Electrochemistry Test

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Electrochemistry Test - 43

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Question 1
1 / -0

A gas $$X$$ at $$1\ atm$$ is bubbled through a solution containing a mixture of $$1\ M$$ $${ Y }^{ - }$$ and $$1\ M$$ $${ Z }^{ - }$$ ions at $$25^{\circ} C$$. If the reduction potential of $$Z > Y > X$$, then _____________.

A
Y will oxidize X and not Z

B
Y will oxidize Z and not X

C
Y will oxidize both X and Z

D
Y will reduce both X and Z

Solution

The order of reduction potential is given as :
$$$Z>Y>X$$$
Higher reduction potential means self-reduction of the species, but it oxidizes other species. This means a species with higher reduction potential acts as an oxidizing agent for species with lower reduction potential.

Lower reduction potential means self-oxidation of the species, but it reduces other species. This means a species with lower reduction potential acts as a reducing agent for species with higher reduction potential.

Thus, $$Y$$ will oxidize $$X$$ $as it has a higher reduction potential than $$X$$,$ but not $$Z$$ $because $$Y$$ has a lower reduction potential than $$Z$$$ .

Hence, option (A) is correct.
Question 2
1 / -0

Standard reduction potentials of the half reactions are given below
$$\displaystyle F_{2(g)} + 2e^- \rightarrow 2F^{-}_{(aq)}$$; $$E^o = +2.85V$$
$$\displaystyle Cl_{2(g)} + 2e^- \rightarrow 2Cl^{-}_{(aq)}$$; $$E^o = +1.36V$$
$$\displaystyle Br_{2(l)} + 2e^- \rightarrow 2Br^{-}_{(aq)}$$; $$E^o = +1.06V$$
$$\displaystyle I_{2(s)} + 2e^- \rightarrow 2I^{-}_{(aq)}$$; $$E^o = +0.53V$$
The strongest oxidising and reducing agents respectively are:

A
$$F_2$$ and $$I^-$$

B
$$Br_2$$ and $$Cl^-$$

C
$$Cl_2$$ and $$Br^-$$

D
$$Cl_2$$ and $$I_2$$

Solution

Because of high standard reduction potential of $$F_{2}$$, it reduces to $$F^{-}$$. So it oxidises other elements strongly. So, it is the strongest agent.
$$I_{2}$$ has the lowest standard reduction potential here. It oxidises itself strongly and reduces others. So, it is a strong reducing agent.
Question 3
1 / -0

Four faradays of electricity were passed through $${AgNO}_3$$, $${CdSO}_4$$, $${AICI}_3$$ and $${PbCI}_4$$ kept in four vessels using insert electrodes. The ratio of moles of Ag, Cd, AI and Pb deposited will be ?

A
12 : 4 : 6 : 3

B
1 : 2 : 3 : 4

C
12 : 6 : 4 : 3

D
4 : 3 : 2 : 1

Solution

$$AgNO_3\longrightarrow Ag^++NO^-_3\\CdSO_4\longrightarrow Cd^{2+}+{(SO_4)}^{2-}\\AlCl_3\longrightarrow Al^{3+}+3Cl^-\\PbCl_4\longrightarrow Pb^{4+}+4Cl^-$$
It requires $$1F$$ of current to deposit $$1$$ mol of $$Ag$$.
Similarly $$2F$$ of current to deposit $$1$$ mol of $$Cd.$$
$$3F$$ of current to deposit $$1$$ mol of $$Al.$$
$$4F$$ of current to deposit $$1$$ mol of $$Pb.$$
When $$4$$ faradays of electricity were passed,
moles of $$Ag$$ diposited$$=\cfrac{4}{1}=4mol. $$
moles of $$Cd$$ diposited$$=\cfrac{4}{2}=2mol. $$
moles of $$Al$$ diposited$$=\cfrac{4}{3}mol. $$
moles of $$lead$$ diposited$$=\cfrac{4}{4}=1mol. $$
$$therefore$$ Ratio of moles of $$Ag,Cd,Al,Pb$$ will be $$4:2:\cfrac{4}{3}:1$$
$$=12:6:4:3.$$
Question 4
1 / -0

The time required for a current of 3 ampere to decompose electrolytically 18 g of $$H_{2}O$$ is:

A
18 hour

B
36 hour

C
9 hour

D
18 second

Solution

$$\theta =nF$$
$$n \rightarrow$$ No. of moles
$$F \rightarrow$$ Faraday's constant
$$\theta \rightarrow$$ charge
Moles of water $$=\cfrac { 18g }{ 18g/mol } =1$$ mole
$$\theta=F$$
$$\theta=I\times t$$
$$I=3$$ amp
$$t=\cfrac { 96500 }{ 3 } $$
$$t=8.935\sim 9$$hour
Question 5
1 / -0

The time required to produce $$2F$$ of electricity with current of 2.5 amperes is:

