either reversible or irreversible

neither reversible or irreversible

MCQ Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 10

Heated saw dust catches fire when a drop of concentrated nitric acid is added to it. This is due to

0%
dehydration
0%
oxidation
0%
reduction
0%
dehydrogenation

Passage of three faradays of charge through an aqueous solution of

$AgNO_3, CuSO_4, Al(NO_3)_3$ and

$NaCl$ will deposit metals at the cathode in the molar ratio of:

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

Explanation

$AgNO_3+e^- \longrightarrow Ag+NO_3^-$

$1$ mole of

$AgNO_3$ required

$1$ mole of

$e^-$ or

$1$ Faraday of electricity.

Thus

$3F$ is used for

$3$ mole of

$AgNO_3$ .

$CuSO_4+2e^- \longrightarrow Cu+SO_4^{2-}$

$1$ mole of

$CuSO_4$ required

$2$ Faraday of charge.

Thus

$3$ Faraday of charge will deposit

$\cfrac {3}{2}$ mole of

$CuSO_4$ .

$Al(MO_3)_3+3e^-$ &

$Al+ 3NO^-_3$

$3F$ of charge is required for

$1$ mole of

$Al(NO_3)_3$ .

$NaCl+e^- \longrightarrow Na+Cl^{-}$

$1F$ of charge required for

$1$ mole of

$NaCl$ .

Thus,

$3F$ of charge is required for

$3$ mole of

$NaCl$ .

The molar ratio will be:

$3: \cfrac {3}{2} :1 :3$

or,

$6 : 3:2:6$ .

Which of the following forms vortex ring?

0%
$P_2O_5$
0%
$PH_3$
0%
$NH_3$
0%
None of the above

On passing electricity through dil.

${ H }_{ 2 }{ SO }_{ 4 }$ solution, the amount of substance liberated at the cathode and anode are in the ratio :

0%
$16:1$
0%
$1:16$
0%
$1:8$
0%
$8:1$

Explanation

Electrolysis of dilute sulphuric acid:-

At cathode-

$2{H}^{+} + 2{e}^{-} \longrightarrow {H}_{2}$

At anode-

$4{OH}^{-} + \longrightarrow 2{H}_{2}O + 2{O}_{2} + 4{e}^{-}$

Overall reaction-

${H}_{2}O \longrightarrow 2{H}_{2} + {O}_{2}$

$\therefore$ Amount of substance liberated at cathode = 2 moles of

${H}_{2} = 4g$

Also, amount of substance liberated at anode = 1 mole of

${O}_{2} = 32g$

$\therefore$ Ratio of the amount of substance liberated at the cathode and anode =

$4 : 32= 1 : 8$

How many kgs of silver would be deposited at cathode if

$0.0342 dm^3$ of oxygen measured at NTP is liberated at the anode when silver nitrate solution is electrolyzed between platinum electrodes?(Given : Atomic masses of Ag = 108, O =16)

0%
0.22 gm
0%
0.44 gm
0%
0.66 gm
0%
None of the above

Explanation

$AgNO_3\rightarrow Ag^++NO_3^-$

$2H_2O\rightarrow AH^++O_2+Ae^-$

For 1 mole

$O_2 \Rightarrow 4F$ change released

At NTP

$1$ mole

$O_2=22.4L$

$1L=\dfrac{1}{22.4}$ mole

$O_2$

$0.0342L=\dfrac{0.0342}{22.4}=0.00153$ mole

$O_2$

For

$0.00153$ mole

$\Rightarrow 4\times 0.00153F$

$\Rightarrow 0.00611F$

$\Rightarrow 0.00611\times 96500$

$\Rightarrow 589.34C$

$96500C$ or

$1F$ give

$\theta$

$\Rightarrow 1$ mole

$Ag =108g\, Ag$

$589.34C$ give

$\theta$

$\Rightarrow \dfrac{589.34\times 108}{96500}$

$\Rightarrow 0.66\ gm$

For the half reaction

$B(s)$

$\rightarrow$

$B^{2+} + 2e^-$

${ E }_{ 1 }^{ \circ }=-0.44V$
and

$B^{2+}$

$\rightarrow$

$B^{3+} + e^-$

${ E }_{ 2 }^{ \circ }=1.3V$
What is

${E}_{3}^{\circ}$ for the reaction

$3e^- + B^{+3 }$

$\rightarrow$ B(s):

