MCQ Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 3

The amount of metal deposited at cathode on passing an electric current of 0.75 amperes in aqueous ferric sulphate solution for 30 minutes will be:(atomic wt. of Fe =56 )

0%
0.00435 g
0%
0.261 g
0%
0.783 g
0%
0.522 g

The charge of an electron is

$- 1.6 \times {10^{ - 19}}C.$ The value of free charge on

$L{i^ + }$ ion will be

0%
$3.6 \times {10^{ - 19}}C$
0%
$1 \times {10^{ - 19}}C$
0%
$1.6 \times {10^{ - 19}}C$
0%
$2.6 \times {10^{ - 19}}C$

Which of the following statements are correct about voltaic cell?
(I) Chemical energy is converted into electrical energy.
(II) Electrons flow from anode to cathode.
(III) Oxidation occurs at anode.
(IV) Anode is negative and cathode is positive.

0%
I and III
0%
I, II and III
0%
I, III and IV
0%
I, II, III and IV

Number of Faradays involved in the net reaction of Lead accumulator is __________.

0%
1
0%
0.5
0%
2
0%
cannot be predicted

Given:

$2Zn + O_2 → 2ZnO; ΔG° = -616 J$

$2Zn + S_2 → 2ZnS; ΔG° = -293 J$

$S_2 + 2O_2 → 2SO_2; ΔG° = -408 J$

Find

${ G }^{ \circ }$ for the following reaction:

$2ZnS+{ 3O }_{ 2 }\rightarrow 2ZnO+2S{ O }_{ 2 }$

0%
-731 J
0%
-1317J
0%
+731 J
0%
+1317 J

The process in which any electrolyte gets decomposed when electricity is passed through it, is called

0%
electrolysis
0%
decomposition
0%
dissociation
0%
splitting

The process of separating ions is called________.

0%
Electroplating
0%
Electrolysis
0%
Electric refilling
0%
All of these

Which of the following statement is false about electrolysis?

0%
The most commonly occurring chemical reaction in electrolysis is the 'chemical decomposition reaction'.
0%
When an iron rod is to be electroplated with chromium, the chromium electrode is used as an anode, and the iron rod is used as a cathode.
0%
During electrolysis of water, ${H_2}$ accumulates at the cathode and ${O_2}$ accumulates at the anode.
0%
When electrolysis is used in metallurgical applications, the concentration of electrolyte changes.

For the cell

$Hg/Hg_{ 2 }Cl/Cl^{ - }(0.1M)//{ Cl }^{ - }(0.001M)/{ Cl }_{ 2 },Pt,{ E }^{ 0 }$ cell is 1.10 V.
Hence

$E_{cell}$ is:

0%
1.1591 V
0%
-1.1591 V
0%
1.0409 V
0%
-1.0409

The magnitude of charge present on one azide ion in coulombs is:

0%
$96500$
0%
$4.8 \times 10^{19}$
0%
$4.8 \times 10^{-19}$
0%
$1.6 \times 10^{-19}$

Explanation

The charge of one electron is

$-1.602 \times { 10 }^{ -19 }\ \text C$ .

The azide ion

${ N }_{ 3 }^{ - }$ has unit negative charge which is equal to the charge on one electron.

Hence, the magnitude of charge on one azide ion is equal to

$1.602 \times { 10 }^{ -19 }\ \text C$ .

Option D is correct.

${ V }_{ 1 }\ L$ of solution A (Resistance = 50 ohm) is mixed with

${ V }_{ 2 }\ L$ of solution B ( Resistance = 100 ohm) Resistance of final solution is 80 \ ohm then

$\dfrac { { V }_{ 2 } }{ { V }_{ 1 } }$ will be (using same cell):

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

1 ampere current is passed for 60 seconds into an electrolytic cell. Number of electrons that pass through the solution is :

0%
$6.0\times 10$ $^{23}$
0%
$1.2 \times 10$ $^{24}$
0%
$37.5 \times 10$ $^{19}$
0%
$7.48\times 10$ $^{21}$

Explanation

$Total \ charge =t\times I= 60\times 1=60\ C$

$No. \ of\ e^- = \dfrac{total\ charge}{charge\ on\ 1 e^-}$

$=\dfrac{60}{1.6\times 10^{-19}}$

$=37.5\times 10^{19} \ electrons$

Option C is correct.

