Chemistry · Physics

Electrochemistry and Current Electricity

272 Questions

Review fundamental concepts of electrochemistry and current electricity through these practice questions. The set includes numerical problems on electrolysis, drift velocity, and electrode potentials. These topics frequently appear in engineering, CSIR NET, and state PSC prelims examinations for thorough preparation.

Electrolysis calculationsElectrode potentialNernst equationDrift velocityFuel cellsMolar conductivity

Electrochemistry and Current Electricity Questions

Multiple choice bio-chemistry biological oxidation classical idea of redox reactions oxidation reduction redox reaction

In acid medium, the standard reduction potential of $NO$ converted to ${ N } _{ 2 }O$ is $1.59 V$. Its standard potential in alkaline medium would be:

  1. $-1.59 V$
  2. $-0.764 V$
  3. $0.764 V$
  4. $0.062 V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$NO \longrightarrow N _2O$      $E^o= 1.59V$

The reaction in acidic medium,
$2NO+2H^+ \longrightarrow N _2O+H _2O$;   $E^o=1.59$
In basic medium,
$2NO+H _2O \longrightarrow N _2O+2OH^-$;   Let $E^o=x$
Now, in alkaline medium,
$H _2O \rightleftharpoons H^++OH^-$
$\Rightarrow K=10^{-14}$
or, $2H _2O \rightleftharpoons 2H^++2OH^-$,   $K= 10^{-28}$
We know, $E^o= \cfrac {0.059}{n}\log K$
$=\cfrac {0.059}{2}\log 10^{-28}= 0.826$
For standard potential,
$E^o _{acidic}+E^o _{basic}=E^o _{cell}$
$\Rightarrow 1.59+x= 0.826$
$\Rightarrow x= 0.764$

Multiple choice bio-chemistry biological oxidation classical idea of redox reactions oxidation reduction redox reaction

A solution contains ${Fe}^{+2},{Fe}^{+3}$ and ${I}^{-}$ ions. This solution was treated with iodine at ${35}^{o}C$. Then the favourable redox reaction is: 

(Given that ${ E } _{ { Fe }^{ +3 }/{ Fe }^{ +2 } }^{ o }=+0.77V;\quad { E } _{ { I } _{ 2 }/{ I }^{ - } }^{ o }=0.536V$)

  1. ${I} _{2}$ will be reduced to ${I}^{-}$

  2. there will be no redox reaction
  3. ${I}^{-}$ will be oxidised to ${I} _{2}$
  4. ${Fe}^{+}$ will be oxidised to ${Fe}^{+3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ 2I }^{ - }\rightarrow { I } _{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$

${ E } _{ Oxidation }^{ 0 }=-0.536V$

${ Fe }^{ 3+ }+e^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$

$E^{ 0 } _{ Reduction }=-0.77V$
----------------------------------------------------------------------------------------------------------------------
${ 2Fe }^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I } _{ 2 }$

$E^{ 0 }={ E } _{ oxidation }^{ 0 }+E _{ Reduction }^{ 0 },+ve$

Hence the reaction will take place.

${ 2I }^{ - }\rightarrow { I } _{ 2 }+{ 2e }^{ - }(oxidation\quad half-reaction)$

${ E }^{ 0 } _{ Oxidation }=-0.536V$

${ Fe }^{ 3+ }+{ e }^{ - }\rightarrow { Fe }^{ 2+ }(reduction\quad half-reaction)$

$E _{ Reduction }^{ 0 }=-0.77V$
------------------------------------------------------------------------------------------------------------------------
$2Fe^{ 3+ }+{ 2I }^{ - }\rightarrow { 2Fe }^{ 2+ }+{ I } _{ 2 };$

${ E }^{ 0 }={ E } _{ oxidation }^{ 0 }+{ E } _{ reduction }^{ 0 };+ve$

Multiple choice botany cell theory, cell specialization, and cell replacement number and shape of cells unicellular and multicellular organisms diversity and classification in animals

