MCQ Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 14

The electrolysis of a solution resulted in the formation of

$H_{2}(g)$ at the cathode and

$O_{2}(g)$ at the anode. The solution is :

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$AgCl(aq)$
0%
highly concentrated $H_{2}SO_{4}(aq)$
0%
highly concentrated $HCl$ solution
0%
$CuCl_{2}(aq)$

An electric current is passed through a copper voltameter and a water voltameter connected in series. If the copper of the copper voltameter now weights 16 mg less, hydrogen liberated at the cathode of the water voltameter measures at STP about:

0%
$4.0ml$
0%
$5.6ml$
0%
$6.4ml$
0%
$8.4ml$

Coulomb is equal to ........

0%
ampere $\times$ second
0%
ampere $\times$ minute
0%
watt $\times$ second
0%
volt $\times$ second

The current strength of

$3.863\ amp$ was passed through molten calcium oxide for

$41$ minutes and

$40$ seconds. The mass of calcium in grams deposited at the cathode is: (Atomic mass of Ca is

$40\ g/mol, 1\ f=96500\ C)$

0%
$4$
0%
$2$
0%
$6$
0%
$8$
0%
$1$

Consider the half-cell reaction(s)

${Cu}^{2+}+{e}^{-}\rightarrow {Cu}^{+};{E}^{o}=0.15V$

${Cu}^{2+}+2{e}^{-}\rightarrow {Cu};{E}^{o}=0.33V$

${E}^{o}$ for the half-cell reaction:

${Cu}^{+}+{e}^{-}\rightarrow Cu$ will be:

0%
$0.48V$
0%
$0.18V$
0%
$0.31V$
0%
$0.51V$

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half-cell reactions and their standard potentials are given below:

$MnO_{4}^{-}(aq)+8H^{+}(aq)+5e^{-}\rightarrow Mn^{2+}(aq)+4H_{2}O(l);$

$E^{\circ}= 1.51V$

$Cr_{2}O_{7}^{2-}(aq)+14H^{+}(aq)+6e^{-}\rightarrow 2Cr^{3+}(aq)+7H_{2}O(l);$

$E^{\circ}= 1.38V$

$Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq);$

$E^{\circ}= 0.77V$

$Cl_{2}(g)+2e^{-}\rightarrow 2Cl^{-}(aq);$

$E^{\circ}= 1.40V$

Identify the only incorrect statement regarding the quantitative estimation of aqueous

$Fe(NO_{3})_{2}$

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$MnO_{4}^{-}$ can be used in aqueous $HCl$
0%
$Cr_{2}O_{7}^{2-}$ can be used in aqueous $HCl$
0%
$MnO_{4}^{-}$ can be used in aqueous $H_{2}SO_{4}$
0%
$Cr_{2}O_{7}^{2-}$ can be used in aqueous $H_{2}SO_{4}$

If one mole electrons is passed through the solutions of

$AlCl_{3}, AgNO_{3}$ and

$MgSO_{4}$ , in what ratio

$Al, Ag$ and

$Mg$ will be deposited at the electrodes?

0%
$3 : 6 : 2$
0%
$2 : 6 : 3$
0%
$1 : 2 : 3$
0%
$3 : 2 : 1$

Explanation

From the second law of electrolysis,

$n_{eq}(Al) = n_{eq}(Ag) = n_{eq}(Mg)$ ---- 1

$n_{eq} = moles \times n-factor$

Now,

$Al^{+3} + 3e^{-}\rightarrow Al$

$Ag^{+} + e^{-} \rightarrow Ag$

$Mg^{+2} + 2e^{-} \rightarrow Mg$

So, ratio will be

$\dfrac {1}{3} mol\ Al : 1 mol\ Ag : \dfrac {1}{2} mol\ Mg = 2 : 6 : 3$ .

The standard reduction potential of the

$Ag^+|Ag$ electrode at 298 K is 0.80 V. The solubility product of AgI is

$6.4 \times 10^{-17}$ at 298 K. (2.303RT/F = 0.06). The potential of

$I^{-} (0.04 \, M)|AgI|Ag$ electrode at 198 K is

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$-0.088\,V$
0%
$+0.088\,V$
0%
$-0.172\,V$
0%
$+0.172\,V$

Which of the following statement (s) differentiate between electrochemical cell and electrolytic cell?

