Electrochemistry Test

Result

Electrochemistry Test - 47

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Which of the following metal do no evolve $${ H }_{ 2 }$$ gas from dil. acid?

A
$$A\left( { E }_{ { A }^{ + }/A }^{ o }=0.25V \right) $$

B
$$B\left( { E }_{ B/{ B }^{ + } }^{ 0 }=+0.22V \right) $$

C
$$C\left( { E }_{ { C }^{ + }/C }^{ 0 }=0.30V \right) $$

D
$$D\left( { E }_{ D/{ D }^{ + } }^{ 0 }=+0.44V \right) $$
Solution
If the standard electrode potential of an electrode is greater than zero then its reduced form is more stable compared to hydrogen gas.
Similarly, if the standard electrode potential is negative then hydrogen gas is more stable than the reduced form of the species.
Here the standard electrode potential is reduction potential and hence option A doesn't give $$H_2$$ gas when reacted with dilute acid.
Question 2
1 / -0

How many electrons flow when a current of 5 amperes passed through a solution for 193 seconds?

A
$$6.022\times { 10 }^{ 23 }$$

B
$$6.022\times { 10 }^{ 22 }$$

C
$$6.022\times { 10 }^{ 25 }$$

D
$$6.022\times { 10 }^{ 21 }$$

Solution

Current, I = 5 A
Time, t = 193 sec
We know that,
Q = It
Plug the values,we get
= 5 A × 193 s
    = 965 Coulombs
Number of electrons = total charge / charge on 1 electrons
                                    = $$\dfrac{965}{1.6 \times {10} ^ {-19}}$$
                                    = $$6.022 \times {10} ^{21}$$ electrons
Question 3
1 / -0

If equal quantities of electricity are passed through three voltmeter containing $$FeSO_4$$, $$Fe_2(SO_4)_3$$ and $$Fe(NO_3)_2$$, then which of the following is not true?

A
Amount of iron deposited in $$FeSO_4$$ and $$Fe_2(SO_4)_3$$ is equal

B
Amount of iron deposited in $$Fe(NO_3)_3$$ is $$\dfrac{2}{3}$$ of the amount of iron deposited in $$FeSO_4$$

C
Amount of iron deposited in $$Fe_2(SO_4)_3$$ and $$Fe(NO_3)_3$$ are equal

D
Same gas will evolve in all three cases at anode

Solution

(1) $$FeSO_4\rightarrow Fe^{2+}+SO_4^{2-}$$

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

$$\therefore 2F$$ charge is needed for $$1$$ mole deposition of $$Fe$$ .

(2) $$Fe_2(SO_4)_3\rightarrow 2Fe^3+3SO_4^{2-}$$

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

$$3F \rightarrow 1$$ mole
$$2F$$ charge is needed for $$\cfrac {2}{3}$$ mole deposition of $$Fe$$.

The gas liberated in all these cases is the same and that is $$H_2$$ gas.

Therefore, option A is correct.
Question 4
1 / -0

The passage of a constant current through a solution of dilute $$H_2SO_4$$ with 'Pt' electrode liberated $$336cm^2$$ of a mixture of $$H_2$$ and $$O_2$$ at S.T.P. The quantity of electricity that was passed is:

A
96500 C

B
1965 C

C
1930 C

D
0.01 Faraday

Solution

$$2H^+ + 2e \rightarrow H_2(8)$$
$$\overset{\circleddash }{40H} \rightarrow O_2(8)+H_2O(8)+4e^-$$
For ; $$1\ F \rightarrow H_2$$ liberated $$=\dfrac{22.4}{2}=11.2\ lit$$
$$O_2$$ liberated $$=\dfrac{22.4}{4}=5.6\ lit$$
$$1\ F \rightarrow $$ volt mixture $$=(11.2+5.6)\,lit$$
$$=16.8\ lit$$
$$16.8\,lit \rightarrow 1\ F$$
$$1\, lit \rightarrow \dfrac{1}{16.8}\ F$$
$$1000\ cc\rightarrow \dfrac{1}{16.8}\ F$$
$$16.8 \, lit \rightarrow 1\ F$$
$$1 \, lit \rightarrow \dfrac{1}{16.8}\ F$$
$$1000\ cc \rightarrow \dfrac{1}{16.8}\ F$$
$$1\ cc \rightarrow \dfrac{1}{16.8 \times 1000}\ F$$
$$336\ cc \rightarrow \dfrac{336}{16.8 \times 1000}\ F$$
$$=\dfrac{20}{1000}\ F$$
$$=\dfrac{20}{1000}\times 96500\ C$$
$$=1930\ C$$
Question 5
1 / -0

The equilibrium constant for the following reaction at $$25^o$$C is $$2.9\times 10^9$$. Calculate standard voltage of the cell.
$$Cl_{2(g)}+Br^-_{(aq)}\rightleftharpoons 2Cl^-_{(aq)}$$

A
$$0.18$$V

B
$$0.22$$V

C
$$0.26$$V

D
$$0.28$$V

Solution

$$K_{eq}= 2.9 \times 10^9$$
At equilibrium, $$\Delta G^o=-RT lnK_{eq}$$
$$\therefore -nFE^o=-RTlnK_{eq}$$
$$\therefore E^o=\cfrac {RTlnK_{eq}}{nF}=\cfrac {8.314 \times 298 \times ln(2.9\times 10^9)}{2 \times 96500}$$
$$\therefore E^o=0.28V$$
Therefore, standard voltage of the cell is $$0.28V$$
Question 6
1 / -0

A solution of $$Ni(NO_3)_2$$ is electrolyzed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?