The maximum work that a standard Daniell cell can do before it stops working is
$[1 \,F = 96487\, C\, mol^{-1}\, and\, E^o = 1.1 V]$

  1. $212.271\, kJ \,mol^{-1}$
  2. $21.227\, kJ \,mol^{-1}$
  3. $721.221 \,kJ \,mol^{-1}$
  4. $71.222 \,kJ \,mol^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The maximum work (Gibbs free energy change) is calculated as W = nFE. For a Daniell cell, n=2. W = 2 * 96487 * 1.1 = 212271.4 J/mol, which is 212.271 kJ/mol.

Multiple choice chemistry states of matter: gaseous and liquid states gay lussac's law gas laws states of matter

The ratio of volumes of $H _2$ and $O _2$ liberated on electrolysis of water is

  1. $1 : 2$
  2. $1 : 3$
  3. $2 : 1$
  4. $3 : 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$H _{2}O \rightarrow H^{+} +OH^{-}$
Cathode  :  4$OH^{-} \rightarrow 2H _{2}O +O _{2}+4e^{-} $
Anode :  4$H^{+} +4e^{-} \rightarrow 2H _{2} $
The ratio of volumes of H$ _{2}$ and O$ _{2}$ is $2 : 1$.

Multiple choice physics some natural phenomena lightning conductors lightning safety advanced of lightning

The electric charge for electrode deposition of one gram equivalent of a substance is :

  1. one amp/sec

  2. 96,500 /sec

  3. one amp / hour

  4. 96,500 C
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to Faraday's laws of electrolysis, the charge required to deposit one gram equivalent of any substance is equal to one Faraday. One Faraday is approximately 96,500 Coulombs of electric charge.

Multiple choice chemistry nature of things objects that float or sink substances that sink or float soluble and insoluble substances

The EMF of the cell: Ag, AgCl in $0.1 M - KCl \parallel$ satd. $ NH _{4}NO _{3}\parallel 0.1 M - Ag NO _{3},$ Ag is 0.42 V at $ 25^{\circ}C.$ 0.1 M-KCl is 50% dissociated and $0.1 M - Ag NO _{3}$ is 40% dissociated. The solubility product of $ AgCl $ is $ (2.303 RT/F = 0.0)$

  1. $1.0 \times 10^{-10} $
  2. $ 2.0 \times 10^{-9} $
  3. $ 1.0 \times 10^{-9}$
  4. $ 2.0 \times 10^{-10} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the Nernst equation and the given dissociation constants, one can calculate the concentration of Ag+ ions and Cl- ions to find the solubility product Ksp = [Ag+][Cl-].

Multiple choice physics dynamics - explaining motion exponentiation index notation and products of prime factors powers

1 MeV is equal to

  1. $1.6\times 10^{13} Joules$
  2. $1.6\times 10^{13} cal.$
  3. $1.6\times 10^{13} ergs$

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

An electronvolt is a unit of energy equal to the work done on an electron accelerated through a potential difference of $1$ volt.


$1 eV=1.6\times10^{-19}\ J$ 

$\therefore1\,MeV=1.6\times 10^{-19}\times10^{6}\,Joules$

$1\ MeV=1.6\times10^{-13}\ J$

Multiple choice chemistry electrolysis electrolytes and non-electrolytes introduction to electrolysis chemical reactions

Saturated solution of $KNO _3$ with agar-agar is used to make salt bridge because :

  1. size of $K^+$ is greater than that of $NO _3^-$
  2. velocity of $NO _3^-$ is greater than that of $K^+$
  3. Velocities of both $K^+$ and $NO _3^-$ are nearly the same
  4. both velocity and sizes of $K^+$ and $NO _3^-$ ions are same
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Because velocity of $k^+$ & $NO _{3}^{-}$ ions same & it so not happens one start getting accumulate which have higher velocity and block the test tube.