For hydrogen-oxygen fuel cell at 1 atm and 298 K

${ H }_{ 2 }(g)+\frac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow { H }_{ 2 }O(l),{ \Delta G }^{ \circ }=-240kJ$

${ E }^{ \circ }$ FOR THE CELL IS APPROXIMATELY,

0%
2.48 V
0%
1.24V
0%
2.5V
0%
1.26 V

In an experiment of electroplating, m grams silver deposits at cathode when 4A current is passing for 2 min. If 6A current is passed for 40 s in the same experiment, silver deposited on cathode will be

0%
m/4
0%
m/2
0%
2m
0%
4m

How can four resistors of resistances

$4\Omega,8\Omega,12\Omega$ and

$24\Omega$ be connected to get the highest resistance and lowest resistance respectively:

0%
All in series, all in parallel
0%
All in parallel, all in series
0%
$24\Omega$ and $12\Omega$ in series with the rest two in parallel, $4\Omega$ and $8\Omega$ in parallel with the rest two in series
0%
$4\Omega$ and $8\Omega$ in series with the rest two in parallel, $12\Omega$ and $24\Omega$ in parallel with the rest two in series

The dissociation constant for

$CH_{3}COOH$ is

$1.8\times 10^{-5}$ at

$298\ K$ . The electrode potential for the half-cell:

$Pt|H_{2}\ (1\ bar)\|0.5\ M\ CH_{3}COOH$ , at

$298\ K$ is:

$(log\ 2 = 0.3\ ;\ log\ 3 = 0.48\ ;\ 2.303\ RT/F = 0.06)$

0%
$-0.3024\ V$
0%
$-0.1512\ V$
0%
$+0.3024\ V$
0%
$+0.1512\ V$

$Zn_{(s)} | Zn^{2+}_{(aq)}(1M) || Ni^{2+}_{(aq)}(1M) |Ni_{(s)}$

Which is incorrect for the above given cell?

0%
Electrochemical cell
0%
Voltaic cell
0%
Galvanic cell
0%
Daniel cell

Explanation

Cell reaction in Daniel cell is:

$Zn(s) + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu(s)$ .

So, D is the correct option.

The correct cell diagram for the following reaction and

$E^{\circ}$ for the cell is:

$2AgBr(s)+H_{2}(g)\rightarrow 2Ag(s)+2H^{+}+2Br^{-}$ ;

$E_{AgBr|Ag|Br^{-}}^{^{\circ}}=+0.10V$

0%
$(Pt)H_{2}|H^{+}||Br^{-}|AgBr|Br_{2}(Pt)\ ;\ E^{\circ}=0.10V$
0%
$(Pt)H_{2}|H^{+}||Br^{-}|AgBr|Br_{2}(Pt)\ ;\ E^{\circ}=-0.10V$
0%
$(Pt)Br_{2}|AgBr|Br^{-}||H^{+}|H_{2}(Pt)\ ;\ E^{\circ}= 0.10 V$
0%
$(Pt)Br_{2}|AgBr|Br^{-}||H^{+}|H_{2}(Pt)\ ;\ E^{\circ}= -0.10V$

A volume of

$100\ ml$ of a buffer of

$1 M \ NH_{3}$ and

$1M \ NH_{4}^{+}$ is placed in two half-cells connected by a salt bridge. A current of

$1.5\ A$ is passed through the cell for

$20$ minutes. If only the electrolysis of water takes place and the electrode reactions are;

Right Electrode:

$2H_{2}O +O_{2}+4e \rightarrow 4 OH^{-}$

Left Electrode:

$2H_{2}O \rightarrow 4H^{+}+O_{2}+4e,$

then, the

$pH$ of:

0%
right electrode will increase
0%
left electrode will increase
0%
both electrodes will increase
0%
both electrodes will decrease

Among the following, identify the correct statement?

0%
Chloride ion is oxidized by $O_{2}$
0%
$Fe^{2+}$ is oxidized by iodine
0%
Iodide ion is oxidized by chlorine
0%
$Mn^{2+}$ is oxidized by chlorine

What is the amount of chlorine evolved, when 2A of current is passed for 30 minutes in an aqueous solution of NaCl?