0%
$0.86 V$
0%
$0.14 V$
0%
$-0.14 V$
0%
$-0.28V$

Explanation

$B\left( s \right) \rightarrow { B }^{ 2+ }+{ 2e }^{ - }\quad \quad { E }_{ 1 }^{ 0 }=-0.44V$

$\underline { { B }^{ 2+ }\rightarrow { B }^{ 3+ }+{ e }^{ - }\quad \quad { E }_{ 2 }^{ 0 }=+1.3V }$

$B\left( s \right) \rightarrow { B }^{ 3+ }+{ 3e }^{ - }\quad ;\quad { E }_{ 3 }^{ 0 }=1.3-0.44V$

$=0.86V$

How much electricity in terms of Faraday is required to produce 40.0 g of

$Al$ from molten

$Al_2O_3$ ?

0%
1 Faraday
0%
2 Faraday
0%
3.2 Faraday
0%
4.44 Faraday

Explanation

Cathode:

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

Anode:

$2O^{2-}+4 e^- \rightarrow O_2$

Overall equation:

${2Al_2O_3}_{(l)}\rightarrow 4Al_{(l)}+{3O_2}_{(g)}$

$3$ Faraday of electricity is required to produce

$1$ mole of

$Al$ .

Now,

$40g$ of

$Al=\cfrac {40}{27}$ mole

$=1.48$ moles of

$Al$

$\therefore 1.48$ moles of

$Al$ can be produced by

$1.48\times 3$ Faraday i.e

$4.44$ Faraday of electricity.

Which of the following will be most easily corroded in moist air?

0%
$Zn$
0%
$Fe$
0%
$Ni$
0%
$Sn$

Explanation

When iron (

$Fe$ ) is exposed to moist air, it reacts with oxygen to form rust.

The formation of a rust is a very complex process, which is thought to begin with the oxidation of iron to ferrous ( iron"+2") ions.

Iron metal reacts in moist air by oxidation to give a hydrated iron oxide. This does not protect the iron surface to further reaction since it flakes off, exposing more iron metal to oxidation. This process is called rusting.

Which among the following cell cannot be regenerated?

0%
$Ni-Cd$ cell
0%
Leclanche cell
0%
All of the given
0%
Lead storage cell

When an electric current is passed through acidified water,

$112\ ml$ of hydrogen gas at

$STP$ collects at the cathode in

$965$ seconds. What is the current passed, in ampere?

0%
$1.0$
0%
$0.5$
0%
$0.1$
0%
$2.0$

Explanation

$\textbf{Step 1: Calculation weight of hydrogen in given volume}$

At STP, volume of one mole of hydrogen gas is

$22.4\ L=22400\ mL$

Weight of one mole of of

$H_2=2\ g$

$22400\ mL\ of\ H_2=2\ g$ .

$\therefore 112\ mL\ of\ H_2=$

$\dfrac{2\times 112}{22400}=0.01\ g$

$\textbf{Step 2: Calculation for amount of current}$

${ 2H }^{ + }+{ 2e }^{ - }\rightarrow { H }_{ 2 }$

According to Faraday's law,

$nF=1\ mole$

where,

$\text{F = Faraday's constant =96500 C}$

$\text{n = number of electrons = 2}$

$\therefore \text{Charge required to deposit 2g of hydrogen} = 2\times 96500$

$\therefore \text{Charge required to deposit 0.01g of hydrogen} = \dfrac{2\times 96500\times 0.01}{2}=965C$

$\text{Current}=\dfrac{Charge}{time}$

$\text{Current}=\dfrac{965}{965}=1A$

So, the current passed is 1 A.

Hence, the correct option is A.

Find the charge of

$48 g$ of

${ Mg }^{ 2+ }$ ions in coulombs.

