ + }$ and ${ SO } _{ 4 }^{ 2- }$ are $x$ and $yS\ { cm }^{ 2 }\ { mol }^{ -1 }$. Hence, equivalent conductivity of ${ H } _{ 2 }{ SO } _{ 4 }$ is:

  1. $2x+\cfrac { y }{ 2 } $
  2. $x+\cfrac { y }{ 2 } $
  3. $\cfrac { x }{ 2 } +y$
  4. $\cfrac { x }{ 2 } +\cfrac { y }{ 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equivalent conductance can be defined as the conductance produce by an equivalent of ion present in the solution.

For ${ H }^{ + }$, no.of ${ e }^{ - }$ change $=1$
For ${ SO } _{ 4 }^{ 2- }$, no.of ${ e }^{ - }$ change $=2$
Thus equivalent conductance of ${ H } _{ 2 }{ SO } _{ 4 }$
${ H } _{ 2 }{ SO } _{ 4 }$ $\rightleftharpoons 2{ H }^{ + }+{ SO } _{ 4 }^{ 2- }$
Eg Conductance $=\cfrac { 2[Ionic\quad conductance\quad of\quad { H }^{ + }] }{ nf } +\cfrac { Ionic\quad Conductance\quad of\quad { SO } _{ 4 }^{ 2- } }{ nf } $
Eg. Conductance $=2\times \cfrac { x }{ 1 } s{ cm }^{ 2 }+\cfrac { y }{ 2 } s{ cm }^{ 2 }$
Eg. Conductance $=2x+\cfrac { y }{ 2 }$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

At infinite dilution equivalent conductances of $B{ a }^{ +2 }$ & $C{ 1 }^{ - }$ ions are $127$ & $76\ oh{ m }^{ -1 }c{ m }^{ -1 }\ e{ q }^{ -1 }$ respectively. Equivalent conductance of $BaCl _{ 2 }$ at infinite dilution is:

  1. $139.5$
  2. $101.5$
  3. $203$
  4. $279$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

${ \Lambda  } _{ eq }^{ \infty  }\left( { Ba }^{ 2+ } \right) =127{ \Omega  }^{ -1 }{ cm }^{ -1 }{ eq }^{ -1 }\ { \Lambda  } _{ eq }^{ \infty  }\left( { Cl }^{ - } \right) =76{ \Omega  }^{ -1 }{ cm }^{ -1 }{ eq }^{ -1 }\ { \Lambda  } _{ eq }^{ \infty  }\left( { Ba }{ Cl } _{ 2 } \right) ={ \Lambda  } _{ eq }^{ \infty  }\left( { Ba }^{ 2+ } \right) +2{ \Lambda  } _{ eq }^{ \infty  }\left( { Cl }^{ - } \right) \ \qquad \qquad \quad =127+76\times 2\ \qquad \qquad \quad =279{ \Omega  }^{ -1 }{ cm }^{ -1 }{ eq }^{ -1 }$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

If equivalent conductance of 1M benzoic acid is $12.8\ { ohm }^{ -1 }{ cm }^{ 2 }$ and if the conductance of benzoate ion and ${ H }^{ -1 }$ ion are 42 and $288.42\ { ohm }^{ -1 }{ cm }^{ 2 }$ respectively. Its degree of dissolution is:

  1. 39%

  2. 3.9%

  3. 0.35%

  4. 0.039%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Lambda ^{ 0 } _{ m\left( C _{ 6 }H _{ 5 }COOH \right)  }={ \Lambda  } _{ (C _{ 6 }{ H } _{ 5 }CO{ O }^{ - }) }^{ 0 }\quad +{ \Lambda  }^{ 0 } _{ \left( { H }^{ + } \right)  }$ 

$\ \quad \quad \quad \quad \quad \quad \quad \quad =42+288.42=330.42\ now\quad we\quad have,\ \alpha =\frac { { \Lambda  } _{ m }^{ c } }{ { \Lambda  } _{ m }^{ 0 } } =\frac { 12.8 }{ 330.42 } =3.9$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

The molar conductivity of cation and anion of salt $BA$ are $180$ and $220\ mhos\ cm^{2} mol^{-2}$ respectively. The molar conductivity of $BA$ at infinite dilution is:

  1. $90\ mhos \ cm^{2} mol^{-1}$
  2. $110\ mhos \ cm^{2} mol^{-1}$
  3. $400\ mhos \ cm^{2} mol^{-1}$
  4. $200\ mhos \ cm^{2} mol^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Molar conductivity at infinite dilution is the sum of the limiting molar conductivities of the cation and anion. For salt BA, we add the cation conductivity (180) and anion conductivity (220) to get 400 mhos cm² mol⁻¹. This additive property holds because at infinite dilution, ions behave independently without interionic interactions. The unit should be mol⁻¹, not mol⁻² as written in the question (a minor typographical error that doesn't affect the calculation).

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

An alkali which dissociates partially on passage of an electric current is:

  1. sodium hydroxide

  2. nickel metal

  3. ammonium hydroxide

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

We have,

Sodium chloride is an ionic compound which is already present in the ionic form and completely dissociated on the passage of electric current,
Nickel and copper metal have higher reduction potential so, reduction takes place on passage of electric current,
ammonium hydroxide is a weak base it is partially dissociated on the passage of electric current. 

Multiple choice imperfections in solids solid state the solid state chemistry

If 1 mole of NaCl is doped with $10^{-3}$ mole of $SrCl _2$. What is the number of cationic  vacancies per mole of NaCl ?

  1. $10^{-3}\, mole ^{-1}$
  2. $6.02\times 10^{18}\, mole ^{-1}$
  3. $10^{50}\, mole ^{-1}$
  4. $6.02\times 10^{20}\, mole ^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When NaCl is doped with SrCl2, each Sr2+ ion replaces two Na+ ions to maintain electrical neutrality, creating one cationic vacancy per Sr2+ ion added. Thus, 10^-3 mole of SrCl2 introduces 10^-3 moles of cationic vacancies, which when multiplied by Avogadro's number gives 6.02 x 10^20 per mole.