Electrochemistry Test

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Electrochemistry Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

When water is added to an aqueous solution of an electrolyte, what is the change in specific conductivity of the electrolyte?

A
Conductivity decreases

B
Conductivity increases

C
Conductivity remains same

D
Conductivity does not depend on number of ions
Solution
Factors affecting electrolytic conductance are:
Concentration of ions: The sole reason for the conductivity of electrolytes is the ions present in them. The conductivity of electrolytes increases with an increase in the concentration of ions as there will be more charge carriers if the concentration of ions is more and hence the conductivity of electrolytes will be high.

Nature of electrolyte: Electrolytic conduction is significantly affected by the nature of electrolytes. The degree of dissociation of electrolytes determines the concentration of ions in the solution and hence the conductivity of electrolytes. Substances such as $$CH_3COOH$$, with a small degree of separation, will have less number of ions in the solution and hence their conductivity will also below, and these are called weak electrolytes. Strong electrolytes such as $$KNO_3$$ have a high degree of dissociation and hence their solutions have a high concentration of ions and so they are good electrolytic conductance.

Temperature: Temperature affects the degree to which an electrolyte gets dissolved in solution. It has been seen that higher temperature enhances the solubility of electrolytes and hence the concentration of ions which results in an increased electrolytic conduction.

When water is added to an aqueous solution the number of ions per unit volume decreases i.e. the concentration of ions decreases and hence thereby conductivity gets decreased.
Question 2
1 / -0

The given figure shows the corrosion of iron in atmosphere.

Fill in the blanks by choosing an appropriate option.
At a particular spot of an object made of iron, $$\underline{(i)}$$ of iron to ferrous ion takes place and that spot behaves as $$\underline{(ii)}$$. Electrons released at $$\underline{(i)}$$ spot move through the metal and go to another spot on the metal and reduce oxygen in presence of $$H^+$$. This spot behave as $$\underline{(iii)}$$. The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust, $$\underline{(iv)}$$ and with further production of $$\underline{(v)}$$ ions.

A
(i)- Oxidation, (ii)- Anode, (iii)-Cathode, (iv)- $$Fe_2O_3.xH_2O$$, (v)-Hydrogen

B
(i)- Reduction, (ii)-Cathode, (iii)-Anode, (iv)-$$Fe_3O_4$$, (v)-Hydrogen

C
(i)-Oxidation, (ii)-Cathode, (iii)-Anode, (iv)-$$Fe_2O_3xH_2O$$, (v)-Hydrogen

D
(i)-Oxidation, (ii)-Anode, (iii)-Cathode, (iv)-$$Fe_2O_3\cdot H_2O$$, (v)-Ferrous

Solution

At a particular spot of an object made of iron, oxidation of iron to ferrous ion takes place and that spot behaves as an anode.

Electrons released at oxidation spot move through the metal and go to another spot on the metal and reduce oxygen in the presence of H$$^+$$. This spot behaves as a cathode. The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust, $$Fe_2O_3.xH_2O$$ and with further production of hydrogen ions.

$$Fe\longrightarrow \underbrace{Fe^{2+}+2e^-}_{(Ferrous)}$$ (Oxidation)

Oxidation takes place at the anode whereas reduction takes place at the cathode.

$$4Fe^{2+}+3O_2+2xH_2O\longrightarrow \underbrace{2Fe_2O_3\cdot xH_2O}_{Rust}$$

Hydrogen ions are produced.
Question 3
1 / -0

Which of the following reactions does not take place during rusting?

A
$$H_2CO_3\rightleftharpoons 2H^++CO^{2-}_3$$

B
$$4Fe^{2+}+O_{2(dry)}\rightarrow Fe_2O_3$$

C
$$4Fe^{2+}+O_2+4H_2O\rightarrow 2Fe_2O_3+8H^+$$

D
$$Fe_2O_3+xH_2O\rightarrow Fe_2O_3\cdot xH_2O$$
Solution
Rust is an iron oxide, a usually red oxide formed by the redox reaction of iron and oxygen in the presence of water or air moisture.
The following redox reaction also occurs in the presence of water and is crucial to the formation of rust:
$$4Fe^{2+}+O_2\rightarrow 4Fe^{3+}+2O^{2-}$$
But in the given reaction $$Fe^{2+}$$ is reacted with dry oxygen. And hence option B is wrong.
Various methods of prevention of rust are-
Oiling - for example bicycle chains
Greasing - for example nut and bolts
Painting - for example car body panels
Galvanising is a method of rust prevention. The iron or steel object is coated in a thin layer of zinc.
Question 4
1 / -0