A
21.4 hr

B
1200 min

C
50000 s

D
1.5 hr

Solution

1 electrons carry charge $$= 1.6\times 10^{-19}C$$
1 mole of electrons contain $$= 6.023 \times 10^{23}$$ electrons
1 mole of electrons carry charge $$= 1.6 \times 10^{-19} \times 6.023 \times 10^{23}$$
$$= 96500 C = 1 $$ Faraday

Thus 2 Faraday $$= \dfrac{96500}{1 }\times 2 = 193000C$$
$$Q = I \times t$$

$$ t = \dfrac{193000}{2.5}$$

$$t = 77200sec = 1286 min = 21.4 hour$$

$$t = 22hr$$

$$\therefore$$ option A is correct
Question 6
1 / -0

Which of the following is not essential for rust to form?

A
Water

B
Iron

C
Oxygen

D
Carbon dioxide

Solution

The presence of water and oxygen is essential for the rusting of iron. Iron does not rust in presence of carbon dioxide.
Thus, the correct option is $$D$$.
Question 7
1 / -0

Question No.60
Indicate the wrong statement according to Valence bond theory:

A
A sigma bond is stronger then $$\pi$$ - bond

B
p-orbitals always have only sidewise overlapping

C
s-orbitals never form $$\pi$$ - bonds

D
There can be only one sigma bond between two atoms

Solution

By Faraday second law, if the same amount of charge is passing through the solution connected in series for the same time then the amount of substances deposited is directly proportional to their equivalent weight. In dilute $${ H }_{ 2 }{ SO }_{ 4 }$$, hydrogen is present at cathode and oxygen is present at the anode in their molecular form. Weight deposited is directly proportional to their equivalent weight.
So, the mass of hydrogen at the cathode: the mass of oxygen at anode = 2: 16 = 1: 8
So, the correct answer is option A.
Question 8
1 / -0

Spacecraft like the satellites that circle the earth and are used in television transmission have electric batteries that supply power for various purposes. These batteries contain ion-exchange resins instead of a liquid electrolyte solution because:

A
of extreme cold conditions

B
of lack of oxygen

C
a high electric voltage is required

D
of zero gravity

Solution

because in the working of the satellites we require high voltage, that why we are ion -exchange resins in batteries.
Question 9
1 / -0

Which of the following metals has been used for building boats because it has resistance to corrosion by sea water?

A
Copper

B
Nickel

C
Fungsten

D
Titanium

Solution

Titanium is used for building boats.
Inert metals which do not react with gases in air seen as oxygen are found to be non-corrosive.
It is corrosion resistance because it forms stable, protective, oxide film when $$Ti$$ is exposed to oxide film.
Question 10
1 / -0

Read the passage given below and answer the question:
Chemical reactions involve the interactions of atoms and molecules. A large number of atoms/molecules (approximately $$6.023 \times 10^{23}$$) are present in few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry, and radiochemistry. the following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A 4.0 molar aqueous solution of NaCl is prepared and 500 mL of this solution is electrolyzed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na = 23, Hg = 200; 1 faraday =96500 coulombs).
The total charge (coulombs) required for complete electrolysis is:

A
24125

B
48250

C
96500

D
193000

Solution

$$2Cl^{\circleddash}\longrightarrow Cl_2+2e^-$$
Moles of $$Cl^{\circleddash}=$$ Moles of $$NaCl=4\times 0.5=2$$ $$mol$$
$$2$$ $$mol$$ of $$Cl^{\circleddash}$$ required $$2$$ $$mol$$ $$F$$ of electricity
$$\therefore$$ Total charge $$=2\times 96500=193000$$ $$C$$

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Electrochemistry Test - 43

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Electrochemistry Test - 43

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Question 1

Y will oxidize X and not Z

Y will oxidize Z and not X

Y will oxidize both X and Z

Y will reduce both X and Z

Solution

Question 2

$$F_2$$ and $$I^-$$

$$Br_2$$ and $$Cl^-$$

$$Cl_2$$ and $$Br^-$$

$$Cl_2$$ and $$I_2$$

Solution

Question 3

12 : 4 : 6 : 3

1 : 2 : 3 : 4

12 : 6 : 4 : 3

4 : 3 : 2 : 1

Solution

Question 4

18 hour

36 hour

9 hour

18 second

Solution

Question 5

21.4 hr

1200 min

50000 s

1.5 hr

Solution

Question 6

Water

Iron

Oxygen

Carbon dioxide

Solution

Question 7

A sigma bond is stronger then $$\pi$$ - bond

p-orbitals always have only sidewise overlapping

s-orbitals never form $$\pi$$ - bonds

There can be only one sigma bond between two atoms

Solution

Question 8

of extreme cold conditions

of lack of oxygen

a high electric voltage is required

of zero gravity

Solution

Question 9

Copper

Nickel

Fungsten

Titanium

Solution

Question 10

24125

48250

96500

193000

Solution

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