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 types of cells electricity physics

A $100$ watt, $110$ volt lamp is connected in series with an electrolytic cell containing $CaSO _4$ solution, the weight of Cd deposited by the current for $10$ hrs is (Sl.wt. of Cd=112.4)

  1. $19.06$g
  2. $38.12$g
  3. $1.906$g
  4. $3.812$g
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice types of cells electricity physics

What is the e.m.f. of a dry cell ?

  1. $1.5 V$
  2. $2.5 V$
  3. $1.1 V$
  4. $-1.1 V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
A common dry cell is the zinc–carbon battery, sometimes called the dry Leclanché cell, with a nominal voltage of 1.5 volts, the same as the alkaline battery (since both use the same zinc–manganese dioxide combination).Hence the correct option A.
Multiple choice types of cells electricity physics

Electro chemical equivalent of substance is 0.0006735 ; its eq. wt. is :

  1. 65

  2. 67.35

  3. 130

  4. cannot be calculated
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electrochemical equivalent (Z) is related to equivalent weight (E) by the formula Z = E / F, where F is Faraday's constant (approximately 96500 C/mol). E = Z * F = 0.0006735 * 96500 = 64.99, which rounds to 65.

Multiple choice types of cells electricity physics

The common dry cell produces a voltage of

  1. $1.5 V$
  2. $4 V$
  3. $2 V$
  4. $3 V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A common dry cell is zinc-carbon cell with a normal voltage of 1.5 V , same as the alkaline cell.

Multiple choice types of cells electricity physics

A cell constructed by coupling a standard copper electrode and a standard magnesium electrode has emf of $2.7$ volts. If the standard reduction potential of copper electrode of $+0.34\ V$, that of magnesium electrode is

  1. $+3.04\ V$
  2. $-3.04\ V$
  3. $+2.36\ V$
  4. $-2.36\ V$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The cell EMF is E_cell = E_cathode - E_anode. Given E_cell = 2.7 V and E_copper = +0.34 V (cathode), then 2.7 = 0.34 - E_magnesium. Solving for E_magnesium gives 0.34 - 2.7 = -2.36 V.

Multiple choice chemistry p- block elements-i study of aluminium chloride study of aluminium structure of some compounds

${Al} _{2}{O} _{3}$ is reduced by electrolysis at low potential and high currents. If $4.0\times {10}^{4}$ amperes of current is passed through molten ${Al} _{2}{O} _{3}$ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency, At. mass of $Al=27g{mol}^{-1}$)

  1. $1.3\times {10}^{4}g$
  2. $9.0\times {10}^{3}g$
  3. $8.1\times {10}^{4}g$
  4. $2.4\times {10}^{5}g$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$Al _2O _3(l) \rightarrow2Al^{3+}(l) + 3O^{2-}(l) $

Reduction equation of $Al^{3+}: $
$Al^{3+}(aq) + 3e^- \rightarrow Al(s) $

As,

Ampere = Coulomb / sec 

Time unit should be second. 
$6 hr \times 60 \dfrac{min}{ hr} \times 60 \dfrac{sec}{ min} = 21600 sec $
Coulomb = Ampere x second $= 40000 \times 21600 = 8.64\times10^8 $

1 mol $e^-$ carries a charge of 1 faraday, or 96485 coulombs. 
Then the mole of $e^-$ used in the electrolysis will be: 

$\dfrac{8.64\times10^8   \text{coulombs}}{ 96485 -\text{coulombs/mol}}- of -e^- = 8954.8  \text{mol of} -e^- $

According to the reduction equation, 3 moles of e- are needed to reduce 1 mol of Al^3+ to Al metal. 
Then , the mole of Al to be deposited at the cathode will be; 
$\dfrac{8954.8 \text{mol e-} }{ \dfrac{3 -\text{mol of e-}}{\text{mol Al}}} = 2985\text{ mol Al} $

Molar mass of $Al : 26.98 g/mol $
Mass of Al produced$: 2985 mol \times 26.98 g/mol = 80535 g $
$Al = 8.1\times10^4 g$

Multiple choice botany plant water relation diffusion pressure deficit other pressure components osmosis in plants

Which of the following equation is correct for incipient plasmolysed cell ?

  1. $DPD=OP-TP$
  2. $OP=TP$
  3. ${ \Psi } _{ w }={ \Psi } _{ s }+{ \Psi } _{ p }$
  4. ${ \Psi } _{ w }={ \Psi } _{ s }-{ \Psi } _{ p }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Diffusion pressure deficit - It is described as the ability of water to move is used to measure in terms of diffusion pressure. 


(C) $\Psi _w = \Psi _s + \Psi _p$

The point during which the protoplast is just pulled or moved away from the cell wall is called incipient plasmolysis.

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= $10000K$