Chemistry · Physics

Electrochemistry and Current Electricity

224 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 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  $

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

The number of free electrons per $10$ mm ordinary copper wire is about $2\times 10^{21}$. The average drift speed of the electrons is $0.25$ mm current flowing is:

  1. $0.8$ A
  2. $8$ A
  3. $80$ A
  4. $5$ A
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,

Number of electron, $n=2\times {{10}^{21}}$

Average drift speed, $0.25\,mm/s$

$ Q=ne $

$ Q=2\times {{10}^{21}}\times 1.6\times {{10}^{-19}} $

$ Q=320\,C $

Since,

$ s=\dfrac{D}{T} $

$ T=\dfrac{D}{s} $

$ T=\dfrac{10}{0.25}=40 $

So, current

$ I=\dfrac{Q}{T} $

$ I=\dfrac{320}{40} $

$ I=8\,A $

Multiple choice physics electric current drift velocity and mobility drift speed drift velocity & mobility

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

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

Mobility mu = v_d / E. E = V / L = 200 / 2 = 100 V/m. mu = 0.5 / 100 = 0.005 = 5 x 10^-3 m^2/Vs.

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

Copper contains $8.4\times 10^{28}$ free electrons$/m^3$. A copper wire of cross-sectional area $7.4\times 10^{-7}m^2$ carries a current of $1$A. The electron drift speed is approximately.

  1. $10^{-8}$m/s
  2. $10^3$m/s
  3. $1$m/s
  4. $10^{-3}$m/s
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$v=\dfrac{I}{ne}=\dfrac{1}{7.4\times 10^{-7}}\times 8.4\times 10^{28}$
$=\dfrac{10^{-2}}{8.4\times 1.6\times 7.4}=10^-3$.

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

a current 10 ampere is maintained in a conductor of cross section of $ 10^{-4}m^2 $. if the electron density is $ 9 \times 10^{28} m^{-3} $ , what is the drift velocity of free electrons?

  1. $ 6.9 \times 10^{-6} ms^{-1} $
  2. $ 6.9 \times 10^{-4} ms^{-1} $
  3. $ 6.9 \times 10^{5} ms^{-1} $

  4. none of thses
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

v_d = I / (neA). I = 10, n = 9e28, A = 1e-4, e = 1.6e-19. v_d = 10 / (9e28 * 1.6e-19 * 1e-4) = 10 / (14.4e5) = 10 / 1440000 = 6.94e-6 m/s.

Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

How many minutes will it take to plate out $5.0g$ of $Cr$ form a $Cr _2(SO _4) _3$ solution using a current of $1.50A$?

  1. $254$
  2. $309$
  3. $152$
  4. $103$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using Faraday's law, mass = (I * t * M) / (n * F). For Cr2(SO4)3, Cr is in the +3 oxidation state (n=3). Molar mass of Cr is 52g/mol. t = (5.0 * 3 * 96500) / (1.5 * 52) = 18557 seconds. 18557 / 60 = 309 minutes.