0%
1.32gm
0%
4.56gm
0%
9.81gm
0%
12.6gm

Explanation

Total Charge Supplied(Q)

$= I\times t$

= 2 x 30 x60

= 3600 C

$NaCl\rightarrow{Na}^{+} + {Cl}^{-}$

96500 C liberate

${Cl}_{2}$ = Eq.Wt of

${Cl}_{2}$ = 35.5 g

3600 C liberates

${Cl}_2$ =

$\dfrac{35.5}{96500} \times 3600$ = 1.32 gm

Which of the following does not evolve oxygen at anodes when electrolysis is carried?

0%
dil, ${ H }_{ 2 }{ SO }_{ 4 }$ with Pt electrode
0%
Fused NaOH with Pt electrode
0%
Acidic water with Pt electrode
0%
dil, ${ H }_{ 2 }{ SO }_{ 4 }$ with Cu electrode

Explanation

In the electrolysis of dil.

$H_2SO_4$ using Cu electrodes, the reaction at anode occurs as:

$\rightarrow Cu^2$

$^+$ +

$e^-$

The electrode is not inert an so participates in the reaction. In all other reaction, inert electrode i.e. Pt is used. Thus oxygen is evolved in these reactions.

Therefore D is the correct solution.

$9$ gm

$H_2O$ is electrolysed completely with the current of

$50\%$ efficiency then?

0%
$96500$ charge is required
0%
$2\times 96500$ C charge is required
0%
$5.6$ L of $O_2$ at STP will be formed
0%
$11.2$ L of $O_2$ at STP will be formed

Explanation

Number of moles of water=

$\cfrac {9}{18}=0.5$

Balanced equation for electrolysis of water

$2H_2O\longrightarrow 2H_2+O_2$

$\therefore 2$ moles of

$H_2O$ dissociates to produce

$1$ mole of

$O_2$ at STP.

$\therefore 0.5$ moles of

$H_2O$ produces

$0.25$ moles of

$O_2$ at STP.

$\therefore 0.25$ moles of

$O_2$ will occupy=

$0.25 \times 22.4=5.6 litres$ .

for the cell reaction

$Cu^{2+}(C_1aq)+Zn(s)=Zn^{2+}(C_2aq)+Cu(s)$
of an electrochemical cell, the change in free energy at a given temperature is a function of

0%
$\ln (C_1)$
0%
$\ln (C_2)$
0%
$\ln(C_1+C_2)$
0%
$\ln (C_2 /C_1)$

Explanation

Answer is option (D)

We know

$ΔG=nFE$

Again according to Nernst equation,

$E=\dfrac{0.059}{n}$

$log \dfrac{[Zn^{2+} ]}{[Cu^{2+} ]}$

$at$ 25^o$$ C.

$ΔG$ (free energy change) in an electrochemical cell at a given temperature is a function of

$\dfrac{[Zn^{2+} ]}{ [Cu^{2+}]}$

$=ln\dfrac{C_2}{C_1}$

Lead pipes are corroded quickly by

0%
Dil. $H_{2}SO_{4}$
0%
Conc. $H_{2}SO_{4}$
0%
Acetic acid
0%
Water

Explanation

Organic acids dissolve lead in the presence of oxygen. So, lead pipes are corroded quickly by acetic acid.

$Pb+2CH_{3}COOH+\dfrac{1}{2}O_{2}\rightarrow Pb(CH_{3}COO)_{2}+H_{2}O$

Which of the following is correct expression for electrode potential of a cell ?

0%
$E=E^0-\dfrac{RT}{nF}\ln \dfrac{[product]}{[reactant]}$
0%
$E=E^0+\dfrac{RT}{F}\ln \dfrac{[product]}{[reactant]}$
0%
$E=E^0-\dfrac{RT}{nF}\ln \dfrac{[reactant]}{[product]}$
0%
$E=E^0-\dfrac{RT}{F}\ln \dfrac{[product]}{[reactant]}$

The main reason for not using a mercury electrolytic cell in NaOH manufacture is that

0%
Hg is toxic
0%
Hg is liquid
0%
Hg has a high vapour pressure
0%
Hg is a good conductor of electricity

In which cell the free energy of a chemical reaction is directly converted into electricity

0%
Lenclanche cell
0%
Concentration cell
0%
Fuel cell
0%
Lead storage battery

$E^o_{Cell}$ for some half-cell reactions are given below. On the basis of these mark the correct answers.

