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 metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

A dilute solution of $H _2SO _4$ was electrolyzed by passing a current of 2 amp. The time required for formation of 0.5 mole of oxygen is:

  1. 26.8 hours

  2. 13.4 hours

  3. 6.7 hours

  4. 28.6 hours
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let current of $2$ amp is passed through the solutions for $t$ seconds
$\therefore$   Charge passed $=2\times t$

$\therefore$   moles of electrons passed $=\dfrac { 2t }{ 96500 } $
At anode :

${ 2OH }^{ \left( - \right)  }\rightarrow 1/2{ O } _{ 2 }+{ H } _{ 2 }O+{ 2e }^{ - }$
$\therefore$   moles of ${ O } _{ 2 }$ released at anode $=\dfrac { 2t }{ 96500 } \times \dfrac { 1 }{ 4 } $

$\therefore$   $\dfrac { 2t }{ 96500 } \times \dfrac { 1 }{ 4 } =0.5$
$\Rightarrow t=96500$ secs $=26.8$ hours

$\therefore$   The time required for formation of $0.5$ mole of ${ O } _{ 2 }$ is $26.8$ hours.

Hence, the correct option is A.
Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

A current being passed for two hour through a solution of an acid liberating 11.2 litre of oxygen at NTP at anode. What will be the amount of copper deposited at the cathode by the same current when passed through a solution of copper sulphate for the same time?

  1. 16 g

  2. 63 g

  3. 31.5 g

  4. 8 g
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solution:- (B) $63 \; g$

At STP,
$1$ mole of oxygen gas $= 32 \; g = 22.4 \; L$
Therefore,
$11.2 \; L$ of ${O} _{2} = \cfrac{32}{22.4} \times 11.2 = 16 \; g$
Therefore,
$\cfrac{{M} _{Cu}}{{M} _{{O} _{2}}} = \cfrac{{E} _{Cu}}{{E} _{O}}$
$\Rightarrow \cfrac{{M} _{Cu}}{16} = \cfrac{\left( \cfrac{63}{2} \right)}{\left( \cfrac{16}{2} \right)}$
$\Rightarrow {M} _{Cu} = 63 \; g$
Hence the amount of copper deposited at the cathode is $63 \; g$.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

A current of 9.65 Ampere flowing for 10 minutes deposits 3 g of metal which is monovalent, the atomic mass of metal is a:

  1. 10

  2. 50

  3. 30

  4. 96.5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Solution:- (B) $50$

Weight of metal deposited $= 3 \; g$
Quantity of electricity passed $\left( q \right)$-
$q = I \times t$
Given:-
$I = 9.65 A$
$t = 10 \; min = 10 \times 60 = 600 \; sec$
$\therefore q = 9.65 \times 600 = 5790 \; C$
Now, as we know that,
Eq. wt. of metal $= \cfrac{\text{Weight of metal}}{q} \times F$
$\Rightarrow$ Eq. wt. of metal $= \cfrac{3}{5790} \times 96500 = 50 \; g$
As the metal is monovalent, atomic weight of metal will be equal to its equivalent weight.
Therefore,
Atomic weight of metal $= 50 \; g$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Resistance of decimolar solution is 50 ohm. If electrodes of surface area 0.0004 $m^2$ each are placed at a distance of 0.02 m then conductivity of solution is : 

  1. $1 \,s\,cm^{-1}$
  2. $0.01 \,s\,cm^{-1}$
  3. $0.001 \,s\,cm^{-1}$
  4. $10 \,s\,cm^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Conductivity k = (1 / R) * (l / A). Resistance R = 50 ohm, distance l = 0.02 m = 2 cm, area A = 0.0004 m^2 = 4 cm^2. Thus k = (1 / 50) * (2 / 4) = 0.02 * 0.5 = 0.01 S cm^-1.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

In passing $3\ F$ of electricity through three electrolytic cells connected in series containing $Ag^{\oplus}, Ca^{2+}$, and $Al^{3+}$ ions, respectively. The molar ratio in which the three metal ions are liberated at the electrodes is:

  1. $1 : 2 : 3$
  2. $2 : 3 : 1$
  3. $6 : 3 : 2$
  4. $3 : 4 : 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For all the solution $Q=i\times t$ is same,