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Spontaneous or non-spontaneous nature of the chemical process.
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Chemical reactions occurring at the electrodes.
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Positive and negative nature of anode.
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Dependence on Faraday's law.

Two electrochemical cells are assembled in which the following reactions occur:

$V^{2+}+VO^{2+}+2H^{+} \rightarrow 2V^{3+}+H_{2}O\ ;\ E_{cell}^{\circ} = 0.616\ V$

$V^{3+}+Ag^{+}+H_{2}O \rightarrow VO^{2+}+Ag(s)+2H^{+}\ ;\ E_{cell}^{\circ} = 0.439\ V$

$E_{Ag^{+}|Ag}^{\circ} = 0.799\ V,$ what is

$E_{V^{3+}|V^{2+}}^{\circ}?$

0%
$-0.256\ V$
0%
$+0.256\ V$
0%
$+1.854\ V$
0%
$-1.854\ V$

From the following

$E^{\circ}$ values for the half-cells:

$(i)\ D\rightarrow D^{2+}+2e^{-};E^{\circ}=-1.5V$

$(ii)\ B^{+}+e^{-}\rightarrow B;E^{\circ}=-0.5V$

$(iii)\ A^{3-}\rightarrow A^{2-}+e^{-};E^{\circ}=1.5V$

$(iv)\ C^{2+}+e^{-}\rightarrow C^{+};E^{\circ}= +0.5V$

Which combination of two half-cells would result in a cell with the largest potential?

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$(i)$ and $(iii)$
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$(i)$ and $(iv)$
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$(iii)$ and $(iv)$
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$(ii)$ and $(iv)$

Statement

$I$ : Electrolysis of

$CuCl_{2}(aq)$ gives 1 mole of Cu and 1 mole of

$Cl_{2}$ by the passage of suitable charge.
Statement

$II$ : Equal equivalents of Cu and

$Cl_{2}$ are formed during the passage of same charge.

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If both statements are CORRECT, and Statement $II$ is the CORRECT explanation of Statement $I$ .
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If both statements are CORRECT, and Statement $II$ is NOT the CORRECT explanation of Statement $I$ .
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If Statement $I$ is CORRECT, but Statement $II$ is INCORRECT.
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If Statement $I$ is INCORRECT,but Statement $II$ is CORRECT.

Statement

$I$ : An electrochemical cell can be set up only when the redox reaction is spontaneous.
Statement

$II$ : A reaction is spontaneous if free energy change at constant temperature and pressure is negative.

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If both statements are , and Statement $II$ is the explanation of Statement $I$ .
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If both statements are , and Statement $II$ is NOT the explanation of Statement $I$ .
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If Statement $I$ is , but Statement $II$ is Incorrect.
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If Statement $I$ is , but Statement $II$ is Correct.

Pick up the false statement(s):

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The net chemical change in a galvanic cell reaction is always a redox reaction.
0%
In a galvanic cell made of cobalt and cadmium electrodes, the cobalt electrode acts as the anode.
0%
Standard potential increases with increasing concentration of the electrolyte.
0%
Calomel electrode is a reference electrode having $0.00$ volt potential.

Which of the following solutions have highest resistance?

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1N - NaCl
0%
0.05 N - NaCl
0%
2N - NaCl
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0.1 N - NaCl

The graph represent a current-voltage behaviour of water. Choose the correct option:

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Ohm's law is obeyed
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Electrolysis in general do not obey ohm's law
0%
Dissociation takes place at E, and it obeys Ohm's law thereafter
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Ohm's law is not valid for low voltage

Which of the following statements does not differentiate between electrochemical cell and electrolytic cell?

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Spontaneous or non-spontaneous nature of the chemical process.
0%
Chemical reactions occurring at the electrodes
0%
Positive and negative nature of anode.
0%
EMF measurement.

Statement

$II$ : In an electrochemical cell, anode and cathode are, respectively, negative and positive electrode.
Statement

$II$ : At anode, oxidation takes place and at cathode reduction takes place.