A
1.622 gm

B
1.722 gm

C
1.822 gm

D
1.922 gm

Solution

$$Ni^{2+} + 2e^- \rightarrow Ni$$
$$t=20 \ min=20 \times 60 \ sec$$
$$n=2$$
$$i=5 \ A$$
Atomic Mass of $$Ni=58.7$$

$$m= \cfrac {Atomic \ Mass}{n \times F} \times i \times t$$

$$= \cfrac {58.7}{2 \times 96500} \times 20 \times 60 \times 5$$

$$= \cfrac {352200}{2 \times 96500}=1.824$$

The mass of $$Ni$$ deposited at the cathode is $$1.824 \ g$$
Question 7
1 / -0

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

A
dehydration

B
oxidation

C
reduction

D
dehydrogenation

Solution
Question 8
1 / -0

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:

A
$$1:2:3:1$$

B
$$6:3:2:6$$

C
$$6:3:0:0$$

D
$$3:2:1:0$$

Solution

$$AgNO_3+e^- \longrightarrow Ag+NO_3^-$$
$$1$$ 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, $$6 : 3:2:6$$ .
Question 9
1 / -0

On passing electricity through dil.$${ H }_{ 2 }{ SO }_{ 4 }$$ solution, the amount of substance liberated at the cathode and anode are in the ratio :

A
$$1:8$$

B
$$8:1$$

C
$$16:1$$

D
$$1:16$$

Solution

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 = $$4 : 32= 1 : 8$$
Question 10
1 / -0

Consider the reaction : $$Cr_2O^{2-}_{7} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$$.

What is the quantity of electricity in coulombs needed to reduce 1 mole of $$Cr_2O^{2-}_{7}$$ ?

A
$$579000\ C$$

B
$$679000\ C$$

C
$$879000\ C$$

D
$$979000\ C$$

Solution

$${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+14{ H }^{ + }+6{ e }^{ - }\rightarrow 2{ Cr }^{ 3+ }+7{ H }_{ 2 }O$$

For complete reduction of $$1$$ mole of $${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- },6{ e }^{ - }$$ is needed.

Hence, Quantity of electricity $$=6\times96500=579000$$ $$C$$.

Hence, option A is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Electrochemistry Test - 47

Result

Electrochemistry Test - 47

Score

-

Rank

-

SHARING IS CARING

Question 1

$$A\left( { E }_{ { A }^{ + }/A }^{ o }=0.25V \right) $$

$$B\left( { E }_{ B/{ B }^{ + } }^{ 0 }=+0.22V \right) $$

$$C\left( { E }_{ { C }^{ + }/C }^{ 0 }=0.30V \right) $$

$$D\left( { E }_{ D/{ D }^{ + } }^{ 0 }=+0.44V \right) $$

Solution

Question 2

$$6.022\times { 10 }^{ 23 }$$

$$6.022\times { 10 }^{ 22 }$$

$$6.022\times { 10 }^{ 25 }$$

$$6.022\times { 10 }^{ 21 }$$

Solution

Question 3

Amount of iron deposited in $$FeSO_4$$ and $$Fe_2(SO_4)_3$$ is equal

Amount of iron deposited in $$Fe(NO_3)_3$$ is $$\dfrac{2}{3}$$ of the amount of iron deposited in $$FeSO_4$$

Amount of iron deposited in $$Fe_2(SO_4)_3$$ and $$Fe(NO_3)_3$$ are equal

Same gas will evolve in all three cases at anode

Solution

Question 4

96500 C

1965 C

1930 C

0.01 Faraday

Solution

Question 5

$$0.18$$V

$$0.22$$V

$$0.26$$V

$$0.28$$V

Solution

Question 6

1.622 gm

1.722 gm

1.822 gm

1.922 gm

Solution

Question 7

dehydration

oxidation

reduction

dehydrogenation

Solution

Question 8

$$1:2:3:1$$

$$6:3:2:6$$

$$6:3:0:0$$

$$3:2:1:0$$

Solution

Question 9

$$1:8$$

$$8:1$$

$$16:1$$

$$1:16$$

Solution

Question 10

$$579000\ C$$

$$679000\ C$$

$$879000\ C$$

$$979000\ C$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.