(a)

$H^- (aq) + e^- \rightarrow \frac{1}{2} H_2 (g); E^o_{Cell} = 1.23V$

(b)

$2H_2O(l) \rightarrow O_2(g) + 4H^- (aq) + 4e^- ; E^o_{Cell} = 1.23V$

(c)

$2SO^{2-}_4(aq) \rightarrow S_2O^{2-}_8(aq) + 2e^-; E^o_{Cell} = 1.96V$

0%
In dilute sulphuric acid solution, hydrogen will be reduced at cathode.
0%
In concentrated sulphuric acid solution, water will be oxidised at anode.
0%
In dilute sulphuric acid solution, water will be oxidised at anode.
0%
In concentrated sulphuric acid solution, $SO^{2-}_4$ ion will be oxidized to tetrathionate ion at anode.

Explanation

In the electrolysis of dil. H

$_2$ SO

$_4$ , above three reaction takes place. Oxidation half reaction occurs at anode, lower value of standard reduction potential will be preferred. At cathode hydrogen ion, will be converted into hydrogen.

$E{^{o}}_{Fe^{2+}/Fe}=x_{1}V$ and

$E{^{o}}_{Fe^{3+}/Fe^{2+}}=x_{2}V$ , then

$E{^{o}}_{Fe^{3+}/Fe}$ will be :

0%
$(2x_{1}+x_{2})V$
0%
$(3x_{2}-x_{1})V$
0%
$(2x_{1}+x_{2})/3V$
0%
$(x_{1}+2x_{2})/3V$

Explanation

$Fe^{3+} \rightarrow Fe^{2+} + e^- \rightarrow x_2$

$Fe^{2+} \rightarrow Fe + 2e^- \rightarrow x_1$

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

$Total\: \Delta G = \text{Sum of } \Delta G \text{ of the}\ 2\ factors$

$\therefore -3 \times F \times E^{\circ}_{cell} =-2x_1 \times F -x_2\times F$

$\therefore 3E^{\circ}=2x_1 + x_2$

$\therefore E^{\circ}=\dfrac{2x_1 + x_2}{3}$

Hence, the correct option is

$\text{C}$

Rusting of iron is catalysed by which of the following:

0%
$Fe$
0%
$Zn$
0%
O $_{2}$
0%
$H^{+}$

Explanation

The rusting or iron occurs in acidic medium. Hence it is catalyzed by

${ H }^{ + }$ ions.

Hence, correct option is D.

In the electrolysis of $NiSO_{4}$ using Nickel electrodes the reaction that takes place at anode is:

0%
$Ni^{+2}+2e^{-}\rightarrow Ni$
0%
$2H^{+}+2e^{-}\rightarrow H_{2}$
0%
$Ni^{+2}\rightarrow Ni^{+}+e^{-}$
0%
$Ni\rightarrow Ni^{+2}+2e^{-}$

Explanation

During electrolysis of nickel sulphate using nickel electrode, at anode, oxidation of nickel occurs. Nickel loses two electrons to form

${ Ni }^{ 2+ }$ ion.

The following are some statements about electrolysis:

I) Involves flow of electrons from anode to cathode

II) Involves no chemical reaction at electrodes

III) Oxidation occurs at anode and reduction at cathode

IV) Anions migrate towards cathode

0%
Only I and III are correct
0%
Only I and II are correct
0%
Only I, II and III are correct
0%
All are correct

$6.24 \times 10^{19}$ electrons is equal approximately to:

0%
10 coulombs
0%
96500 coulombs
0%
one electron volt
0%
0.1F

Explanation

Since

$6.24\times { 10 }^{ 18 }electrons=1coulomb\\ \therefore \quad 6.24\times { 10 }^{ 19 }electrons=10coulomb\\ \qquad \qquad \qquad \qquad \qquad \left( \dfrac { 6.24\times { 10 }^{ 19 } }{ 6.24\times { 10 }^{ 18 } } \right)$

A photo-cell employs photoelectric effect to convert:

0%
change in the frequency of light into a change in electric voltage.
0%
change in the intensity of illumination into a change in photoelectric current.
0%
change in the intensity of illumination into a change in the work function of the photocathode.
0%
change in the frequency of light into a change in the electric current.

Fill in the blanks:

The filament resistance of bulb is ......, to its resistance when it is not glowing.

0%
greater
0%
lower
0%
equal
0%
none of above.

