Electrochemistry Test

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Electrochemistry Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

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

A
10 coulombs

B
96500 coulombs

C
one electron volt

D
0.1F

Solution

Since $$6.24\times { 10 }^{ 18 }electrons=1coulomb\\ \therefore \quad 6.24\times { 10 }^{ 19 }electrons=10coulomb\\ \qquad \qquad \qquad \qquad \qquad \left( \dfrac { 6.24\times { 10 }^{ 19 } }{ 6.24\times { 10 }^{ 18 } } \right) $$
Question 2
1 / -0

In electro-chemical corrosion of metals, the metal undergoing corrosion:

A
Becomes anode

B
Becomes cathode

C
Becomes inert

D
None is correct

Solution

During corrosion, the oxidation of metal to metal ions occur. Hence the metal undergoing corrosion acts as anode.
Hence, correct answer is option $$A$$.
Question 3
1 / -0

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

A
$$Ni^{+2}+2e^{-}\rightarrow Ni$$

B
$$2H^{+}+2e^{-}\rightarrow H_{2}$$

C
$$Ni^{+2}\rightarrow Ni^{+}+e^{-}$$

D
$$Ni\rightarrow Ni^{+2}+2e^{-}$$

Solution

During electrolysis of nickel sulphate using nickel electrode, at anode, oxidation of nickel occurs. Nickel loses two electrons to form $$ { Ni }^{ 2+ } $$ ion.
Question 4
1 / -0

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

A
Only I and III are correct

B
Only I and II are correct

C
Only I, II and III are correct

D
All are correct

Solution

During electrolysis, an electric current passes through the electrolyte. This is accompanied by a chemical change. Hence option (II) is incorrect.
During electrolysis, anions migrate towards anode and cations migrate towards cathode.
Hence option (IV) is incorrect.
Question 5
1 / -0

An iron rod is immersed in $$KCl$$ solution such that half its length is exposed to air and the other half immersed in $$KCl$$ solution. The part corroded faster is __________.

A
part of the rod exposed to air

B
part of the rod immersed in $$KCl$$ solution

C
both A & B

D
none of the above

Solution
Question 6
1 / -0

A photo-cell employs photoelectric effect to convert:

A
change in the frequency of light into a change in electric voltage.

B
change in the intensity of illumination into a change in photoelectric current.

C
change in the intensity of illumination into a change in the work function of the photocathode.

D
change in the frequency of light into a change in the electric current.

Solution

In photoelectric effect when monochromatic radiations of suitable frequency fall on the photo-sensitive plate called cathode, the photoelectrons are emitted which get accelerated towards the anode. These electrons flow in the outer circuit resulting in the photoelectric current.
Using the incident radiations of a fixed frequency, it is found that the photoelectric current increases linearly with the intensity of incident light. Hence, a photocell employs photoelectric effect to convert the change in the intensity of illumination into a change in photoelectric current.
Question 7
1 / -0

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.

A
I and III

B
I, II and III

C
I, III and IV

D
I, II, III and IV

Solution

The voltaic cells convert chemical energy into electrical energy. Thus, a spontaneous chemical reaction generates chemical energy.

Through external circuit, electrons flow from anode to cathode.

Oxidation occurs at anode and reduction occurs at the cathode.

Since, during oxidation the ions supply electrons to anode, the anode is a negative electrode. Similarly, the cathode is a positive electrode.

Option D is correct.
Question 8
1 / -0

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

A
$$6.0\times 10$$$$^{23}$$

B
$$1.2 \times 10$$$$^{24}$$

C
$$37.5 \times 10$$$$^{19}$$

D
$$7.48\times 10$$$$^{21}$$

Solution

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

$$No. \ of\ e^- = \dfrac{total\ charge}{charge\ on\ 1 e^-}$$$$=\dfrac{60}{1.6\times 10^{-19}}$$

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

Option C is correct.
Question 9
1 / -0

In corrosion of iron :

A
an electrochemical reaction (galvanic cell) is formed in which Fe acts as anode and cathode is where $$O_{2}$$ is reduced

B
electrons flow from anode to cathode through the metal while ions flow through the waterdroplets

C
dissolved $$O_{2}$$ oxidises $$Fe^{2+}$$ to $$Fe^{3+}$$before it is deposited as rust $$(Fe_{2}O_{3}.H_{2}O)$$

D
all of the above take place

Solution

The corrosion of iron is an electrochemical process.
The anode region and cathode region are present on iron.
$$Iron(II)$$ is oxidized to $$Iron(II)$$ 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.
Question 10
1 / -0

In producing chlorine through electrolysis $$100$$ watt power at 125 V is being consumed. How much chlorine per minute is liberated ? Electrochemical Equivalent of chlorine is $$0.367$$ $$\times$$ 10$$^{-6}$$ kg/coulomb:

A
$$21.3$$ mg

B
$$17.6$$ mg

C
$$24.3$$ mg

D
$$13.6$$ mg

Solution

Given value in this solution:

Power consumed in electrolysis $$= 100\ W$$

Current in flow in this solution, $$i= \dfrac{100}{125} = 0.8\ A$$

According to first law of Faraday's law
$$M = ZQ$$ and $$ Q = it$$

So,
$$ M = Zit$$

$$Z= 0.367 \times{10}^{-6};\ i= 0.8\ A ; t=60\ s$$

$$M= 0.367 \times 10^{-6}\times0.8 \times 60$$
$$= 17 . 6 \times{10}^{-6}$$ kg
$$ = 17.6\ mg$$

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Electrochemistry Test - 21

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Electrochemistry Test - 21

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Question 1

10 coulombs

96500 coulombs

one electron volt

0.1F

Solution

Question 2

Becomes anode

Becomes cathode

Becomes inert

None is correct

Solution

Question 3

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

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

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

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

Solution

Question 4

Only I and III are correct

Only I and II are correct

Only I, II and III are correct

All are correct

Solution

Question 5

part of the rod exposed to air

part of the rod immersed in $$KCl$$ solution

both A & B

none of the above

Solution

Question 6

change in the frequency of light into a change in electric voltage.

change in the intensity of illumination into a change in photoelectric current.

change in the intensity of illumination into a change in the work function of the photocathode.

change in the frequency of light into a change in the electric current.

Solution

Question 7

I and III

I, II and III

I, III and IV

I, II, III and IV

Solution

Question 8

$$6.0\times 10$$$$^{23}$$

$$1.2 \times 10$$$$^{24}$$

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

$$7.48\times 10$$$$^{21}$$

Solution

Question 9

an electrochemical reaction (galvanic cell) is formed in which Fe acts as anode and cathode is where $$O_{2}$$ is reduced

electrons flow from anode to cathode through the metal while ions flow through the waterdroplets

dissolved $$O_{2}$$ oxidises $$Fe^{2+}$$ to $$Fe^{3+}$$before it is deposited as rust $$(Fe_{2}O_{3}.H_{2}O)$$

all of the above take place

Solution

Question 10

$$21.3$$ mg

$$17.6$$ mg

$$24.3$$ mg

$$13.6$$ mg

Solution

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