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 conductivity and its types electrochemistry

The specific conductivity of $0.1$ $N$ $KCl$ solution at $20^0$C is $0.0212$ $ohm^{-1} cm^{-1}$. The solution was found to offer resistance of $55$ ohms. Find the cell constant of the conductivity cell.

  1. $2.25$
  2. $1.166$
  3. $1.936$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Specific conductance $K=0.0212 \Omega ^{-1}cm^{-1}$

Normality = $0.1 N$, Resistance = $55\Omega $

Cell constant $G^*$ = conductivity $\times $ Resistance

$G^*=K\times R$

$=0.0212\times 55$

$=1.166 cm^{-1}$

Option B
Multiple choice conductivity and its types electrochemistry

What would be the equivalent conductivity of a cell in which $0.5$ N salt solution offers a resistance of $40$ ohm whose electrodes are $2$ cm apart and $5$ $cm^{2}$ in area?

  1. $10$ $ohm^{-1}$ $cm^2$ $eq^{-1}$
  2. $20$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$
  3. $30$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$
  4. $25$ $ohm^{-1}$ $cm^{2}$ $eq^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,
$R=40 ohm ,l=2cm,A=2cm^{2}$

We know the relation,

$\kappa=\dfrac{1}{R}\times\dfrac{l}{A}=\dfrac{1}{40}\times\dfrac{2}{5}=\dfrac{1}{100} ohm^{-1} m^{-1}$


Now we also know the relation,

$\Lambda _{eq}=\dfrac{\kappa\times1000}{N}=\dfrac{1}{100}\times\dfrac{1000}{0.5}=20 ohm^{-1} cm^{2} eq^{-1}$

Hence, option B is correct.

Multiple choice conductivity and its types electrochemistry

The equivalent conductivity conductivity of 1M ${{\text{H}} _{\text{2}}}{\text{S}}{{\text{O}} _{\text{4}}}$ solution would be if specific conductance is ${\text{26}} \times {\text{1}}{{\text{0}}^{ - 2}}{\text{S}}\,{\text{c}}{{\text{m}}^{ - 1}}$.

  1. $1.3 \times {10^2}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{e}}{{\text{q}}^{ - 1}}$
  2. $1.6 \times {10^2}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,$
  3. $13\,{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$
  4. $1.3 \times {10^3}{\text{S}}\,{\text{c}}{{\text{m}}^2}\,{\text{mo}}{{\text{l}}^{ - 1}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Normality = Molarity $ \times 2\,{\text{factor}}$ 

$ = 1 \times 2 = 2{\text{N}}$     $2{{\text{H}}^ + } + S{\text{c}}{{\text{m}}^2}$
$\Delta eq = \dfrac{{\kappa  \times 1000}}{N} = \dfrac{{2.6 \times {{10}^{ - 2}} \times 1000Sc{m^{ - 12 = 2}}}}{2}$  $1Lt = {10^3}c{m^3}$
$ = 1.3 \times 10/{10^3}c{m^3}$
$ = 1.3 \times {10^{ - 1 + 3}}Sc{m^2} = 1.3 \times {10^2}Sc{m^2}e{q^{ - 1}}$

Multiple choice conductivity and its types electrochemistry

The resistance of $0.2\ M$ solution of an electrolyte is $50\ \Omega$.The specific conductance of the solution is $1.3\ S\ m^{-1}$. If the resistance of the $0.4\ M$ solution of the same electrolyte is $260\ \Omega$, its molar conductivity is :

  1. $62.5\ S\ m^{2} mol^{-1}$
  2. $6250\ S\ m^{2} mol^{-1}$
  3. $6.25\ \times10^{-4}S\ m^{2} mol^{-1}$
  4. $625\times10^{-4}\ S\ m^{2} mol^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Molar conductivity is calculated as (kappa * 1000) / M. First, find the cell constant (G*) using the 0.2 M solution: G* = kappa * R = 1.3 * 50 = 65 m^-1. For the 0.4 M solution, kappa = G* / R = 65 / 260 = 0.25 S/m. Molar conductivity = kappa / concentration = 0.25 / 400 (converting 0.4 M to 400 mol/m^3) = 6.25 * 10^-4 S m^2 mol^-1.