A current of $$1.40$$ ampere is passed through $$500$$mL of $$0.180$$M solution of zinc sulphate for $$200$$ seconds. The molarity of $$Zn^{2+}$$ ions after deposition of zinc is:

A
$$0.154$$M

B
$$0.177$$M

C
$$2$$M

D
$$0.180$$M

Solution

$$Zn^{2+}+2e^-\longrightarrow Zn$$
$$Faraday=\cfrac{Charge}{96500}=\cfrac{1.40\times 200}{96500}=2.90\times 10^{-3}$$ $$mol$$
$$\therefore Zn$$ deposited $$=\cfrac{2.90\times 10^{-3}}{2}=1.45\times 10^{-3}$$ $$mol$$
Now, $$mol$$ of $$Zn^{2+}$$ initially $$=0.180\times 0.5$$
$$Molarity\times V=0.09$$ $$mol$$
$$mol$$ of $$Zn^{2+}$$ left after deposition $$=0.09-0.00145=0.08855$$ $$mol$$
$$Molarity=\cfrac{Moles}{Volume}=\cfrac{0.08855}{0.5}=0.177$$ $$M$$
Hence option (B) is correct.
Question 5
1 / -0

When an aqueous solution of $$AgNO_3$$ is electrolysed between platinum electrodes, the substances liberated at anode and cathode are:

A
Silver is deposited at cathode and $$O_2$$ is liberated at anode

B
Silver is deposited at cathode and $$H_2$$ is liberated at anode

C
Hydrogen is liberated at cathode and $$O_2$$ is liberated at anode

D
Silver is deposited at cathode and Pt is dissolved in electrolyte

Solution

At cathode:
The reaction with a higher value of $$E^{0}$$ takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
$$Ag^{+}_{(aq)}+e^{-}\rightarrow Ag_{(s)};E^{0}-0.80V$$
At anode:
Since Pt electrodes are inert, the anode is not attacked by NO3- ions. Therefore, OH - or NO3- ions can be oxidized at the anode. But OH - ions having a lower discharge potential and get preference and decompose to liberate O2.
$$OH^{-}\rightarrow OH+e^{-}$$
$$4OH^{-}\rightarrow 2H_2O+O_2$$

Hence option (A) is correct
Question 6
1 / -0

During electrolysis of a solution of $$AgNO_3$$, $$9650$$ coulombs of charge is passed through the solution. What will be the mass of silver deposited on the cathode?

A
$$108$$g

B
$$10.8$$g

C
$$1.08$$g

D
$$216$$g

Solution

According to this Faraday's first law, the chemical deposition due to flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.
The given reaction is,
$$Ag^{+}+e^{-}\rightarrow Ag$$
Amount of Ag deposited by 1F(96500) is 108g
So, amount of Ag deposited by $$9650C=\dfrac{108}{96500}\times 9650=10.8g$$
Question 7
1 / -0

How many coulombs of electricity is required to reduce $$1$$ mole of $$Cr_2O^{2-}_7$$ in acidic medium?

A
$$4\times 96500$$ C

B
$$6\times 96500$$ C

C
$$2\times 96500$$ C

D
$$1\times 96500$$ C

Solution

In acidic medium dichromate ions undergoes reduction and the reaction is,
$$Cr_2O_7+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_2O$$
So, 1 mole of $$Cr_2O^{2-}_{7}$$ ion requires 6 moles of electrons i.e. $$6\times96500$$C of electricity
Question 8
1 / -0

Which of the statements about solutions of electrolysis is not correct?

A
Conductivity of solution depends upon size of ions

B
Conductivity depends upon viscosity of solution

C
Conductivity does not depend upon solvation of ions present in solution

D
Conductivity of solution increases with temperature
Solution
Electrolysis is the process of passing an electric current through a substance in order to produce chemical changes in the substance.
Solvation is the process in which molecules of a solvent attract the particles of a solute.
The main forces in solvation are ion-dipole and hydrogen bonding attractions. It is the main reason why solutes dissolve in solvents.
During electrolysis,conductivity depends upon solvation of ions present in solution. Greater the solvation of ions, lesser is the conductivity.
Question 9
1 / -0

How many electrons are there in one coulomb of electricity?

