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
  1. Sodium and chloride ions

  2. Sodium and hydroxide ions

  3. Hydrogen and chloride ions

  4. H2O molecules

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

In conductometric titration of HCl against NaOH, at the equivalence point (B) and beyond, the solution contains Na+ and Cl- ions (from the neutralization reaction) and excess OH- ions. The conductance at B is primarily due to Na+ and Cl- ions present as products of the neutralization. Initially, H+ and Cl- contribute; after neutralization, Na+ and Cl- remain.

Multiple choice
  1. Na2SO4.10.H2O

  2. Na(NH4)HPO4

  3. Na2CO3.10H2O

  4. Na2SO4

  5. Na2SO4.7H2O

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

Molecular formula of Glauber’s salt is Na2SO4.10.H2O. Dissolving salt cake (Na2SO4) in water and crystallisation at 32 oC yields Glauber’s salt.

Multiple choice
  1. increases, decreases

  2. decreases, increases

  3. increases, remains constant

  4. remains constant, decreases

  5. none of these

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

The initial solubility is due to the hydrated salt. Below 305K, Na2SO4.10H2O crystalises out. Above 305K the anhydrous salt separates out due to its negative temperature co-efficient of solubility.

Multiple choice
  1. The solutions in both the bottles will conduct electricity.

  2. Only the solution in bottle A will conduct electricity.

  3. Only the solution in bottle B will conduct electricity.

  4. The solutions in both the bottles will not conduct electricity.

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

The solution in bottle A is acidic and the solution in bottle B is basic. Both acids and bases dissociate into ions in water. So, both the solutions conduct electricity when an electric current is passed through them.

Multiple choice
  1. increase

  2. decrease

  3. first decrease then increase

  4. remains constant.

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

This option is correct and can be explained as follows: When NaCl(s) is added to the system it dissociates as NaCl(s)    Na+(aq)  +  Cl-(aq)The Cl- ions obtained from the dissociation of solid NaCl will combine with Ag+ ions to precipitate AgCl. Hence the equilibrium will be shifted in forward direction in accordance with Le-Chatelier's principle and more Ag+ and SO42- ions will be formed resulting in consumption of Ag2SO4 and hence its concentration will decrease.

Multiple choice
  1. eq. wt. of Cu2S is M1/8

  2. eq. wt. of CuS is M2/4

  3. eq. wt. of Ba(MnO4)2 is M3/5

  4. Cu2S and CuS, both have same equivalent in mixture

  5. none of these

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

The required reaction is, MnO4 +  8H+ +5e  →  Mn2+ + 4H2O Cu+ – e → Cu2+ S2– – 6e → S4+ Therefore, eq. wt. of Cu2S = (M1 / ((1*2) + 6)) = M1 / 8 Hence, this is a correct option.  

Multiple choice
  1. a reversible hydrogen electrode

  2. a rotating ring-disk electrode

  3. A rotating disk electrode

  4. a glass electrode

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

A glass electrode is a type of ion-selective electrode made of a doped glass membrane that is sensitive to a specific ion. It is an important part of the instrumentation for chemical analysis and physico-chemical studies.

Multiple choice
  1. The dropping mercury electrode

  2. A dynamic hydrogen electrode

  3. A fluoride selective electrode

  4. Gas diffusion electrodes

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

A fluoride selective electrode is a type of ion selective electrode sensitive to the concentration of the fluoride ion. A common example is the lanthanum fluoride electrode.

Multiple choice
  1. a nickel-cadmium battery

  2. a nickel hydrogen battery

  3. a nickel-metal hydride cell

  4. a nickel-iron battery

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

The nickel-iron battery (NiFe battery) is a storage battery having a nickel(III) oxide-hydroxide cathode and an iron anode, with an electrolyte of potassium hydroxide. The active materials are held in nickel-plated steel tubes or perforated pockets.

Multiple choice
  1. A flow battery

  2. A frog battery

  3. A water-activated battery

  4. Zinc-air batteries

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

A water-activated battery is a disposable reserve battery that does not contain an electrolyte and hence produces no voltage until it is soaked in water for several minutes.

Multiple choice
  1. Lithium tetrachloroaluminate

  2. Lithium tetrakis(pentafluorophenyl)borate

  3. Lithium tetramethylpiperidide

  4. Lithium triethylborohydride

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

Lithium tetrachloroaluminate is an inorganic compound, a tetrachloroaluminate of lithium, with the formula LiAlCl4. Solution of lithium tetrachloroaluminate in thionyl chloride is the liquid cathode and electrolyte of some lithium batteries, e.g. lithium-thionyl chloride cell.

Multiple choice
  1. 4 mL

  2. 5 mL

  3. 6 mL

  4. 7 mL

  5. 8 mL

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

[Cr(H2O)5Cl]Cl2 + 2AgNO3 → 2AgCl + [Cr(H2O)5Cl](NO32
Number of ionisable chloride ions in complex [Cr(H2O)5Cl]Cl2 = 2 Millimoles = Molarity x Volume (mL) x 2 = 0.02 x 30 x 2 = 1.2 Therefore, required Ag+ ions = 1.2 millimoles According to Millimoles = Molarity x V (mL) 1.2 = 0.2 x VV = 6 mL