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If both statements are CORRECT, and Statement $II$ is the CORRECT explanation of Statement $I$ .
0%
If both statements are CORRECT, and Statement $II$ is NOT the CORRECT explanation of Statement $I$ .
0%
If Statement $I$ is CORRECT, but Statement $II$ is INCORRECT.
0%
If Statement $I$ is INCORRECT, but Statement $II$ is CORRECT.

${ E }_{ cell }^{ o }$ for the reaction

$Co(s)+{ Ni }^{ 2+ }\quad \longrightarrow { Co }^{ 2+ }+Ni(s)$ is

$+0.03$ volt. If cobalt metal is added to an aqueous solution having

$\left[ { Ni }^{ 2+ } \right] =1M$

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the reaction will not proceed in the forward direction at all
0%
the displacement of ${Ni}^{2+}$ from solution by $Co$ will go to completion
0%
the displacement of ${Ni}^{2+}$ from solution by $Co$ will proceed to a considerable extent, but the reaction will stop before the ${Ni}^{2+}$ is completely displaced
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only the reverse reaction will occur

Resistance of 0.2 M solution of an electrolyte is 50

$\Omega$ . The specific conductance of the solution of 0.5 M solution of the same electrolyte is 1.4 S

$m^{-1}$ and resistance of the same solution of the electrolyte is 280

$\Omega$ . The molar conductivity of 0.5 M solution of the electrolyte is ?

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$5\times 10^{-4}$
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$5\times 10^{-3}$
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$5\times 10^{3}$
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$5\times 10^{2}$

Explanation

We have

$κ= \frac{1}{R}\; X \;Cell \;constant$ .

For 0.02M solution :

$1.4\;Sm^{−1} = \frac{1}{50\;ohm.}\;X\; Cell \;constant$

or Cell constant =

$(1.4\;Sm^−1)(50\;ohm) =70m^{−1}$

For 0.5M solution

$κ=(\frac{1}{280ohm})(70\;m^{−1})=0.25\;Sm^{−1}$

We have

$Λ_m=\frac{κ}{Molarity}$

$\frac{0.25\;Sm^{−1}}{0.5\;molL^{−1}}$ )

$5\;×\;10^{−4}Sm^2mol^{−1}$ [Option (A)]

Corrosion is an electrochemical process. It involves:

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Loss of electrons by iron, $Fe\longrightarrow Fe^{2+}+2e$ . i.e iron acts as an anode
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Impurities act as a cathode. Electrons are used in forming hydroxyl ions
$H_2O+O+2e \longrightarrow 2OH^-$
0%
Ferrous ions are oxidised to ferries ions in presence of dissolved oxygen.
$2Fe^{2+}+O+H_2O\longrightarrow 2Fe^{3+}+2OH^-$
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All the above reactions

$\cfrac { 1 }{ 2 } { H }_{ 2 }(g)+AgCl(s)={ H }^{ + }(aq)+{ Cl }^{ - }(aq)+Ag(s)$ occurs in the galvanic cell:

0%
$Ag/AgCl(s)|KCl(sol)\parallel Ag{ NO }_{ 3 }(sol)|Ag\quad$
0%
$Pt/{ H }_{ 2 }(g)|HCl(sol)\parallel Ag{ NO }_{ 3 }(sol)|Ag$
0%
$Pt/{ H }_{ 2 }(g)|HCl(sol)\parallel AgCl(s)|Ag$
0%
$Pt/{ H }_{ 2 }(g)|KCl(sol)\parallel AgCl(s)|Ag$

Explanation

$\cfrac { 1 }{ 2 } { H }_{ 2 }(g)+AgCl(s)={ H }^{ + }(aq)+{ Cl }^{ - }(aq)+Ag(s)$

$Pt/{ H }_{ 2 }(g),HCl(sol)\parallel AgCl(s)/{ Ag }^{ 2+ }$

Hence, option C is correct.

In the galvanic cell,

$Cu|{ Cu }^{ 2+ }(1M)\parallel { Ag }^{ + }(1M)|Ag$ , the electrons will travel in the external circuit:

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from $Ag$ to $Cu$
0%
from $Cu$ to $Ag$
0%
electrons do not travel in the external circuit
0%
none of these

Which of the following facts about the chemical cell and concentration cell is correct?

