Electrochemistry Test

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Electrochemistry Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

How many kgs of silver would be deposited at cathode if $$0.0342 dm^3$$ of oxygen measured at NTP is liberated at the anode when silver nitrate solution is electrolyzed between platinum electrodes?(Given : Atomic masses of Ag = 108, O =16)

A
0.22 gm

B
0.44 gm

C
0.66 gm

D
None of the above

Solution

$$AgNO_3\rightarrow Ag^++NO_3^-$$
$$2H_2O\rightarrow AH^++O_2+Ae^-$$
For 1 mole $$O_2 \Rightarrow 4F$$ change released
At NTP
$$1$$ mole $$O_2=22.4L$$
$$1L=\dfrac{1}{22.4}$$ mole $$O_2$$
$$0.0342L=\dfrac{0.0342}{22.4}=0.00153$$ mole $$O_2$$
For $$0.00153$$ mole $$\Rightarrow 4\times 0.00153F$$
$$\Rightarrow 0.00611F$$
$$\Rightarrow 0.00611\times 96500$$
$$\Rightarrow 589.34C$$
$$96500C$$ or $$1F$$ give $$\theta$$ $$\Rightarrow 1$$ mole $$Ag =108g\, Ag$$
$$589.34C$$ give$$\theta$$ $$\Rightarrow \dfrac{589.34\times 108}{96500}$$
$$\Rightarrow 0.66\ gm$$
Question 2
1 / -0

For the half reaction
$$B(s)$$ $$\rightarrow$$ $$B^{2+} + 2e^-$$ $${ E }_{ 1 }^{ \circ }=-0.44V$$
and
$$B^{2+}$$ $$\rightarrow$$ $$B^{3+} + e^-$$ $${ E }_{ 2 }^{ \circ }=1.3V$$
What is $${E}_{3}^{\circ}$$ for the reaction
$$3e^- + B^{+3 }$$ $$\rightarrow $$ B(s):

A
$$0.86 V $$

B
$$0.14 V$$

C
$$-0.14 V $$

D
$$-0.28V$$

Solution

$$B\left( s \right) \rightarrow { B }^{ 2+ }+{ 2e }^{ - }\quad \quad { E }_{ 1 }^{ 0 }=-0.44V$$
+ $$\underline { { B }^{ 2+ }\rightarrow { B }^{ 3+ }+{ e }^{ - }\quad \quad { E }_{ 2 }^{ 0 }=+1.3V } $$
$$B\left( s \right) \rightarrow { B }^{ 3+ }+{ 3e }^{ - }\quad ;\quad { E }_{ 3 }^{ 0 }=1.3-0.44V$$
$$=0.86V$$
Question 3
1 / -0

How much electricity in terms of Faraday is required to produce 40.0 g of $$Al$$ from molten $$Al_2O_3$$?

A
1 Faraday

B
2 Faraday

C
3.2 Faraday

D
4.44 Faraday

Solution

Cathode: $$Al^{3+}+3e^- \rightarrow Al$$
Anode: $$2O^{2-}+4 e^- \rightarrow O_2$$
Overall equation: $${2Al_2O_3}_{(l)}\rightarrow 4Al_{(l)}+{3O_2}_{(g)}$$
So $$3$$ Faraday of electricity is required to produce $$1$$ mole of $$Al$$.
Now, $$40g$$ of $$Al=\cfrac {40}{27}$$ mole$$=1.48$$ moles of $$Al$$
$$\therefore 1.48$$ moles of $$Al$$ can be produced by $$1.48\times 3$$ Faraday i.e $$4.44$$ Faraday of electricity.
Question 4
1 / -0

Which of the following will be most easily corroded in moist air?

A
$$Zn$$

B
$$Fe$$

C
$$Ni$$

D
$$Sn$$

Solution

When iron ( $$Fe$$ ) is exposed to moist air, it reacts with oxygen to form rust.

The formation of a rust is a very complex process, which is thought to begin with the oxidation of iron to ferrous ( iron"+2") ions.

Iron metal reacts in moist air by oxidation to give a hydrated iron oxide. This does not protect the iron surface to further reaction since it flakes off, exposing more iron metal to oxidation. This process is called rusting.
Question 5
1 / -0

When $$96500$$ coulombs of electricity are passed through barium chloride solution, the amount of barium deposited will be:

A
$$0.5\ mol$$

B
$$1.0\ mol$$

C
$$1.5\ mol$$

D
$$2.0\ mol$$

Solution

$${ BaSO }_{ 4 }\rightleftharpoons { Ba }^{ 2+ }+{ SO }_{ 4 }^{ 2- }$$
$$\therefore$$ No. of electrons involved $$=n=2$$
$$2\times 96500$$ Coulomb deposits one mole of barium
$$\therefore 96500$$ Coilomb deposits $$0.5$$ mole of barium.
Question 6
1 / -0

When an electric current is passed through acidified water, $$112\ ml$$ of hydrogen gas at $$STP$$ collects at the cathode in $$965$$ seconds. What is the current passed, in ampere?