0%
$2.4\times { 10 }^{ 23 }C$
0%
$6.82\times { 10 }^{ 5 }C$
0%
$3.86\times { 10 }^{ 5 }C$
0%
$1.93\times { 10 }^{ 5 }C$

Explanation

Molar mass of

${Mg}^{+2} = 24 g$

Given mass of

${Mg}^{+2} = 48 g$

No. of moles of

${Mg}^{+2} = \cfrac{\text{Given mass}}{\text{Molar mass}} = \cfrac{48}{24} = 2 \text{ moles}$

1 mole of

${Mg}^{+2}$ = 2 moles of electrons

$\therefore$ 2 moles of

${Mg}^{+2}$ = 4 moles of electrons

$\therefore$ No. of electrons = no. of moles of electrons

$\times$ Avogadro's constant

$\Rightarrow$ No. of electrons

$= 4 \times 6.023 \times {10}^{23} = 2.4 \times {10}^{24}$

As we know that,

Charge on 1 electron

$= 1.6 \times {10}^{-19} C$

$\therefore$ Total charge on 48 g

${Mg}^{+2} = \text{no. of }{e}^{-}s \times \text{charge on 1 } {e}^{-} = \left( 2.4 \times {10}^{24} \right) \times \left( 1.6 \times {10}^{-19} \right) = 3.84 \times {10}^{5} C \approx 3.86 \times {10}^{5} C$

Hence, the correct answer is

$3.86 \times {10}^{5} C$ .

The specific conductance of

$0.1N$

$KCl$ solution at

${23}^{o}C$ is

$0.012{ohm}^{-1}{cm}^{-1}$ . The resistance of cell containing the solution at the same temperature was found to be

$55$ ohm. The cell constant will be:

0%
$0.142{cm}^{-1}$
0%
$0.66{cm}^{-1}$
0%
$0.918{cm}^{-1}$
0%
$1.12{cm}^{-1}$

Explanation

$K=0.012{ r }^{ -1 }{ cm }^{ -1 }$

$R=55ohm$

$K=K\times G$ where K is cell constant

$0.012=K\times \dfrac { 1 }{ 55 }$

$K=0.012\times 55$

$K=0.66{ cm }^{ -1 }$

Excess of silver nitrate solution is added to

$100 \,mL$ of

$0.01 \,M$ pentaaqua chloro chromium (III) chloride solution. The mass of silver chloride obtained in grams is:

[Atomic mass of silver is

$108$ ]

0%
$287 \times 10^{-3}$
0%
$143.5 \times 10^{-3}$
0%
$143.5 \times 10^{-2}$
0%
$287 \times 10^{-2}$

Explanation

${ \left[ Cr\left( { H }_{ 2 }{ O }_{ 5 } \right) Cl \right] }Cl_{ 2 }$ is the reagent

i.e.

$2{ Cl }^{ \left( - \right) }$ can be combined with

$Ag$ to give

$AgCl$

mass of

$AgCl$

$ppt$ ,

no. of moles of

${ Cl }^{ \left( - \right) }=\dfrac { 2\times 100\times 0.01 }{ 1000 } =0.002$

$\therefore$ moles of

$AgCl$

$ppt=0.002$

$\therefore$ mass of

$AgCl=0.002\times 143.5$

$=2.87\times { 10 }^{ -3 }g$

According to electrochemical theory of corrosion involves:

0%
cathodic dissolution of metal
0%
cathodic deposition of metal
0%
anodic dissolution of metal
0%
anodic deposition of metal

Explanation

In corrosion

$Fe^{2+}$ get oxidised to

$Fe^{3+}$ which takes place on anode surface.

Anode:

$Fe(s)\rightarrow { Fe }^{ 2+ }+2{ e }^{ - }$

Cathod:

${ O }_{ 2 }+4{ H }^{ + }+4{ e }^{ - }\rightarrow { 2H }_{ 2 }O$

Overall:

$2Fe(s)+{ O }_{ 2 }+4{ H }^{ + }\rightarrow 2{ Fe }^{ 2+ }+{ H }_{ 2 }O$

${ 4Fe }^{ 2+ }+{ O }_{ 2 }+(2+4x){ H }_{ 2 }O\rightarrow { 2Fe }_{ 2 }{ O }_{ 3 }.x{ H }_{ 2 }O+4{ H }^{ + }$

So,

$Fe^{3+}$ get deposited on metal surface.

$10800C$ of electricity on passing through the electrolyte solution deposited

$2.977$ of metal with atomic mass

${ 106.4gmol }^{ -1 }$ the charge on the metal cation is:

0%
$+4$
0%
$+3$
0%
$+2$
0%
$+1$

Explanation

$10800C\rightarrow 2.977g$ of metal deposited.