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Chemical cell is an electrolytic cell whereas concentration cell is a galvanic cell
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Chemical cell has an overall cell reaction whereas the concentration cell has no overall reaction.
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Two half cells of both the chemical and concentration cells are chemically different
0%
${E}_{cell}$ equations (Nernst equation) of both the cells have the therm ${E}_{cell}^{o}$

One coulomb is the charge of

0%
$1$ mole of electrons
0%
$\dfrac{1}{96500}$ mole of electrons
0%
$\% 500$ moles of electrons
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$none\ of\ these$

Two electrolytic cells containing

$CuSO_{4}$ and

$AgNO_{4}$ respectively are connected in series and a current is passed through them unit 1 mg of copper is deposited in the first cell. The amount of silver deposited in the second cell during this time is approximately
[ Atomic weight of copper and silver are respectively 63.57 and 107.88]

0%
$3.4 mg$
0%
$5.1 mg$
0%
$6.8 mg$
0%
$1.7 mg$

Explanation

$\dfrac{m_{1}}{m_{2}} = \dfrac{E_{1}}{E_{2}}$ (By faraday law for same current and time)
Where

$E_{1}$ and

$E_{2}$ are the chemical equivalents and

$m_{1}$ and

$m_{2}$ are the masses of copper and silver respectively.

$E = \dfrac{Atomic\, weight}{Valency} E_{1} = \dfrac{63.57}{2} = 31 . 79$ and

$E_{2} = \dfrac{107.88}{1} =107 . 88$

$\therefore \dfrac{1mg}{m_{2}} = \dfrac{31.79}{107.88} \Rightarrow m_{2} = \dfrac{107.88}{31.79}mg = 3.4 mg$

Resistance of a voltameter is 2

$\Omega$ , it is connected in series to a battery of 10 V through a resistance of 3

$\Omega$ . In a certain time mass deposited on cathode is 1 gm. Now the voltameter and the 3

$\Omega$ resistance are connected in parallel with the battery. Increase in the deposited mass on cathode in the same time will be

0%
0
0%
1.5 gm
0%
2.5 gm
0%
2 gm

Explanation

Remember mass of the metal deposited on cathode depends on the current through the voltameter and not on the current supplied by the battery. Hence by using

$m =Zit$ , we can say

$\dfrac{m_{Parallel}}{m_{Series}}=\dfrac{i_{Parallel}}{i_{Series}}\Rightarrow m parallel = \dfrac{5}{2}\times 1=2.5 gm.$
1727701_1817154_ans_7178f67fff894b03bdc17588380bb08e.png

1727701_1817154_ans_7178f67fff894b03bdc17588380bb08e.png

The amount of charge required to liberate 9 gm of aluminium ( atomic weight = 27 and valency = 3 ) in the process of electrolysis is ( Faraday's number = 96500 coulombs/gm equivalent)

0%
321660 coulombs
0%
69500 coulombs
0%
289500 coulombs
0%
96500 coulombs

Explanation

The atomic mass of

$Al=27 g/mol$

$9g$ of

$Al=\dfrac{9}{27}=0.333 moles$

1 mole

$Al= 3$ moles electrons.

$0.333$ moles

$Al=0.333\times 3=1$ mole electrons

1 mole electrons =1 faraday of electricity

$=96500 C.$

Hence, option (D) is the correct answer.

$E_n = -\dfrac{313.6}{n^2}$ , if the value of

$E_i = -38.84$ to which value 'n' corresponds?

0%
2
0%
4
0%
1
0%
3

Explanation

Given :-

$E_n = \dfrac{-313.6}{n^2}$

To calculate :- The value of n if the value of

$E_i = -38.84 kcal \ mol^{-1}$

Sol :-

$E_i = -38.84$

$-38.84 = \dfrac{-313.6}{n^2}$

$n^2 = \dfrac{313.6}{38.84}$

$n^2 = 8$

$n = 2.82 \approx 3$

So, the correct option is (D) 3

If redox reaction takes place in cell then electromotive force (e.m.f) of cell will be:

0%
positive
0%
negative
0%
zero
0%
one

Which of the following mixtures represents rusting of iron?

