Chemistry

Electrochemistry and Solutions

206 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 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 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 $Q=20F$
$\Rightarrow Q=20\times 96500$
$\Rightarrow Q=1.93\times10^6 C$

Multiple choice oxoacids of halogens p- block elements-ii p-block elements the p-block elements chemistry

Which of the following forms are the most basic in 0.1 M solution?

  1. $\mathrm{NaCl}$
  2. $\mathrm{NaOCl}$
  3. $\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\mathrm{O} _{2}$
  4. $\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{l}\mathrm{O} _{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$NaCl^{-1}$
$NaOCl^{+1}$ - most basic as Cl is in +1 oxidation state and on hydrolysis it gives weak acid HOCl. Thus it is most basic among the given compounds.
$\overset{+3}{NaClO _{2}}$
$\overset{+5}{NaClO _{3}}$

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 solubility of $AgCl$ in $NaCl$ solution is less than that in pure water, because of the  ________.

  1. solubility product of $AgCl$ is less than of $NaCl$

  2. common ion effect
  3. both $A$ and $B$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since, $NaCl$ is soluble to a very significant extent, when $AgCl$ is added to $NaCl$ solution, the common ion $[Cl^-]$ increases in the solution. To have the solubility product or $K _{sp}$ of $AgCl$ constant, $[Ag^+]$ will decrease or $AgCl$ will percipitate out from the solution. This is common ion effect. Hence solubility of $AgCl$ in $NaCl$ solution will be less than that in pure water.

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

100 mL of 20.8% $BaCl _2$ solution and 50 mL of 9.8% $H _2SO _4$ solution will form $BaSO _4$
$(Ba=137, Cl=35.5, S=32, H=1, O=16)$
$BaCl _2+H _2SO _4\rightarrow BaSO _4+2HCl$

  1. 23.3 g

  2. 11.65 g

  3. 30.6 g

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$100ml$ of $20.8$% $BaCl _2$ solution= $20.8g$  $BaCl _2$

$50ml$ of $9.8$% $H _2SO _4$ solution= $4.9g$  $H _2SO _4$
Reaction: $BaCl _2+H _2SO _4\longrightarrow BaSO _4\downarrow +2HCl$
           $208 g mol^{-1}$   $98 g mol^{-1}$      $233 g mol^{-1}$
$\therefore 98g$ $H _2SO _4$ reacts with $208g$ $BaCl _2$
$4.9g$ $H _2SO _4$ reacts with $\cfrac {208}{98}\times 4.9=10.4 g$ $BaCl _2$
$98g$ $H _2SO _4$ will produce $233g$ $BaSO _4$
$\therefore 4.9g$ $H _2SO _4$ will produce= $\cfrac {233}{98}\times 4.9=11.65g$ $BaSO _4$

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 addition of NaCl to AgCl decreases the solubility of AgCl because ________.

  1. Solubility product decreases

  2. Solubility product remains constant.

  3. solution becomes unsaturated

  4. solution becomes super saturated.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

NaCl is highly soluble and when it is added to AgCl it decreases the solubility of AgCl because of common ion $Cl^-$ and solution become super saturated.

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

What is $[{ NH } _{ 4 }^{ + }]$ in a solution containing 0.02M ${ NH } _{ 3 }$ (${ K } _{ b }={ 1.8\times 10 }^{ -5 }$) and 0.01M KOH?



  1. ${ 1.8\times 10 }^{ -5 }$
  2. ${ 9\times 10 }^{ -6 }$
  3. ${ 3.6\times 10 }^{ -5 }$

  4. NONE OF THE ABOVE
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a solution of NH3 (weak base) and KOH (strong base), the concentration of OH- is dominated by the strong base (0.01M). Using the Kb expression, Kb = [NH4+][OH-] / [NH3], we get 1.8e-5 = NH4+ / 0.02. Solving for [NH4+] gives 3.6e-5 M.

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

$As _2S _3$ solution has negative charge, capacity to precipitate is highest in:

  1. $AlCl _3$
  2. $Na _3PO _4$
  3. $CaCl _2$
  4. $K _2SO _4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solution:- (A) $Al{Cl} _{3}$

According to Hardy-Schulze rule, more is the valence of effective ion, greater is its coagulating power.
Hence ${As} _{2}{S} _{3}$ precipitate the most in $Al{Cl} _{3}$.

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 solubility of CaF$ _2$ (K$ _{sp} = 5.3\times 10^{-9}$) in $0.1$ M solution of NaF would be : (Assume no reaction of cation/anion) .

  1. $5.3 \times 10^{-10}$ M
  2. $5.3 \times 10^{-8}$ M
  3. $5.3 \times 10^{-7}$ M
  4. $5.3 \times 10^{-11}$ M
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$(C)\ 5.3\times 10^{-7}m$

$CaF _2\rightleftharpoons Ca^{2+}+2F^-$

$K _{sp}=[Ca^{}2+][F^-]^2=S(S+0.1)^2=S\times 0.1^2=5.3\times 10^{-9}$

Note: $S<<0.1$ so, $S+ 0.1 \approx 0.1$ 

$\Rightarrow S=5.3\times 10^{-7}\ M$ 

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 percentage of pyridine $\left( {{C _5}{H _5}N} \right)$ that forms pyridinum ion $\left( {{C _5}{H _5}{N^ + }H} \right)$ in a $0.10M$ aqueous pyridine solution $\left( Given - {{K _b}, for \  {C _5}{H _5}N = 1.7 \times {{10}^{ - 9}}} \right)$ is    

  1. $0.0060\% $
  2. $0.013\% $
  3. $0.77\% $
  4. $1.6\% $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a weak base B, Kb = C * alpha^2 / (1 - alpha). Since alpha is very small, Kb = C * alpha^2. Here, 1.7e-9 = 0.1 * alpha^2, so alpha^2 = 1.7e-8, and alpha = 1.3e-4. The percentage is 1.3e-4 * 100 = 0.013%.

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

Assertion: Due to common ion effect, the solubility of $HgI _2$ is expected to be less in an aqueous solution of KI than in water. But $HgI _2$ dissolves in an aqueous solution of KI to form a clear solution.
Reason: $I^{\circleddash}$ ion is highly polarisable.

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Assertion is incorrect but Reason is correct

  5. Both Assertion and Reason are incorrect
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Due to common ion effect, the solubility of $HgI _2$ is expected to be less in an aqueous solution of KI than in water as
$HgI _2 + KI \longrightarrow K _2[HgI _4]$.
since $I^{\circleddash}$ ion is large sized and therefore is highly polarisable.
But (R) is not the correct explanation of (A)

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

Why only ${As}^{+3}$ gets precipitated as ${As} _{2}{S} _{3}$ and not ${Zn}^{+2}$ as $ZnS$ when ${H} _{2}S$ is passed through an acidic solution containing ${As}^{+3}$ and ${Zn}^{+2}$?

  1. Solubility product of ${As} _{2}{S} _{3}$ is less than that of $ZnS$
  2. Enough ${As}^{+3}$ are present in acidic medium

  3. Zinc salt does not ionise in acidic medium

  4. Solubility product changes in presence of an acid
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The species having minimum value of ${K} _{sp}$ will get precipitated first of all because ionic product will exceed the solubility product of such species.

${K} _{sp}$ of ${As} _{2}{S} _{3}$ is less than $ZnS$. In acid medium ionisation of ${H} _{2}S$ is suppressed (common ion effect) and ${K} _{sp}$ of $ZnS$ does not exceed.