$96500C\rightarrow \cfrac { 2.977 }{ 10800 } \times 96500\\ \quad \quad \quad =26.600g$

Charge on metal cation is

$=\cfrac { 106.4 }{ 26.6 }$

$=+4$

Number of coulombs required to liberate 0.5 mol of

${ O }_{ 2 }$ is:

0%
19300
0%
193000
0%
96500
0%
9650

Explanation

Formation of

${ O }_{ 2 }$ is

${ 4e }^{ - }$ transfer.

i.e.

$4\times 96500C\rightarrow 1 \ mol$ of

${ O }_{ 2 }$

$\therefore 0.5\ mol{ O }_{ 2 }\rightarrow \dfrac { 1 }{ 2 } \times 4\times 96500$

$=193000C$

If the density of copper is

$8.94 g/{ cm }^{ 3 }$ , the number of faradays required to plate an area

$(10cm\times 10cm)$ of thickness of

${ 10 }^{ -2 }$ cm using

${ CuSO }_{ 4 }$ solution as electrolyte is:

0%
$0.1 F$
0%
$0.28 F$
0%
$0.4 F$
0%
$0.5 F$

Explanation

Vol. of plate

$=10\times 10\times \dfrac { 1 }{ 100 } =1\ cm^3$

$\therefore$ density

$=\dfrac { mass }{ volume }$

$\therefore$ mass of

$Cu$ to be deposited

$=8.94g$

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

moles of

$Cu=\dfrac { 8.94 }{ 63.5 }$

Faraday required

$=\dfrac { 2\times 8.94 }{ 63.5 } =0.28F$

The ratio of volume of

$H_2$ and

$Cl_2$ evolved at NTP by electrolysis of aqueous solution of

$HCl$ is:

0%
$1:3$
0%
$3:5$
0%
$1:1$
0%
$1:7$

Explanation

Answer is option

$C.$

Electrolysis of

$HCl-$

At cathode,

$2H^++2e^-\longrightarrow H_{2(g)}$

At anode,

$2Cl^-\longrightarrow Cl_{2(g)}+2e^-$

Theoretically the gas volume ratio is

$1:1$

as overall reaction is given by,

$2H^+_{(aq)}+2Cl^-_{(aq)}\longrightarrow H_{2(g)}+Cl_{2(g)}$

A current of 9.95 amperes flowing for 10 minutes, deposits 3g of metal. An equivalent weight of the metal is:

0%
$12.5$
0%
$18.5$
0%
$21.5$
0%
$48.5$

Explanation

$\because W=z.i.t\rightarrow 3=\cfrac{E}{F}\times 9.95\times 10\times 60$

$\therefore\quad E={\left(\cfrac{3\times 96500}{9.95\times 10\times 60}\right)}$

$=48.5$

How many faradays are needed for reduction of 2.5 moles of

$Cr_2O_7^{2-}$ into

$Cr^{3+}$ ?

0%
15
0%
12
0%
6
0%
3

Explanation

${ n }_{ factor }=2\times (3)=6\left( for2{ Cr }^{ +6 }\rightarrow 2{ Cr }^{ +3 } \right)$

$\therefore$ Number of equivalent

$=\left( { n }_{ factor }\times no. of\quad moles \right)$

$=(6\times2.5)$

$=15\quad equivalent$

Or,

$15 faraday$ .

The standard electrode potential of the half cells is given below:

$Zn^{2+}+2e^{-}\rightarrow Zn; E=-0.76 V$

$Fe^{2+}+2e^{-}\rightarrow Fe; E=-0.44 V$
The emf of the cell

$Fe^{2+}=Zn^{-}\rightarrow Zn^{2+}+Fe$

0%
$1.54 V$
0%
$-1.54 V$
0%
$0.32V$
0%
$+0.19 V$

Explanation

${ E }_{ cell }^{ 0 }=\left( { E }_{ { Fe }^{ 2+ }|Fe }^{ 0 }+{ E }_{ Zn|{ Zn }^{ 2+ } }^{ 0 } \right)$

$=\left( -0.44 V + 0.76 \right)$

$E^0=+0.32 Volt$

Amount of charge is required to convert

$17\ g\ H_{2}O_{2}$ into

$O_{2}$ .