A
$$1.0$$

B
$$0.5$$

C
$$0.1$$

D
$$2.0$$

Solution

At $$STP$$, $$22.4\ litres = 1mole\ of\ H_2\ gas = 2g$$ of $$H_2$$ gas.

$$\therefore$$ $$112\ ml\ of\ H_2\ gas\ contains\ \dfrac { 2\times 112 }{ 22.4\times { 10 }^{ 3 } } =0.01g$$ of $$H_2$$ gas

Now,
$${ 2H }^{ + }+{ 2e }^{ - }\rightarrow { H }_{ 2 }$$

According to Faraday's law,
$$nF=1mole$$

where, $$F=$$ Faraday's constant$$=96500\ C$$
& $$n = number of electrons = 2$$

$$\therefore$$ $$1$$ mole deposits $$2\times96500C$$

$$0.01g$$ deposits $$\dfrac { 2\times 96500\times 0.01 }{ 2 } =965\ C$$

$$\therefore$$ Current $$= \dfrac{Charge}{Time}=\dfrac{965}{965}$$

$$= 1 Ampere $$

Hence, option $$(A)$$ is correct.
Question 7
1 / -0

Which among the following cell cannot be regenerated?

A
Lead storage cell

B
$$Ni-Cd$$ cell

C
Leclanche cell

D
All of the given

Solution
Question 8
1 / -0

Find the charge of $$48 g$$ of $${ Mg }^{ 2+ }$$ ions in coulombs.

A
$$2.4\times { 10 }^{ 23 }C$$

B
$$6.82\times { 10 }^{ 5 }C$$

C
$$3.86\times { 10 }^{ 5 }C$$

D
$$1.93\times { 10 }^{ 5 }C$$

Solution

Molar mass of $${Mg}^{+2} = 24 g$$
Given mass of $${Mg}^{+2} = 48 g$$
No. of moles of $${Mg}^{+2} = \cfrac{\text{Given mass}}{\text{Molar mass}} = \cfrac{48}{24} = 2 \text{ moles}$$
1 mole of $${Mg}^{+2}$$ = 2 moles of electrons
$$\therefore$$2 moles of $${Mg}^{+2}$$ = 4 moles of electrons
$$\therefore$$ No. of electrons = no. of moles of electrons $$\times$$ Avogadro's constant
$$\Rightarrow$$ No. of electrons $$= 4 \times 6.023 \times {10}^{23} = 2.4 \times {10}^{24}$$
As we know that,
Charge on 1 electron $$= 1.6 \times {10}^{-19} C$$
$$\therefore$$ Total charge on 48 g $${Mg}^{+2} = \text{no. of }{e}^{-}s \times \text{charge on 1 } {e}^{-} = \left( 2.4 \times {10}^{24} \right) \times \left( 1.6 \times {10}^{-19} \right) = 3.84 \times {10}^{5} C \approx 3.86 \times {10}^{5} C$$
Hence, the correct answer is $$3.86 \times {10}^{5} C$$.
Question 9
1 / -0

The specific conductance of $$0.1N$$ $$KCl$$ solution at $${23}^{o}C$$ is $$0.012{ohm}^{-1}{cm}^{-1}$$. The resistance of cell containing the solution at the same temperature was found to be $$55$$ ohm. The cell constant will be:

A
$$0.142{cm}^{-1}$$

B
$$0.66{cm}^{-1}$$

C
$$0.918{cm}^{-1}$$

D
$$1.12{cm}^{-1}$$

Solution

$$K=0.012{ r }^{ -1 }{ cm }^{ -1 }$$
$$R=55ohm$$
$$K=K\times G$$ where K is cell constant
$$0.012=K\times \dfrac { 1 }{ 55 } $$
$$K=0.012\times 55$$
$$K=0.66{ cm }^{ -1 }$$
Question 10
1 / -0

Which metal is usefull for protection of iron by corrosion?

A
Aluminium

B
Sodium

C
Zinc

D
Copper

Solution

A thin layer of zinc coats other metals such as iron. It protects the iron from corrosion. Also since zinc is a more reactive metal it acts as a sacrificial metal. The oxygen in the air reacts with zinc to form zinc oxide, thus protecting the iron.

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Electrochemistry Test - 48

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Electrochemistry Test - 48

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SHARING IS CARING

Question 1

0.22 gm

0.44 gm

0.66 gm

None of the above

Solution

Question 2

$$0.86 V $$

$$0.14 V$$

$$-0.14 V $$

$$-0.28V$$

Solution

Question 3

1 Faraday

2 Faraday

3.2 Faraday

4.44 Faraday

Solution

Question 4

$$Zn$$

$$Fe$$

$$Ni$$

$$Sn$$

Solution

Question 5

$$0.5\ mol$$

$$1.0\ mol$$

$$1.5\ mol$$

$$2.0\ mol$$

Solution

Question 6

$$1.0$$

$$0.5$$

$$0.1$$

$$2.0$$

Solution

Question 7

Lead storage cell

$$Ni-Cd$$ cell

Leclanche cell

All of the given

Solution

Question 8

$$2.4\times { 10 }^{ 23 }C$$

$$6.82\times { 10 }^{ 5 }C$$

$$3.86\times { 10 }^{ 5 }C$$

$$1.93\times { 10 }^{ 5 }C$$

Solution

Question 9

$$0.142{cm}^{-1}$$

$$0.66{cm}^{-1}$$

$$0.918{cm}^{-1}$$

$$1.12{cm}^{-1}$$

Solution

Question 10

Aluminium

Sodium

Zinc

Copper

Solution

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