A
$$

6.023 \times 10^{23}

$$

B
$$

1.64 \times 10^{ 24}

$$

C
$$

6.24 \times 10^{ 18}

$$

D
$$

6.24 \times 10^{ 24}

$$

Solution

Change on $$1e^-=1.602\times 10^{-19}$$ $$C$$

If total charge $$=1$$ $$C$$

$$\therefore$$ Total no. of $$e^-=\cfrac{1}{1.602\times 10^{-19}}=6.24\times 10^{18}e^-$$

Hence, the correct option is $$(C)$$.
Question 10
1 / -0

The ratio of weights of hydrogen and magnesium deposited by the same amount of electricity from aqueous $${ H }_{ 2 }SO_{ 4 }$$ and fused $$MgSO_{ 4 }$$ are:

A
1:8

B
1:12

C
1:16

D
1:1

Solution

Acc. to Faraday $${ 2 }^{ nd }$$ Law of electrolysis.
$$ \cfrac { { w }_{ 1 } }{ { w }_{ 2 } } =\cfrac { { E }_{ 1 } }{ { E }_{ 2 } } \quad \quad \quad \quad \quad { w }_{ 1 }=$$ At deposited of element H.
$$\cfrac { { w }_{ 1 } }{ { w }_{ 2 } } =\cfrac { 1 }{ 12 } \quad \quad \quad \quad \quad { w }_{ 2 }=$$ At deposited of element Mg.
$${ E }_{ 1 }=$$ eq. at of N.
$$ { E }_{ 2 }=$$ eq. at of Mg.

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Electrochemistry Test - 45

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Question 1

Conductivity decreases

Conductivity increases

Conductivity remains same

Conductivity does not depend on number of ions

Solution

Question 2

(i)- Oxidation, (ii)- Anode, (iii)-Cathode, (iv)- $$Fe_2O_3.xH_2O$$, (v)-Hydrogen

(i)- Reduction, (ii)-Cathode, (iii)-Anode, (iv)-$$Fe_3O_4$$, (v)-Hydrogen

(i)-Oxidation, (ii)-Cathode, (iii)-Anode, (iv)-$$Fe_2O_3xH_2O$$, (v)-Hydrogen

(i)-Oxidation, (ii)-Anode, (iii)-Cathode, (iv)-$$Fe_2O_3\cdot H_2O$$, (v)-Ferrous

Solution

Question 3

$$H_2CO_3\rightleftharpoons 2H^++CO^{2-}_3$$

$$4Fe^{2+}+O_{2(dry)}\rightarrow Fe_2O_3$$

$$4Fe^{2+}+O_2+4H_2O\rightarrow 2Fe_2O_3+8H^+$$

$$Fe_2O_3+xH_2O\rightarrow Fe_2O_3\cdot xH_2O$$

Solution

Question 4

$$0.154$$M

$$0.177$$M

$$2$$M

$$0.180$$M

Solution

Question 5

Silver is deposited at cathode and $$O_2$$ is liberated at anode

Silver is deposited at cathode and $$H_2$$ is liberated at anode

Hydrogen is liberated at cathode and $$O_2$$ is liberated at anode

Silver is deposited at cathode and Pt is dissolved in electrolyte

Solution

Question 6

$$108$$g

$$10.8$$g

$$1.08$$g

$$216$$g

Solution

Question 7

$$4\times 96500$$ C

$$6\times 96500$$ C

$$2\times 96500$$ C

$$1\times 96500$$ C

Solution

Question 8

Conductivity of solution depends upon size of ions

Conductivity depends upon viscosity of solution

Conductivity does not depend upon solvation of ions present in solution

Conductivity of solution increases with temperature

Solution

Question 9

$$6.023 \times 10^{23} $$

$$1.64 \times 10^{ 24} $$

$$6.24 \times 10^{ 18} $$

$$6.24 \times 10^{ 24} $$

Solution

Question 10

1:8

1:12

1:16

1:1

Solution

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$$

6.023 \times 10^{23}

$$

$$

1.64 \times 10^{ 24}

$$

$$

6.24 \times 10^{ 18}

$$

$$

6.24 \times 10^{ 24}

$$