0%
$1\ F$
0%
$2\ F$
0%
$3\ F$
0%
none of these

Explanation

$2{ H }_{ 2 }\overset { \left( -2 \right) }{ { O }_{ 2 } } \longrightarrow 2{ H }_{ 2 }O+{ \overset { \left( 0 \right) }{ O } }_{ 2 }$

Change in electrons

$=2$

No. of moles of

$17$ gms of

${ H }_{ 2 }{ O }_{ 2 }=\dfrac { 17 }{ 34 } =0.5$

Charge of one electrons

$=1.607\times { 10 }^{ -9 }$ coloumbs.

Total no. of electrons transferred

$=0.5\times 2\times 6.023\times { 10 }^{ 23 }$

Total charged required

$=0.5\times 2\times 6.023\times { 10 }^{ 23 }\times 1.607\times { 10 }^{ -19 }$

$=9.6\times { 10 }^{ 4 }$

$=1$ Faraday.

Ans :- Option A.

Given,

$E_{Ag^{+}/Ag}^{\circ} = +0.80V, E_{Co^{2+}/Co}^{\circ} = -0.28 V, E_{Cu^{2+}/Cu}^{\circ} = +0.34 V, E_{Zn^{2+}/Zn}^{\circ} = -0.76V$ ,
Which metal will corrode to greater extent?

0%
$Ag$
0%
$Cu$
0%
$Co$
0%
$Zn$

Identify the correct half cell reaction of the following:

$Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$

0%
$SO_4^{2-} \rightarrow SO_4 + 2e^-$
0%
$Zn^{2+}+2e^-\rightarrow Zn$
0%
$Cu \rightarrow Cu^{2+} +2e^-$
0%
$Cu^{2+} +2e^- \rightarrow Cu$

If a pieces of iron gains

$10\%$ of its mass due to partial rusting into

$Fe_{2}O_{3}$ the percentage of total iron that has rusted is

0%
$3$
0%
$13$
0%
$23.3$
0%
$25.67$

For the half reaction

$B(s) \rightarrow B^{2+} + 2e^{-}$

$E_{1}^{\circ} = - 0.44 V$
and

$B_{2}+\rightarrow B^{3+} + e^{-}$

$E_{2}^{\circ} = 1.3 V$
What is

$E^{\circ}$ for the reaction,

$3e^{-} + B^{+3} \rightarrow B(s)$

0%
$0.86$ V
0%
$0.14$ V
0%
$-0.14$ V
0%
$-0.28$ V

Explanation

$B(s)\rightarrow { B }^{ 2 }+{ 2e }^{ - }\quad \quad \quad { E }_{ 1 }^{ 0 }=-0.44V$

$- i$

${ B }^{ 2+ }\rightarrow { B }^{ 3+ }+{ e }^{ - }\quad \quad \quad { E }_{ 1 }^{ 0 }=1.3V$

$- ii$

${ B }^{ 3+ }+{ 3e }^{ - }\rightarrow B(s)\quad \quad \quad \quad { E }_{ 3 }^{ 0 }=?$

$- iii$

iii

$=- (II+I)$

Or,

$-3\times { E }_{ 3 }^{ 0 }\times F=-\left\{ -1\times 1.3\times F+0.44\times 2\times F \right\}$

$\therefore { E }_{ 3 }^{ 0 }=-0.14V$

The weight in grams of

$O_2$ formed at

$Pt$ anode during the electrolysis of aq.

$K_2SO_4$ solution during the passage of one coulomb of electricity is:

0%
$\dfrac{16}{96800}$
0%
$\dfrac{8}{96500}$
0%
$\dfrac{32}{96500}$
0%
$\dfrac{64}{96500}$

Explanation

Weight of oxygen

$=\dfrac{16}{96500}$

A current

$9.65$ ampere is passed through the aqueous solution

$NaCl$ using suitable electrodes for

$1000\ s$ . The amount of

$NaOH$ formed during electrolysis is:

0%
$2.0\ g$
0%
$4.0\ g$
0%
$6.0\ g$
0%
$8.0\ g$

For the reaction:

$F{ e }^{ 2+ }(aq)+A{ g }^{ + }(aq)\longrightarrow Ag(s)+F{ e }^{ +3 }(aq)$

Predict which change will decrease the cell voltage?