In corrosion of iron :

0%
an electrochemical reaction (galvanic cell) is formed in which Fe acts as anode and cathode is where $O_{2}$ is reduced
0%
electrons flow from anode to cathode through the metal while ions flow through the waterdroplets
0%
dissolved $O_{2}$ oxidises $Fe^{2+}$ to $Fe^{3+}$ before it is deposited as rust $(Fe_{2}O_{3}.H_{2}O)$
0%
all of the above take place

Explanation

The corrosion of iron is an electrochemical process.

The anode region and cathode region are present on iron.

$Iron(II)$ is oxidized to

$Iron(III)$ by oxygen and is then converted to rust. At the cathode, oxygen is reduced.

The electron flow is from anode to cathode through metal.

This is accompanied by the flow of ions through water droplets.

Maintenance-free batteries, now in use, in place of common batteries, have :

0%
Electrodes made of lead-lead oxide
0%
Electrodes made of calcium-containing lead alloy
0%
Non-aqueous solvents as medium
0%
Platinum electrodes

When galvanic cell is started, with passage of time:

0%
spontaneity of the cell reaction decreases; $E_{cell}$ decreases
0%
reaction quotient Q decreases; $E_{cell}$ increases
0%
reaction quotient Q increases; $E_{cell}$ decreases
0%
at equilibrium $Q=K_{c}; E_{cell}=0$

Explanation

The reactants get used up and the products are produced. This means the concentration of reactants decreases and products increases.

We know Q =

$\frac{P_{1}^{n_{1}} * P_{2}^{n_{2}}}{R_{1}^{m_{1}} * R_{2}^{m_{2}}}$

Thus, the value of Q increases.

We know

$\Lambda G = \Lambda G_{o} + RT log(Q)$ =

$-nFE_{cell}$

Thus, the value of

$\Lambda G$ increases and hence spontaneity decreases. Also,

$E_{cell}$ decreases.

The capacity of cell is measured in:

0%
amperes
0%
ampere-hour
0%
watts
0%
watt-hours

The electric current in a chemical cell is due to movement of :

0%
positive hole charges
0%
positive ions only
0%
negative ions only
0%
positive and negative ions

The internal resistance of a dry cell is of the order of:

0%
0.2 to 0.4 Ohm.
0%
1 to 1.4 Ohm.
0%
2 to 5 Ohm.
0%
1 to 15 Ohm.

Explanation

In dry-cell battery, the electrolyte is in the form of a paste. And there is no presence of nickel and iron material. Due to that, the internal resistance of the dry is very less. The value is approximately in between 0.2 to 0.4

$\Omega.$

The secondary battery is such a battery:

0%
which cannot be recharged
0%
which can be recharged
0%
which can be reused after replacing its chemical
0%
which is charged by primary cells

An example of secondary battery cell is:

0%
Edison Alkaline cell.
0%
Daniel cell.
0%
Lachanche cell.
0%
Bunsen cell.

Primary battery is a battery:

0%
which can be recharged.
0%
which cannot be reconditioned by replacing chemical.
0%
which cannot be reused.
0%
which cannot be recharged.

The quantity of electricity needed to liberate

$0.5\ gram$ equivalent of an element is

0%
$48250\ faraday$
0%
$48250\ coulomb$
0%
$19300\ faraday$
0%
$19300\ coulomb$

Explanation

Electricity required,

$Q=$ No. of equiv.

$\times$ 96500 coulombs

$=0.5\times 96500=48250\ C$

Internal resistance of a battery cell increases with :

0%
increase in concentration of electrolyte
0%
increase in distance between two electrodes
0%
increase in area of the plates inside the electrolyte
0%
increase in size of the electrodes

Metals react differently with water. The more vigorous is the reaction, the more is the bubble formation. The following diagram illustrates the reaction of sodium, magnesium and copper with water.
What can be concluded from this diagram?

0%
Magnesium reacts more vigorously with water than sodium and copper.
0%
Sodium reacts more vigorously with water than magnesium and copper.
0%
Sodium reacts more vigorously than copper but less vigorously than magnesium.
0%
Copper reacts more vigorously with water than sodium and magnesium.

The electric charge for electrode deposition of 1 g equivalent of a substance is:

0%
1 ampere per second.
0%
96,500 coulomb per second.
0%
1 ampere for 1 hour.
0%
charge on 1 mole of electrons.

Which of the following is true for electrolytic refining?

0%
Pure metal is used at anode and impure metal at cathode
0%
Pure metal is used at cathode and impure metal at anode
0%
Both can be possible A and B
0%
None of these

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 3 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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