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 chemistry atom fundamental particles of an atom the structure of atoms discovery of subatomic particles

The total number of electrons present in 1.8 g of water is :

  1. $6.023\times 10^{22}$
  2. $10.8576\times 10^{23}$
  3. $10.8576\times 10^{22}$
  4. $6.023\times 10^{23}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that, 1 molecule of water consists of 10 electrons (2 electrons each of two hydrogen atoms and 8 electrons of one oxygen atom).
The number of molecules present in 1.8 g of water.
$18 : g : of : H _2O\xrightarrow[]{contains}6.023\times 10^{23}molecules$
$\Rightarrow1.8 : g : of : H _2O : contains : '! x' : molecules$
$x=\dfrac{1.8\times 6.023\times 10^{23}}{18}$
$=\dfrac{ 6.023\times 10^{23}}{10}=6.023\times 10^{22}molecules$.
$\therefore$ Number of electrons in 1.8 g of water $=6.023\times 10^{22}\times10=6.023\times 10^{23}$

Multiple choice chemistry petrochemicals and polymers addition polymerisation types of polymerisation reactions polymer

Lactic acid, $ HC _{3}H _{5} O _{3},$ produced in 1 g sample of muscle tissue was titrated using phenolphthalein as indicator against $OH^{-}$ ions which were obtained by the electrolysis of water. As soon as $ OH^{-} $ ions are produced, they react with lactic acid and at complete neutralization, immediately a pink colour is noticed. If electrolysis was made for 1158 s using 50.0 mA current to reach the end point, what was the percentage of lactic acid in muscle tissue?

  1. 5.4 %

  2. 2.7%

  3. 10.8%

  4. 0.054%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Moles of OH- = (I * t) / F = (0.05 * 1158) / 96500 = 0.0006 moles. Since lactic acid (C3H6O3) is monoprotic, moles of acid = 0.0006. Mass = 0.0006 * 90 g/mol = 0.054 g. Percentage = (0.054 / 1) * 100 = 5.4%.

Multiple choice chemistry production of metals extraction of aluminium metallurgy of aluminium extraction of metals by electrolysis

Use the relationship $\Delta \,G^{\circ}\,=\,- nFE^{\circ} _{cell}$ to estimate the minimum voltage required to electrolyse $Al _2O _3$ in the Hall-Heroult process.

$\Delta G^{\circ} _{f}(Al _2O _3)\,=\,-1520\,kJ\,mol^{-1}$

$\Delta G^{\circ} _{f}(CO _2)\,=\,-394\,kJ\,mol^{-1}$

The oxidation of the graphite anode to $CO _2$ permits the electrolysis to occur at a lower voltage than if the electrolysis reactions were :

$Al _2O _3\,\rightarrow\,2Al\,+\,3O _2$.

What is the approximate value of low voltage?

  1. 2 V

  2. 3 V

  3. 4V

  4. 5V
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Net reaction in Hall-Heroult process is

$3C\,+\,2Al _2O _3\,\rightarrow\,4Al\,+\,3CO _2$

Or $4Al^{3+}\,+\,12e^{-}\,\rightarrow \,4Al,$

Number of electrons (n) = 12

$\Delta\,G^{\circ }\,=\,3 \Delta\, G^{\circ } _f (CO _2)\,-\,2 \Delta\,G^{\circ } _f(Al _2O _3)$

$=-3\,\times\,394\,-\,2(-1520)$

$= 1858\ kJ$


$\Delta G^{\circ}\,=\, -nFE^{\circ} _{cell}$

$-E^{\circ} _{cell}\,=\, \displaystyle \frac {\Delta G^{\circ}}{nF}\, =\, \displaystyle\frac {1858\,\times\, 1000}{12\, \times\, 96500}$

$= 1.60\ V$

Thus, Hall-Heroult process takes place at lower voltage.

Multiple choice conductivity and its types electrochemistry

What do you mean by equivalent Conductivity?