${ n } _{ Ag }=\cfrac { Q }{ nF } =\cfrac { Q }{ F } ({ Ag }^{ + }+{ e }^{ - }\longrightarrow Ag)$
${ n } _{ Ca }=\cfrac { Q }{ 2F } ({ Ca }^{ 2+ }+2{ e }^{ - }\rightarrow Ca)$
${ n } _{ Al }=\cfrac { Q }{ 3F } ({ Al }^{ 3+ }+3{ e }^{ - }\rightarrow Al)$
$\therefore $ ${ n } _{ Ag }:{ n } _{ Ca }:{ n } _{ Al }=1:\cfrac { 1 }{ 2 } :\cfrac { 1 }{ 3 } =6:3:2$
$\therefore $   Molar ratio$=6:3:2$.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

The charge required for reducing $1$ mole of $MnO^- _4$ to $Mn^{2+}$ is:

  1. $1.93\times 10^5$C
  2. $2.895\times 10^5$C
  3. $4.28\times 10^5$C
  4. $4.825\times 10^5$C
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We will write the reactions and do molar analysis to get clear idea of  the solution.
In the following reaction the oxidation state of Mn is changing from +7 to +2.
$MnO^{-} _{4}+5e^{-}\rightarrow Mn^{2+}$
Here, 5 moles of electrons are needed for reduction of 1 mole of $MnO^{-} _4$ to $ Mn^{2+}$
As, 5 moles of elctrons=5 Faradays
$\implies$Quantity of charge required $=5\times96500=4.825\times10^{5} $ Coulombs

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Fill in the blank with appropriate words.
The electrolytic solution is always neutral because the total charge on $\underline{(i)}$ is equal to $\underline{(ii)}$ on $\underline{(iii)}$. Unlike the metallic conductor, the electrolyte conducts the electric current by virtue of movement of its $\underline{(iv)}$. The property due to which a metal tends to go into solution in term of positive ions is known as $\underline{(v)}$.
(i), (ii), (iii), (iv) and (v) respectively are:

  1. cations, partial charge, anions, electrons, reduction

  2. cations, total charge, anions, ions, oxidation

  3. cations, ionic charge, anions, atoms, dissolution

  4. cations, partial charge, anions, molecules, electrolysis
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

To maintain neutrality, the total charge on cation is equal to the total charge on anion.

Metallic conductor contains ions. Hence, the current is due to the movement of ions.
Oxidation is the loss of an electron. 
Hence, metals undergo oxidation.
$M\longrightarrow M^{2+}+2e^-$

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

A solution containing one mole per litre of each $Cu(NO _3) _2$, $AgNO _3, Hg _2(NO _3) _2$ and $Mg(NO _3) _2$ is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts(reduction potential) are.
$Ag^+/Ag=+0.80, Hg^{2+} _2/2Hg =+0.79$
$Cu^{2+}/Cu=+0.34, Mg^{2+}/Mg =-2.37$
With increasing voltage, the sequence of deposition of metals on the cathode will be:

  1. Ag, Hg, Cu, Mg

  2. Mg, Cu, Hg, Ag

  3. Ag, Hg, Cu

  4. Cu, Hg, Ag
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

 The standard electrode potentials ${\text{E}} _{{{\text{I}} _2}/{1^ - }}^ \circ ,{\text{E}} _{{\text{B}}{{\text{r}}^ - }/{\text{B}}{{\text{r}} _2}}^ \circ $ and ${\text{E}} _{{\text{Fe/F}}{{\text{e}}^{2 + }}}^ \circ $ are respectively $ + 0.54{\text{V,}} - 1.09{\text{V}}$ and 0.44 V. On the basis of the above data which of the following process is nonspontaneous?

  1. ${\text{B}}{{\text{r}} _2} + 2{{\text{I}}^ - }\;{\text{2B}}{{\text{r}}^ - } + {{\text{I}} _2}{\text{ }}$
  2. ${\text{Fe}} + {\text{B}}{{\text{r}} _2}\;{\text{F}}{{\text{e}}^{2 + }} + 2{\text{B}}{{\text{r}}^ - }$
  3. ${\text{Fe}} + {{\text{I}} _2}{\text{ F}}{{\text{e}}^{2 + }} + 2{{\text{I}}^ - }$
  4. ${{\text{I}} _2} + 2{\text{B}}{{\text{r}}^ - }2{{\text{I}}^ - } + {\text{B}}{{\text{r}} _2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

the one which is having higher oxidation potential will undergo oxidation,

in D, $I^-$ is having lower reduction potential in comparison to $Br^-$, so has to undergo oxidation but in option D, it is getting oxidized.
hence, it is a nonspontaneous reaction.