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

Potential difference of $100 V$ is applied to the ends of a copper wire one metre long. Find the ratio of average drift velocity and thermal velocity of electrons at $27^\circ C$. (Consider there is one conduction electron per atom. The density of copper is $9.0 \times 10^3$; Atomic mass of copper is $63.5 g$.
$N _A = 6.0 \times 10^{23}$ per gram-mole, conductivity of copper is $5.81 \times 10^{7} \Omega^{-1}$.$K =1.38 \times 10^{-23} JK^{-1}$).

  1. $3.67 \times 10^{-6}$
  2. $4.3 \times 10^{-6}$
  3. $6 \times 10^{-5}$
  4. $5.6 \times 10^{-6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Drift velocity is given by v_d = (eE tau) / m, and thermal velocity is v_th = sqrt(3kT / m). Using the provided conductivity and density, the ratio is calculated as approximately 3.67 x 10^-6.

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

An electric current of $16A$ exists in a metal wire of cross section ${ 10 }^{ -6 }{ m }^{ 2 }$ and length $1m$. Assuming one free electron per atom. The drift speed of the free electrons in the wire will be:
(Density of metal $=5\times { 10 }^{  }kg/{ m }^{ 3 }$, atomic weight $=60$)

  1. $5\times { 10 }^{ -3 }m/s$
  2. $2\times { 10 }^{ -3 }m/s$
  3. $4\times { 10 }^{ -3 }m/s$
  4. $7.5\times { 10 }^{ -3 }m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We know,


$I=neAv _d$
      where n=electron density,
                  e=electronic charge
                  A= cross section area
                  $v _d$=drift velocity

But, $n=\dfrac{\rho}{Atm. \ Wt}\times N _A$

And $v _d=\dfrac{I}{neA}$

So, $v _d=\dfrac{16\times 60}{5\times 10^4\times N _A\times 1.6\times 10^{-19}\times 10^{-6}}$

Taking $N _A=6\times 10^{23}$

$v _d=\dfrac{120\times 10^{-2}}{6}=2\times 10^{-3}$

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

A copper wire of cross-section $2\ {mm}^{2}$ carries a current of $30\ A$. Calculate the root mean square velocity (thermal velocity) of free electrons at $27^oC$. Also ${v} _{d}$ is very small compared to it.
[Data given: ${ \rho  } _{ { C } _{ 0 } }=8.9\ gm/cc$, Boltzmann constant $(k)=1.38\times {10}^{23}J/K$
${m} _{0}=9.1\times {10}^{-31}kg.{N} _{A}=6.023\times {10}^{23}$ atomic weight of $Cu=63$] 

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics electric current drift velocity and mobility drift speed drift velocity & mobility

The drift velocity of the electron in a copper wire of length 2m under the application of a potential difference of 200 V is $0.5 ms^{-1}$.Their mobility is (in $m^{-2} V^{-1} s^{-1}$)

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

Length ,$d=2m$

potential difference ,$V=200V$
Drift velocity, $v _d=0.5 m/s$
mobility ,$\mu=\dfrac{v _d}{E}$
$=\dfrac{v _d .d}{V}$
$=\dfrac{0.5}{200} \times 2$
$=\dfrac{0.5}{100}=5 \times 10^{-3} m^2 V^{-1} s^{-1}$

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

Current is flowing with a current density $J=480\ amp/cm^{2}$ in a copper wire. Assuming that each copper atom contribution one free electron and gives that  Avogadro number$=6.0\times 10^{23}\ atoms/mole$  Density of copper $=9.0\ g/cm^{3}$ .Atomic weight of copper $=64\ g/mole$ Electronic charge $=1.6\times 10^{-19}$ coulomb. The drift velocity of electrons is:

  1. $1\ mm/s$
  2. $2\ mm/s$
  3. $0.5\ mm/s$
  4. $0.36\ mm/s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that,

 Current density $J=480\,amp/c{{m}^{2}}$

Avogadro number $N=6.0\times {{10}^{23}}\,atoms/mole$

Density of copper $\rho =9.0\,g/c{{m}^{3}}$
Atomic weight of copper $m=64\,g/mole$