  1. It is defined as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte in solution.

  2. It is defined as the conducting power of all the ions produced by dissolving ten gram equivalent of an electrolyte in solution.

  3. It is defined as the conducting power of all the ions produced by dissolving hundred gram equivalent of an electrolyte in solution.

  4. It is defined as the conducting power of all the ions produced by dissolving thousand gram equivalent of an electrolyte in solution.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equivalent Conductivity is defined as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte in solution. It is expressed as and is related to specific conductance as. (M is Molarity of the solution)

Multiple choice conductivity and its types electrochemistry

The resistance of $1\ N$ solution of $CH _{3}COOH$ is $250\ ohm$ when measured in a cell of cell constant $1.15\ cm^{-1}$. The equivalent conductance will be:

  1. $4.6\ ohm^{-1} cm^{2} eq^{-1}$
  2. $9.2\ ohm^{-1} cm^{2} eq^{-1}$
  3. $18.4\ ohm^{-1} cm^{2} eq^{-1}$
  4. $0.023\ ohm^{-1} cm^{2} eq^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$K = C \times \dfrac {l}{A} = \dfrac {1}{250}\times 1.15 = 4.6\times 10^{-3}\ ohm^{-1} cm^{-1}$
$\wedge _{e} = K\times \dfrac {1000}{N} = 4.6\times 10^{-3} \times \dfrac {1000}{1} = 4.6\ ohm^{-1} cm^{2} eq^{-1}$.

Multiple choice conductivity and its types electrochemistry

${\text{N}}{{\text{a}} _{\text{3}}}{\text{Al}}{{\text{F}} _{\text{6}}}\,\,$ is added to $\,{\text{A}}{{\text{l}} _{\text{2}}}{{\text{O}} _{\text{3}}}$

  1. Improve the electrical conductivity of the cell

  2. Increases rate of production

  3. Increases the melting point

  4. Decrease the electrical conductivity
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Cryolite (Na3AlF6) is added to alumina (Al2O3) in the Hall-Heroult process primarily to lower the melting point of the mixture and improve electrical conductivity.

Multiple choice conductivity and its types electrochemistry

The equivalent conductance of $0.02$ M acctic acid  $1.62.*{10^{ - 3}}$.  Degree of ironisation $'a'$ of $C{H _3}COOH$ is:
$({x _H} = 349.83oh{m^{ - 1}}and\lambda C{H _3}CO{O^ - } = 40.89ohm{s^{ - 1}}$

  1. $0.01$
  2. $0.02$
  3. $0.03$
  4. $0.04$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice conductivity and its types electrochemistry

The resistance of $N/2$ solution of an electrolyte in a cell was found to be $45$ ohm. The equivalent conductivity of a solution, if the electrodes in the cell are $2.2$ cm apart and have an area of $3.8 cm^2$ will be:

  1. $52.72\ S cm^2 eq^{-1}$
  2. $22.57\ S cm^2 eq^{-1}$
  3. $27.52\ S cm^2 eq^{-1}$
  4. $25.72\ S cm^2 eq^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Cell constant = $\dfrac{l}{a}=\dfrac{2.2}{3.8}=0.579 cm^{-1} $

Specific conductance = cell constant $\times$ conductance

Specific conductance = $\dfrac{0.579}{45} $

Equivalent conductance= $\dfrac{0.579}{45} \times \dfrac{1000}{0.5} $

Equivalent conductance = $25.73 Scm^{2} eq^{-1} $

Hence, option D is correct.
Multiple choice conductivity and its types electrochemistry

The specific conductance $(K)$ of an electrolyte of $0.1\ N$ concentration is related to equivalent conductance $(\wedge _{e})$ by the following formula.

  1. $\wedge _{e} = K$
  2. $\wedge _{e} = 10 K$
  3. $\wedge _{e} = 100 K$
  4. $\wedge _{e} = 10000 K$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

we have,

equivalent conductance = $\dfrac{K*1000}{N}$

equivalent conductance= $\dfrac{K*1000}{0.1 N}$

equivalent conductance= $10000K$

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.