Electrochemistry Test

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Electrochemistry Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

In electrolysis of NaCl when Pt electrode is taken then $${ H }_{ 2 }$$ is liberated at the cathode while with Hg cathode it forms sodium amalgam, because:

A
Hg is more inert than Pt

B
more voltage is required to reduce $${ H }^{ + }$$ at Hg than at Pt

C
Na is dissolved in Hg while it does not dissolve in Pt

D
concentration of $${ H }^{ + }$$ ions is larger when Pt electrode is taken

Solution

Sodium chloride in water dissociates as :
$$NaCl \rightleftharpoons Na^+ + Cl^- $$
$$H_2O \rightleftharpoons H^+ + OH^- $$
When electric current is passed through this solution using platinum electrodes, $$Na^+$$ and $$H^+ $$ move towards cathode and $$Cl^-$$ and $$OH^- $$ ions move towards anode.
If mercury is used as a cathode, $$H^+ $$ ions are not discharged at mercury cathode because mercury has high hydrogen overvoltage. $$Na^+$$ ions are discharged at the cathode in preference of $$H^+$$ ions yielding sodium, which dissolves in mercury to form sodium amalgam.
Question 2
1 / -0

How long (approximate) should water be electrolysed by passing through $$100$$ amperes current so that the oxygen released can completely burn $$27.66\ g$$ of diborane?
[Atomic weight of $$B=10.8\ u$$]

A
$$6.4$$ hours

B
$$0.8$$ hours

C
$$3.2$$ hours

D
$$1.6$$ hours

Solution

$$B_2H_6 + 3O_2 \longrightarrow B_2O_3 +3H_2O $$

27.66 gm of $$B_2H_6$$ i.e 1 mole of $$B_2H_6$$ requires 3 moles of $$O_2$$ .

Now oxygen is produced by electrolysis of water.
$$2H_2O \overset{4F}\longrightarrow2H_2+ O_2$$

1 mole of $$O_2$$ is produced by 4F charge therefore 3 mole of $$O_2$$ will produced by 12F charge

By applying ,Q= It

$$12\times96500 =100\times t$$

$$t = \dfrac{12\times96500}{100\times3600}$$ hours

$$t = 3.2 $$ hours

Hence option C is correct option.
Question 3
1 / -0

In the electrolysis of $$AgNO_3$$ solution 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. Molar mass of Ag is $$107.9 g mol^{-1}$$.

A
$$426 ^oC$$

B
$$536^oC$$

C
$$626^o C$$

D
$$736^o C$$

Solution

$$Mass \ Deposited = 0.7 \ g$$
$$n=1$$
$$Molar \ mass = 107.9 \ gmol^{-1}$$
$$\boxed {m= \cfrac {Atomic \ Mass}{n \times F} \times i \times t}$$
$$0.7= \cfrac {107.9}{1 \times 96500} \times i \times t$$
$$\cfrac {0.7 \times 96500}{107.9}= i \times t$$
$$i \times t = 626.04 \ C$$
The quantity of electricity required in Coulomb is $$626 \ ^oC$$
Question 4
1 / -0

Based on the cell notation for spontaneous reaction, at the anode:
$$Ag(s)|AgCl(s)|{ Cl }^{ - }(aq)\parallel { Br }^{ - }(aq)|{ Br }_{ 2 }(I)|C(s)$$

A
$$AgCl$$ gets reduced

B
$$Ag$$ gets oxidized

C
$${ Br }^{ - }$$ gets oxidized

D
$${ Br }_{ 2 }$$ gets reduced

Solution

$$Ag\left( s \right) \left| AgCl\left( s \right) \right| { Cl }^{ \left( - \right) }\left| { Br }^{ - } \right| \left| { Br }_{ 2 } \right| C\left( s \right) $$
Anode : $$Ag+{ Cl }^{ \left( - \right) }\rightarrow AgCl+{ e }^{ - }$$
Cathode : $$\dfrac { 1 }{ 2 } { Br }_{ 2 }+{ e }^{ \left( - \right) }\rightarrow { Br }^{ \left( - \right) }$$
$$\therefore $$ at anode, oxidation takes place.
Hence, $$Ag$$ gets oxidized.
Question 5
1 / -0

$$10800\ C$$ of electricity through the electrolyte deposited $$2.977\ g$$ of metal with atmic mass $$106.4\ g\ mol^{-1}$$. The charge on the metal cation is:

A
$$+4$$

B
$$+3$$

C
$$+2$$

D
$$+1$$

Solution

$$10800 C=2.977g$$

$$96500 C= ?$$
Hence,

$$\dfrac { 96500\times 2.977 }{ 10800 }$$

$$ \\ =26.600g/mol$$

$$\\thus,\quad \dfrac { 106.4g/mol }{ 26.600g/mol }$$

$$\\ \quad \quad \quad \quad =+4$$ is the answer.
Question 6
1 / -0

$$Al_2O_3$$ is reduced by electrolysis at low potentials and high currents. If $$4.0\times 10^4$$ amperes of current is passed through molten $$Al_2O_3$$ for $$6$$ hours, what mass of aluminium is produced?
(Assume $$100\%$$ current efficiency and atomic mass of $$Al$$ $$=27$$ g $$mol^{-1}$$)