0%
Increase the amount of Ag
0%
Decrease the concentration of $A{ g }^{ + }$
0%
Increase the concentration of $F{ e }^{ +3 }$
0%
Both $B$ and $C$

Explanation

$E_{cell}=E^o_{cell}-\cfrac {2.303RT}{nF}\log \cfrac {[Ag][Fe^{+3}]}{[Fe^{2+}][Ag^+]}$

Increasing the concentration of

$Fe^{+3}$ will decrease cell voltage

$(E_{cell})$ .

The electrolysis of aqueous solution of

$Cu{Br}_{2}$ using platinum electrode would leads to:

0%
$Cu$ at anode
0%
${Br}_{2}$ gas at anode and ${O}_{2}$ gas at cathode
0%
$Cu$ at cathode
0%
${H}_{2}$ gas at cathode

Explanation

'Copper

$(Cu)$ at the cathode' is the answer.

Detailed explanation:

Below is the chemical reaction behind the electrolysis of aqueous solution of

$CuBr_2$ ( Copper Bromide) with use of Platinum electrodes (inert).

$CuBr_2 \longrightarrow Cu^{2+} + 2{Br}^-$

Reaction that happens at Anode:

$2{Br}^- \longrightarrow { Br}_2 + 2e^-$

Reaction that happens at Cathode:

$Cu^{2+} + 2e^- \longrightarrow Cu$

Option C is the correct answer.

The number of coulombs required to deposit

$5.4 g$ of Aluminium when the given electrode reaction is represented as

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

0%
$1.83 \times 10^5C$
0%
$57900C$
0%
$5.86 \times 10^5 C$
0%
$3F$

Explanation

According to Faraday's law of electrolysis

$W=\cfrac{Z\times Q}{F}$

$W=$ mass deposited

$=5.4g$

$Z=$ Equivalent mass

$=\cfrac{Molar\quad mass}{e^-\quad accepted}$

$F=$ Faraday constant

$=96500C$

$Q=$ Charge

$5.4g=(\cfrac{27g}{3})\times\cfrac{Q}{96500C}$

$Q=57900C$

Thus,

$57900C$ is required to deposit

$5.4g$ of

$Al$

How many faraday charge is required to convert one mole of

${C}_{6}{H}_{5}NO_{2}$ in to

${C}_{6}H_{5}NH_{2}$ ?

0%
$8$
0%
$12$
0%
$6$
0%
$32$

Given that

${ Cl }_{ 2\left( g \right) }+{ 2e }^{ - }\longrightarrow { 2Cl }^{ - }\left( aq \right)$ is the reduction half-reaction for the overall reaction

$2Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) \longrightarrow 2AgCl\left( s \right)$ .

What is the oxidation half reaction?

0%
$Ag\left( s \right) \longrightarrow Ag^{ + }\left( aq \right) +e^{ - }$
0%
$Ag\left( s \right) +{ Cl }^{ - }\left( aq \right) \longrightarrow Ag\left( s \right) +e^{ - }$
0%
$Ag\left( s \right) +{ Cl }_{ 2 }\left( g \right) +e^{ - }\longrightarrow AgCl\left( s \right) +Cl^{ - }\left( aq \right)$
0%
${ 2Cl }^{ - }\left( aq \right) \longrightarrow { Cl }_{ 2 }\left( g \right) +2e^{ - }$

From the following half-cell reaction and their potential, What is the smallest possible standard e.m.f spontaneous reaction?

${ PO }_{ 4 }^{ 3- } (aq) + 2H_2O(I) + 2e^- \longrightarrow { HPO }_{3}^{2-} + 3OH^-(aq); E^o = -1.05 V$

$PbO _2(s) + H_2O(I) + 2e^- \longrightarrow PbO(S) + 2OH^-(aq); E^o = +0.28 V$

$IO{-}{3}(aq) + 2H_2O(I) + 4e^- \longrightarrow IO^-(aq) + 4OH^-(aq); E^o = +0.56 V$

0%
$+0.00$
0%
$+0.74$
0%
$+0.56$
0%
$+0.28$

The time taken by the galvanic cell which operates almost ideally under reversible conditions at a current of

