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 chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Electrolysis of a solution of $Mn{ SO } _{ 4 }$ in aqueous sulphuric acid is a method for the preparation of $MnO _ 2$. Passing a current of $27A$ for $24$ hours gives $1kg$ of $MnO _2$. The current efficiency in this process is:

  1. $100$%
  2. $95.185$%
  3. $80$%
  4. $82.951$%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The reaction taking place on the given reaction are,

$Mn^{2+}+2H _2O\rightarrow MnO _2+2H^{+}+H _2$

Amount of current is given by,
$m= Z \times I \times t$

$I=m\dfrac{F\times x}{t\times M}$

    $=1000g\times{\dfrac{96500\times 2}{24\times 60 \times 60 \times 86.9}}$

    $\implies I=25.70A$

Now current efficiency$=\dfrac{25.7}{27} \times 100=95.185$ %

Hence,option B is correct answer.

Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Consider the reaction : $Cr _2O _7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+}+7H _2O$


What is the quantity of electricity in coulombs needed to reduced $1\ mol$ of $Cr _2O _7^{2-}$?

  1. $6 \times 10^6C$
  2. $5.79\times 10^5C$
  3. $5.25 \times 10^5C$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$Cr _2O _7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+}+7H _2O$

The valency factor for Chromium is $6$.

Applying faraday's first law-
$ \dfrac{Q} {F} =mole \times V.f.$

$  Q            =mole \times V.f. \times F$ 

$  Q            =1 \times 6 \times 96500$

$  Q            =5.79\times 10^5\ C$
Multiple choice chemistry chemical changes applications of electrolysis electroplating extraction of metals by electrolysis

Calculate the amounts of Na and chlorine gas produced during the electrolysis of fused $NaCl$ by the passage of 1 ampere current for 25 minutes. 

  1. 0.3565 gm and 0.55 gm

  2. 0.3565 gm and 0.66 gm

  3. 0.55 gm and 0.3565

  4. None of above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$i= 1 \ amp$
$t= 25 \ min = 25 \times 60 \ sec$
Molar mass of $Na= 23 \ g \ mol^{-1}$
Molar mass of $Cl _2= 71 \ g \ mol^{-1}$
$Cl^-= \cfrac 12 {(12)}=35.5 \ g$
Formula used, $m= \cfrac {Molar \ mass}{n \times F} \times i \times t$
For $Na$, $n=1$
$m= \cfrac {23}{1 \times 96500} \times 1 \times 25 \times 60$
$=0.3575 \ g$

For $Cl$, $n=1$
$m= \cfrac {35.5}{1 \times 96500} \times 1 \times 25 \times 60$
$=0.5518 \ g$

$\therefore$ The amounts of $Na$ and $Cl$ gas produced are $0.3575 \ g$ and $0.5518 \ g$ respectively.
Multiple choice group 17 elements - trends in chemical properties group 17 elements - properties p- block elements-ii p-block elements chemistry

Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidizing power.

Ion $Cl{ O } _{4}^{-}$ $I{ O } _{4}^{-}$ $Br{ O } _{4}^{-}$
Reduction potential $E^{\circ}$/V $E^{\circ}$= 1.19V $E^{\circ}$= 1.65V $E^{\circ}$= 1.74V
  1. $Cl{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Br{ O } _{4}^{-}$
  2. $I{ O } _{4}^{-}$ > $Br{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$
  3. $Br{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$
  4. $Br{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The reduction potential of the substance is the ability of the substance to be reduced.So as the reduction potential increases reducing ability of the  substance increases.It means oxidizing power of the substance (The ability of the substance to make the other substance to lose the electrons increases which noting but oxidizing power) increases.

So the decreasing order of oxidizing power is $Br{ O } _{4}^{-}$ > $I{ O } _{4}^{-}$ > $Cl{ O } _{4}^{-}$.

Hence option $C$ is correct.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

An acidic solution of Cu+ salt containing 0.4 g of Cu+ is electrolysed until all the Cu is deposited. The electrolysis is continued for 7 more minutes with volume of the solution kept at 100 ml and the current at 1.2 amp. Calculate volume of the gases evolved at NTP during entire electrolysis( atomic weight of Cu= 63.6)

  1. H2 = 68.32 ; O2 = 85.23 ml

  2. H2 = 40.34 ; O2 =70.12 ml

  3. H2= 51.46 ; O2 = 90.12 ml

  4. H2=58.46 ; O2 = 99.68 ml
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is a complex stoichiometry and electrolysis problem. After Cu+ is deposited, the electrolysis of water occurs (2H2O -> 2H2 + O2). Calculating moles of electrons from current and time, then applying Faraday's laws, leads to the volumes of H2 and O2 evolved.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

Given below are the half cell reactions

${ Mn }^{ 2+ }+{ 2e }^{ - }\longrightarrow { Mn }  E=-1.81V$

$2\left( { Mn }^{ 3+ }+{ e }^{ - }\longrightarrow { Mn }^{ 2+ } \right) E=+1.51V$ will be

        




  1. 0.33V ; the reaction will not occur .

  2. 0.33V ; the reaction will occur.

  3. 2.69V ; the reaction will not occur.

  4. 2.69V ; the reaction will occur
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the cell potential, we combine the half-reactions. The overall reaction is 3Mn^2+ -> Mn + 2Mn^3+. The E_cell = E_cathode - E_anode. Using the given values, E_cell = 1.51 - 1.81 = -0.30V. Since E is negative, the reaction is non-spontaneous.

Multiple choice metallic and electrolytic conduction electrolysis chemical reactions electrochemistry chemistry

How much time is required for the complete decomposition of 2 moles of water using a current of 2 ampere?

  1. 26.805 h

  2. 153.61 h

  3. 107.22 h

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Solution:- (D) none of these
$2 {H} _{2}O \longrightarrow 2 {H} _{2} + {O} _{2}$
From the above reaction-
$1$ mole of ${H} _{2}O$ exchanges $2$ moles of electrons, then $2$ moles of ${H} _{2}O$ will exchange $4$ moles of electrons.
From Faraday's law of electrolysis,
$q = nF$
$\Rightarrow i \times t = nF \; \left( \because q = i \times t \right)$
$\Rightarrow t = \cfrac{nF}{i} = \cfrac{4 \times 96500}{2} = 193000 \; s = 53.61 \; hr$
Hence the time required is $53.61$ hours.
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

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