A
$$1.3\times 10^4$$ g

B
$$9.0\times 10^3$$ g

C
$$8.05\times 10^4$$ g

D
$$2.4\times 10^5$$ g

Solution

$$n=3$$ $$ Al^{3+} +3e^- \rightarrow Al$$
$$Molar \ mass = 27 \ gmol^{-1}$$
$$t=6 \ hours=6 \times 60 \times 60 sec$$
$$i=4 \times 10^4 \ A$$
$$\boxed {m= \cfrac {Atomic \ Mass}{n \times F} \times i \times t}$$
$$m= \cfrac {27}{3 \times 96500} \times 4 \times 10^4 \times 6 \times 60 \times 60$$
$$=\cfrac {23328}{3 \times 965} \times 10^4$$
$$= 8.05 \times 10^4 \ g$$
Mass of $$Al$$ produced is $$8.05 \times 10^4 \ g$$
Question 7
1 / -0

If a $$100 \; mL$$ solution of $$0.1M$$. $$HBr$$ is titrated using a very concentrated solution of $$NaOH$$, then the conductivity (specific conductance) of this solution at the equivalence point will be:
(assume volume change is negligible due to the addition of $$(NaOH)$$. Report your answer after multiplying it with 10 in $$Sm^{-1}$$.
[Given: $$\lambda^{\circ}_{(Na^+)} = 8 \times 10^{-3} Sm^2 mol^{-1},
\lambda^{\circ}_{(Br^-)} = 4 \times 10^{-3} Sm^2 mol^{-1}$$]

A
6

B
12

C
15

D
24

Solution
Question 8
1 / -0

The number of coulombs necessary to deposit 1 g of potassium metal (molar mass $$39 g mol^{-1}$$) from $$K^+ ions$$ is:

A
$$96500^oC$$

B
$$1.93 \times 10^5 $$$$^oC$$

C
$$1237$$$$^o C$$

D
$$2474$$$$ ^oC$$

Solution

$$Mass \ Deposited = 1 \ g$$
$$n=1$$ $$(K^+ \ ions)$$; $$ K^+ +e^- \rightarrow K$$
$$Molar \ mass = 39 \ gmol^{-1}$$
$$\boxed {m= \cfrac {Atomic \ Mass}{n \times F} \times i \times t}$$
$$1= \cfrac {39}{1 \times 96500} \times i \times t$$
$$\cfrac {96500}{39}= i \times t$$
$$i \times t = 2474.35 \ C$$
The number of coulombs necessary to deposit $$1 \ g$$ of potassium metal from $$K^+$$ ions is $$2474 \ ^oC$$
Question 9
1 / -0

When $$9.65$$ ampere current was passed for $$1.0$$ hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is:

A
$$109.0\ g$$

B
$$98.1\ g$$

C
$$9.81\ g$$

D
$$10.9\ g$$

Solution
Question 10
1 / -0

The charge required to deposit $$40.5 \,g$$ of $$Al$$ (atomic mass $$= 27.0 \,g$$) from the fused $$Al_2(SO_4)_3$$ is:

A
$$4.34 \times 10^5 C$$

B
$$43.4 \times 10^5 C$$

C
$$1.44 \times 10^5$$

D
None of these

Solution

$$Al^{3+} + 3e^- \rightarrow Al$$
3F 1 mol = 27.0 g
to deposite 27 g required charge = $$3 \times 96,500 c$$
Therefore, to deposite 40.4 g required charge
$$4.34 \times 10^5 C$$

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Electrochemistry Test - 64

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Electrochemistry Test - 64

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Question 1

Hg is more inert than Pt

more voltage is required to reduce $${ H }^{ + }$$ at Hg than at Pt

Na is dissolved in Hg while it does not dissolve in Pt

concentration of $${ H }^{ + }$$ ions is larger when Pt electrode is taken

Solution

Question 2

$$6.4$$ hours

$$0.8$$ hours

$$3.2$$ hours

$$1.6$$ hours

Solution

Question 3

$$426 ^oC$$

$$536^oC$$

$$626^o C$$

$$736^o C$$

Solution

Question 4

$$AgCl$$ gets reduced

$$Ag$$ gets oxidized

$${ Br }^{ - }$$ gets oxidized

$${ Br }_{ 2 }$$ gets reduced

Solution

Question 5

$$+4$$

$$+3$$

$$+2$$

$$+1$$

Solution

Question 6

$$1.3\times 10^4$$ g

$$9.0\times 10^3$$ g

$$8.05\times 10^4$$ g

$$2.4\times 10^5$$ g

Solution

Question 7

6

12

15

24

Solution

Question 8

$$96500^oC$$

$$1.93 \times 10^5 $$$$^oC$$

$$1237$$$$^o C$$

$$2474$$$$ ^oC$$

Solution

Question 9

$$109.0\ g$$

$$98.1\ g$$

$$9.81\ g$$

$$10.9\ g$$

Solution

Question 10

$$4.34 \times 10^5 C$$

$$43.4 \times 10^5 C$$

$$1.44 \times 10^5$$

None of these

Solution

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