${ 10 }^{ -16 }$ A to deliver 1 mole of electron is:

0%
$19.30 \times{ 10 }^{ 20 }s$
0%
$4.825 \times{ 10 }^{ 20 }s$
0%
$9.65 \times{ 10 }^{ 20 }s$
0%
$3.4 \times{ 10 }^{ 11 }s$

Explanation

According to the problem:

Given that:

Current

$= {10^{ - 16}}A$

Charge

$=96548\,C$ (Charge on

$1$ mole of electrons)

Using the formula,

Charge

$(Q)=$ current

$(A) \times$ Time

$(T)$

Therefore,

$T = \left( {\frac{{96548}}{{{{10}^{ - 16}}}}} \right)$

Implies that

$T = 9.65 \times {10^{20}}sec$

When molten lithium chloride

$(LiCl)$ is electrolyzed, lithium metal is formed at the cathode. If current efficiency is

$75\%$ then how many grams of lithium are liberated when

$1930\ C$ of charge pass through the cell? (Atomic weight :

$Li=7$ )

0%
$0.105$
0%
$0.120$
0%
$0.28$
0%
$0.240$

Explanation

Weight of metal deposited

$= \cfrac {ECt}{96500}$ ; Applied charge=

$1930C$

Efficiency=

$75$ %

Effective charge=

$\cfrac {75}{100}(1930)=1447.5C$

$96500C$ liberate

$1$ mole.

$1447.5C$ liberates

$\cfrac {1447.5}{96500}$ moles

$=0.105$ moles .

Which of the following is a strong electrolyte?

0%
$\mathrm { Ca } \left( \mathrm { NO } _ { 3 } \right) _ { 2 }$
0%
HCN
0%
$\mathrm { H } _ { 2 } \mathrm { SO } _ { 3 }$
0%
$\mathrm { NH } _ { 4 } \mathrm { OH }$

Explanation

A solute/ solution that completely ionizes or dissociates in a solution is called as strong electrolyte. Here

$Ca(NO_3)_2$ is a strong electrolyte, as it completely dissociates into

$Ca^{2+}$ and

$NO_3^-$ ions in an aqueous solution.

How many Faraday are required to reduce one mol of

${ MnO }_{ 4 }^{ - }$ to

${ Mn }^{ 2+ }$ ?

0%
1
0%
2
0%
3
0%
5

Explanation

$Mn{O^{ - 4}} = M{n^{ + 2}} + 5{e^ - }$

$Mn$ in $Mn{O_4{^{ -}}}$ has an oxidation state of +7.

Going from +7 to +2 requires an additional 5 electrons

1 mol of $e=1\ F$

5 mol of $e = 5\ F$

A schematic representation of enthalpy changes for the reaction,

${ C }_{ graphite }$ +

$\frac { 1 }{ 2 } { O }_{ 2 }$ (g)

$\rightarrow$ CO (g) is given below. The missing value is

0%
+ 10.5 kJ
0%
- 11.05 kJ
0%
- 110.5 kJ
0%
- 10.5 kJ

Explanation

Since enthalpy is a state function therefore it does not depend on the path . Thats y the enthalpy of direct method and indirect method of formation of product ( i . e here CO2 ) will be same so thats y.....

-393.5 = X + (-283 KJ)

X = -110.5 KJ

Hence, the correct option is

$\text{C}$

Three faradays of electricity are passed through molten

$Al_2O_3$ , aqueous solution of

$CuSO_4$ and molten

$NaCl$ taken in different electrolytic cells. The amount of

$Al$

$Cu$ and

$Na$ deposited at the cathodes will be in the ratio of?

0%
$1$ mole $: 2$ mole $: 3$ mole
0%
$3$ mole $: 2$ mole $: 1$ mole
0%
$1$ mole $: 1.5$ mole $: 3$ mole
0%
$1.5$ mole $: 2$ mole $: 3$ mole

Explanation

$Al_{2}O_{3}$ ,

$Al$ has charge

$3+$

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

Thus, 3 Faraday are required for

$1$ mole. So,

$3F$ will produce 1 mole of

$Al_{2}O_{3}$

$CuSO_{4}$ ,

$Cu$ has charge

$2+$

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

Thus,

$2$ Faraday are required for

$1$ mole. So,

$3F$ will produce

$1.5$ moles of

$CuSO_{4}$

$NaCl$ ,

$Na$ has charge

$1+$

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

Thus,

$1$ Faraday are required for

$1$ mole, So,

$3F$ will produce

$3$ mole of

$NaCl$

Hence, ratio will be

$1$ mole :

$1.5$ mole :

$3$ mole

What is the charge of

$96$ amu of

$S^{2-}$ ?