Multiple choice physics electric current and its effects thermal effect of electric current heating effect of electric current effect of electric current

A copper voltameter is connected in series with a coil of resistance $10 \Omega $. When a steady current is passed through the circuit, $0.297 g$ of copper is found to be deposited at the cathode in $15 min$. The electrochemical equivalent of copper is $3.3 \times 10^{-7} kg C^{-}$. Heat liberated in the coil is:

  1. $900 J$
  2. $90 J$
  3. $9 J$
  4. $9 kJ$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle I = \frac {m}{Zt} = \frac {297 \times 10^{-6}}{3.3 \times 10^7 \times 15 \times 60}=1A$

$H = I^2Rt = 1 \times 10 \times 15 \times 60$
$=9000J = 9kJ$

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

One electron volt is equal to .......................

  1. $\displaystyle 1.6\times 10^{-19}$ Joule
  2. $\displaystyle 16\times 10^{-19}$ Joule
  3. $\displaystyle 1.6\times 10^{-10}$ Joule
  4. $\displaystyle 1.6\times 10^{-9}$ Joule
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

One electron volt is the energy gained by an electron accelerated through a potential difference of 1 volt, which is 1.6 * 10^-19 Joules.

Multiple choice chemistry amines introduction to diazonium salts structure, preparation and physical properties of diazonium salts diazonium salts

In an analytical determination of arsenic, a solution containing arsenious acid, $ H _{3}AsO _{4},$ KI and a small amount of starch is electrolysed. The electrolysis produces free $I _{2}$ from $ I^{-}$ ions and the $ I _{2}$ immediately oxidizes the arsenious acid to hydrogen arsenate ion, $ HAsO _{4}^{2-}.$
$ I _{2}+H _{3}AsO _{3}+H _{2}O \rightarrow 2I^{-} + HAsO _{4}^{2-}+4H^{+}$
When the oxidation of arsenic is complete, the free iodine blue colour. If during a particular run, it takes 96.5 s for a current of 10.0 mA to give an end point (indicated by the blue colour),how many g of arsenic are present in the solution. (As = 75)

  1. $ 6.3 \times 10^{-4} g$
  2. $1.26 \times 10^{-4}g$
  3. $ 3.15 \times 10^{-5}g$
  4. $ 6.3 \times 10^{-5}g$
Reveal answer Fill a bubble to check yourself
D Correct answer
Multiple choice
  1. In electrolysis of water, oxygen gets deposited at the

  1. cathode

  2. anode

  3. at both the electrodes

  4. there is no oxygen released
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In water electrolysis, oxygen gas is produced at the anode (positive electrode) and hydrogen at the cathode (negative electrode). This follows from oxidation-reduction principles: water molecules lose electrons at the anode to form oxygen. The '11.' prefix appears to be a question number artifact.

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Calculate the standard voltage that can be obtained from an ethane oxygen fuel cell at $25^o C$.
$C _2H _6(g) + 7/2O _2(g) \rightarrow 2CO _2(g) + 3H _2O(1); \Delta G^o = -1467 \,kJ$

  1. $+0.91$
  2. $+0.54$
  3. $+0.72$
  4. $+1.08$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The standard cell potential is calculated using E_cell = -Delta G^o / (n F). For the ethane-oxygen fuel cell, burning one mole of C2H6 involves 14 electrons (since carbon goes from -3 to +4, 2 carbons give 14 electrons). Delta G^o = -1467 kJ = -1467000 J. F = 96500 C/mol. E = 1467000 / (14 * 96500) = +1.08 V.

Multiple choice physics static electricity properties of charges charge properties of charge

A gold coin has a charge of $+10^{-4} C$. The number of electrons removed from it is:

  1. $10^6$
  2. $625 \times 10^{12}$
  3. $1.6 \times 10^{-25}$
  4. $1.6 \times 10^{13}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total charge of electrons removed from the coin would be equal to the positive charge acquired by the coin.

Thus, $10^{-4}C=n\times 1.6\times 10^{-19}C$
$\implies n=625\times 10^{12}$