0%
FeO and Fe(OH) $_3$
0%
FeO and Fe(OH) $_2$
0%
Fe $_2$ O $_3$ and Fe(OH) $_3$
0%
Fe $_3$ O $_4$ and Fe(OH) $_2$

A cell reaction would be spontaneous if the cell potential and

$\Delta_{r} G$ are respectively:

0%
positive and negative
0%
negative , negative
0%
zero ,zero
0%
positive , zero

Explanation

For a spontaneous reaction, the standard change in gibbs free energy

$\Delta G^{\circ}$ and the standard cell potential Ecell will be negative and positive respectively.

$\Delta G^{\circ}<0$ then

$E_{\text {cell }}^{\circ}>0$ as

$\Delta C^{\circ}=-n F E_{\text {cele }}^{\circ}$

Thus,

$\Delta \mathrm{G}^{\circ}$ and

$\Delta \mathrm{E}_{\text {cell }}^{\circ}$ have of opposite signs.

So, option A. is the correct answer.

Which of the following statements is correct?

0%
$E_{cell}$ and $\Delta _rG$ of a cell reaction,both are extensive properties.
0%
$E_{cell}$ and $\Delta _rG$ of a cell reaction, both are intensive properties.
0%
$E_{cell}$ is an intensive property, while $\Delta _rG$ of a cell reaction is an extensive property.
0%
$E_{cell}$ is an extensive property, while $\Delta _rG$ of a cell reaction is an intensive property.

Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below:

$MnO^{-}_4(aq)+8H^{+}(aq)+6e^{-} \longrightarrow$

$\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l) \mathrm{E}^{\circ}=1.51 \mathrm{V}$

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+14 \mathrm{H}^{+}(a q)+6 e^{-} \longrightarrow$

$2 \mathrm{Cr}^{3+}(a q)+7 \mathrm{H}_{2} \mathrm{O}(l) ; \mathrm{E}^{\circ}=1 \cdot 38 \mathrm{V}$

$\mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) ; \mathrm{E}^{\circ}=0.77 \mathrm{V}$

$\mathrm{Cl}_{2}(g)+2 e^{-} \rightarrow 2 \mathrm{Cl}^{-}(a q) ; \mathrm{E}^{\circ}=1 \cdot 40 \mathrm{V}$

Identify the only incorrect statement regarding the quantitative estimation of aqueous

$\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}$

0%
$\mathrm{MnO}_{4}^{-}$ can be used in aqueous $HCl$
0%
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ can be used in aqueous $HCl$
0%
$\mathrm{MnO}_{4}^{-}$ can be used in aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$
0%
$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}$ can be used in aqueous $\mathrm{H}_{2} \mathrm{SO}_{4}$

Explanation

$\mathrm{MnO}_{4}^{-}$ will oxidise

$\mathrm{Cl}^{-}$ ion according to the equation,

$2 \mathrm{KMnO}_{4}^{-}+16 \mathrm{H}^{+}+10 \mathrm{Cl}^{-} \rightarrow$

$2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{Cl}_{2}$
The cell corresponding to this reaction is

$\mathrm{Pt}, \mathrm{Cl}_{2}(1 \mathrm{bar})\left|\mathrm{Cl}^{-} \| \mathrm{MnO}_{4}^{-}, \mathrm{Mn}^{2+}, \mathrm{H}^{+}\right| \mathrm{P}_{1}$

$E_{\text {cell }}^{\circ}=1 \cdot 51-1 \cdot 40=0 \cdot 11 \mathrm{V}$
As

$E_{\text {cell }}^{\circ}$ is + ve, the above reaction is feasible and

$\mathrm{MnO}_{4}^{-}$ will oxidise not only

$\mathrm{Fe}^{2+}$ ion but

$\mathrm{Cl}^{-}$ ion also.

Which of the following is displaced by Fe ?

0%
Ag
0%
Zn
0%
Na
0%
None of these

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 14 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 14 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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