0%
$2 C$
0%
$3.2 \times 10^{-19}C$
0%
$9.6 \times 10^{-19}C$
0%
$6C$

Explanation

Atomic mass of Sulfide ion

$= 32\ amu$

Number of sulfide ions when there is a mass of

$96\ amu= 3 \ ions$

Number of sulfide ions

$= 3$

Charge on 1 sulfide ion

$= -2$

The charge on 3 sulfide ions

$= 3 \times (-2) =-6$

The charge of 96 amu sulfide ions is (-6) i.e. charge on 6 electrons.

$\therefore$ charge on 6 electrons

$= 6 \times 1.6 \times 10^{-19}C$

$=9.6 \times 10^{-19}C$

Aluminum is obtained by

0%
Heating alumina with iron
0%
Heating alumina with coke
0%
Electrolysis of alumina solution in molten cryolite
0%
Electrolysis of fused aluminum chloride

How many Faraday charge is required for

${ KMnO }_{ 4 }\rightarrow { Mn }^{ 2+ }$ ?

0%
3 F
0%
2 F
0%
5 F
0%
1 F

Explanation

$MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 6H_2O$

1 mole 5mole 1mole

$Mn^{7+} + 5e^- \longrightarrow Mn^{2+}$

Required charge =

$n\times F$

Where,

$n$ = difference of charge on ions

$F$ = constant=

$96487C$

Oxidation state of Mn is changing from +7 to +2 therefore 5 moles of electron are needed for reduction of 1 mole of

$MnO^-_4 to Mn^{2+}$ .

5 moles of electrons =

$5F$ .

Hence,Required charge =

$5 \times F$

Therefore option C is correct answer.

If the following half cells have the

$E^o$ values as

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

$E^o= +0.77V$ and

${Fe}^{+2}+2 {e}^{-} \longrightarrow {Fe}$ ;

$E^o= -0.44 V$ The

$E^o$ of the half cell is

$Fe^{+3} + 3e^- \rightarrow Fe$ will be:

0%
$0.33$ V
0%
$1.21$ V
0%
$0.04$ V
0%
$0.605$ V

When ammonia is added to a solution, its

$pH$ is raised to

$11$ . Which half-cell reaction is affected by the

$pH$ and by how much?

0%
$E_{oxd}$ will increase by a factor of $0.65$ from $E^{\circ}_{oxd}$
0%
$E_{oxd}$ will decrease by a factor of $0.65$ from $E^{\circ}_{oxd}$
0%
$E_{red}$ will increase by a factor of $0.65$ from $E^{\circ}_{red}$
0%
$E_{red}$ will decrease by a factor of $0.65$ from $E^{\circ}_{red}$

Electrolysis is :

0%
reversible process
0%
irreversible process
0%
either reversible or irreversible
0%
neither reversible or irreversible

Conductivity of water (in

$\Omega ^{-1}\,cm^{-1}$ ) is

0%
0.4
0%
0.04
0%
0.0004
0%
0.00004

Indicator electrode is

0%
SHE
0%
Calomel electrode
0%
Ag/AgCl electrode
0%
Quinhydrone electrode

The amount of chlorine produced per-second through electrolysis in a plate which consumes

$100$ kW power at

$200$ V is: (Given: Electrochemical equivalent of chlorine

$=0.367\times 10^{-3}$ g/C)

0%
$18.35$ g
0%
$1.835$ g
0%
$183.5$ g
0%
$0.1835$ g

The time required to remove electrolytically one-fourth of Ag from 0.2 litre of 0.1 M

$AgNO_{3}$ solution by the current of 0.1 amp is:

0%
160 min
0%
80 min
0%
40 min
0%
320 min

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 10 - MCQExams.com

0:0:3

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Electrochemistry Quiz 10 - MCQExams.com

0:0:3

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.