Electronic charge $e=1.6\times {{10}^{-19}}\,coulomb$

We know that

Current density $J=\dfrac{I}{A}$

Now, the drift velocity is

${{v} _{d}}=\dfrac{J}{ne}....(I)$

We know that,

$n={{N} _{A}}\times \dfrac{1}{m}\times \rho $

Now, put the value of n in equation (I)

  $ {{v} _{d}}=\dfrac{Jm}{{{N} _{A}}e\rho } $

 $ {{v} _{d}}=\dfrac{480\times 64}{6.0\times {{10}^{23}}\times 1.6\times {{10}^{-19}}\times 9} $

 $ {{v} _{d}}=\dfrac{30720}{86.4\times {{10}^{4}}} $

 $ {{v} _{d}}=355.5\times {{10}^{-4}}\,cm/s $

 $ {{v} _{d}}=0.36\,mm/s $

Hence, the drift velocity is $0.36\,mm/s$ 

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

Assume that each atom of copper contributes one free electron. The density of copper is $9g cm^{-3}$ and atomic weight of copper is $63$. If the current flowing through a copper wire of $1mm$ diameter is $1.1 $ ampere, the drift velocity of electrons will be:- 

  1. $0.01 mm s^{-1}$
  2. $0.02 mm s^{-1}$
  3. $0.2 mm s^{-1}$
  4. $0.1 mm s^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Drift velocity v_d = I / (neA). Given density, atomic weight, and current, calculate n = (density * Na) / atomic weight. Then solve for v_d = I / (ne * pi * r^2). The result is 0.1 mm/s.

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

How many electrons should be removed from a coil of mass 1.6 gram so that it may float in an electric field of intensity $10^9 NC^-1$ directed upwards ?

  1. $10^6$
  2. $10^7$
  3. $10^8$
  4. $10^9$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the coin to float, the electric force must balance gravity: qE = mg. Here q = ne, so n = mg / (eE). Using m = 1.6e-3 kg, g = 9.8 m/s^2, E = 10^9, n = (1.6e-3 * 9.8) / (1.6e-19 * 10^9) = 9.8e7, which is approximately 10^8.

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

A current of $1.0A$ exists in a copper wire of cross-section $1.0mm^2$.Assuming one free electron per atom

Calculate drift speed of the free electron in the wire _______(the density of copper is $9000kgm^{2})$

  1. $0.74 mms^{-1}$
  2. $7.4 mms^{-1}$
  3. $74 mms^{-1}$
  4. $0.074 mms^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$i=ne AV _d$
$n=\dfrac{6\times 10^{23}\times 9000}{63.5\times 10^{-3}}$


$=8.5\times 10^{28}m^{-3}$

Hence $vd = \dfrac{i}{neA}$
$=\dfrac{1}{8.2\times 10^{28}\times 1.602\times 10^{19}\times 10^{-3}}$
$=7.4\times 10^{-5}m/s$
$0.07mm/s$

Multiple choice physics electric current, potential difference and resistance drift velocity and mobility drift speed drift velocity & mobility

There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is $ 1 mm^2 $. If the number of free electrons per $ cm^3  is 8.4 \times 10^22 $, then the drift velocity would be  

  1. 1.0 mm/sec

  2. 1.0 m / sec

  3. 0.1 mm/sec

  4. 0.01 mm / sec
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

Current $I=1.344\,A$

Area of cross-section, $A=1\,m{{m}^{2}}=0.01\,c{{m}^{2}}$

Number of free electrons, $n=8.4\times {{10}^{22}}$

Charge on electron $e=1.6\times {{10}^{-19}}\,$

Drift velocity is given by

$ {{v} _{d}}=\dfrac{i}{nAe} $

$ =\dfrac{1.344}{8.4\times {{10}^{22}}\times 0.01\times 1.6\times {{10}^{-19}}} $

$ =\dfrac{1.344}{{{10}^{3}}\times 0.1344} $

$ ={{10}^{-2}}\,cm/\sec  $

$ =1\,mm/\sec  $