Multiple choice chemistry electrolysis electrolytes and non-electrolytes introduction to electrolysis chemical reactions

In which of the following electrolysis, the composition of electrolyte is expected to remain constant under optimum conditions:

  1. Aq. AgNO$ _{3}$ solution between Ag electrodes
  2. Aq. CuSO$ _{4}$ solution between Pt electrodes

  3. Fused NaCl between Pt electrodes
  4. Aqueous AgNO$ _{3}$ solution between Pt electrodes
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In case of Aq.  $AgNO _3$  with Ag electrodes, metal from anode goes to the solution and metal from solution goes to the cathode, hence the composition of electrolyte does not change with time and remain constant.

Hence. option A is correct.

Multiple choice chemistry electrolysis electrolytes and non-electrolytes introduction to electrolysis chemical reactions

Which of the following will form the cathode with respect to iron anond in an electrolytic cell ?

  1. Mg

  2. Al

  3. Cu

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

In an electrochemical cell, the metal with a higher reduction potential (more noble) acts as the cathode, while the metal with a lower reduction potential (more active) acts as the anode. Copper (Cu) has a higher reduction potential than iron (Fe).

Multiple choice chemistry electrolysis electrolytes and non-electrolytes introduction to electrolysis chemical reactions

Two solutions of X and Y electrolytes are taken in two beakers and diluted by adding $500$ mL of water. $\Lambda _m$ of X increases by $1.5$ times while that of Y increases by $20$ times, what could be the electrolytes X and Y?

  1. $X\rightarrow NaCl, Y\rightarrow KCl$
  2. $X\rightarrow NaCl, Y\rightarrow CH _3COOH$
  3. $X\rightarrow KOH, Y\rightarrow NaOH$
  4. $X\rightarrow CH _3COOH, Y\rightarrow NaCl$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

On dilution, the molar conductivity of X increases by 1.5 times and Y by 20 times. Hence, X is a strong electrolyte and Y is a weak electrolyte.


Hence, option B is correct.

Multiple choice physics electrical conductivity of liquids does liquid conduct electricity? liquids conduct electricity do liquids conduct electricity?

Which solution will conduct the current?

  1. Sugar in water

  2. Sugar in ethanol

  3. Iodine in ethanol
  4. $MgCl _2$ in water
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In water magnesium chloride ionizes to form ions and these ions conduct electricity. Sugar and iodine are non electrolytes. They do not dissociate into ions in aqueous solution. Hence, they do not conduct electricity.

Multiple choice chemistry ionic equilibrium introduction to ionic equilibria in solution ionic equilibrium in solution ionisation of weak acids and weak bases

Two electrolytic cells containing molten solutions of Nickel chloride and Aluminium chloride are connected in series. If same amount of electric current is passed through them, what will be the weight of Nickel obtained when $18gm$ of Aluminium is obtained? $\left( Al-27gm/mole,Ni-58.5gm/{ mole }^{ -1 } \right) $

  1. $58.5gm$
  2. $117gm$
  3. $29.25gm$
  4. $5.85gm$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$By Faraday's Second Law$
$\frac{(m)N _{i}}{(m) _{N _{A1}}}=\frac{(E)N _{i}}{(E) _{N _{A1}}}$
$\frac{(m)N _{i}}{18}=\frac{58.5\times 2}{3\times 27}$
$(m){N _{i}}=58.5 g$

Multiple choice types of cells electricity physics

For a cell reaction involing a two-electron change, the standard e.m.f of the cell is fonud to be$0.295V$ at $25^{o}C$. The equilibrium constant of the reaction at $25^{o}C$ will be:

  1. $10$
  2. $1\ \times 10^{10}$
  3. $1\ \times 10^{-10}$
  4. $29.5\ \times 10^{-2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The relationship between the standard cell potential and the equilibrium constant is given by the Nernst equation at 298K as log(K) = (n * E_cell) / 0.0591. Substituting n = 2 and E_cell = 0.295V yields log(K) = 10, which means K equals 10 to the power of 10. Thus, option B is the correct mathematical outcome.