Electrochemistry Test

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Electrochemistry Test - 52

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Question 1
1 / -0

What is the charge of $$96$$ amu of $$S^{2-}$$?

A
$$2 C$$

B
$$3.2 \times 10^{-19}C$$

C
$$9.6 \times 10^{-19}C$$

D
$$6C$$

Solution

Atomic mass of Sulfide ion $$= 32\ amu$$
Number of sulfide ions when there is a mass of $$96\ amu= 3 \ ions$$
Number of sulfide ions $$= 3$$

Charge on 1 sulfide ion $$= -2$$
The charge on 3 sulfide ions $$= 3 \times (-2) =-6$$
The charge of 96 amu sulfide ions is (-6) i.e. charge on 6 electrons.

$$\therefore $$ charge on 6 electrons $$= 6 \times 1.6 \times 10^{-19}C$$
$$=9.6 \times 10^{-19}C$$
Question 2
1 / -0

An electrochemical cell can behave like an electrolytic cell when:

A
$$E_{cell}=0$$

B
$$E_{cell}> E_{ext}$$

C
$$E_{cell}< E_{ext}$$

D
$$E_{cell}= E_{ext}$$
Question 3
1 / -0

Aluminum is obtained by

A
Heating alumina with iron

B
Heating alumina with coke

C
Electrolysis of alumina solution in molten cryolite

D
Electrolysis of fused aluminum chloride

Solution

The extraction is done by electrolysis. The ions in the aluminium oxide must be free to move so that electricity can pass through it. Aluminium oxide has a very high melting point (over 2000C) so it would be expensive to melt it. Aluminium oxide does not dissolve in water, but it does dissolve in molten cryolite.
Question 4
1 / -0

Number of coulombs required to deposit $$90\ g$$ of $$Al$$ when the electrode reaction $$Al^{3+}+3e\rightarrow Al$$ is

A
$$9.65\ \times 10^{6}$$

B
$$8.685\ \times 10^{4}$$

C
$$9.65\ \times 10^{4}$$

D
$$6.955$$

Solution

According to equation 1 mole of $$Al$$ i.e 27 grams of $$Al$$ is deposited by 3 Faraday electricity
So 90 grams of $$Al $$ is deposited by
$$\frac{90\times 3\times 96500}{27}=9650000$$
which can be written as $$9.65\times 10^{6}$$
Question 5
1 / -0

The emf of Deniell cell is 1.1 volt.If the value of Faraday is 96500 coulombs per mole, the change in free energy in kJ is :

A
$$212.30$$

B
$$-212.30$$

C
$$106.15$$

D
$$-106.15$$

Solution
Question 6
1 / -0

If the following half cells have the $$E^o$$ values as
$$Fe^{+3} + e^- \rightarrow Fe^{+2}$$; $$E^o= +0.77V$$ and $${Fe}^{+2}+2 {e}^{-} \longrightarrow {Fe}$$ ; $$E^o= -0.44 V$$ The $$E^o$$ of the half cell is $$Fe^{+3} + 3e^- \rightarrow Fe$$ will be:

A
$$0.33$$ V

B
$$1.21$$ V

C
$$0.04$$ V

D
$$0.605$$ V

Solution
Question 7
1 / -0

When ammonia is added to a solution, its $$pH$$ is raised to $$11$$. Which half-cell reaction is affected by the $$pH$$ and by how much?

A
$$ E_{oxd} $$ will increase by a factor of $$0.65$$ from $$ E^{\circ}_{oxd} $$

B
$$ E_{oxd} $$ will decrease by a factor of $$0.65$$ from $$ E^{\circ}_{oxd} $$

C
$$ E_{red} $$ will increase by a factor of $$0.65$$ from $$ E^{\circ}_{red} $$

D
$$ E_{red} $$ will decrease by a factor of $$0.65$$ from $$ E^{\circ}_{red} $$

Solution
Question 8
1 / -0

Electrolysis is :

A
reversible process

B
irreversible process

C
either reversible or irreversible

D
neither reversible or irreversible

Solution

A reversible process is process in which the system and the surrounding returns back to the original state from the final state when the thermodynamic conditions are reversed.
In case of electrolysis, if the terminals of the battery are reversed then the ions start to flow in the opposite direction as before and the whole process is reversed. Eg: If in the forward process water dissociates in hydrogen and oxygen, then in the reverse process the hydrogen and oxygen thus produced will combine back to form water.
Question 9
1 / -0

Conductivity of water (in $$\Omega ^{-1}\,cm^{-1}$$) is

A
0.4

B
0.04

C
0.0004

D
0.00004

Solution
Question 10
1 / -0

The reaction : $$H_{2}(g)+2Ag(Cl)(s)\rightarrow 2H^{+}(aq)+2Cl^{-}(aq)+2Ag(s)$$ occurs in the galvanic cell

A
$$Ag|AgCl(s)|KCl(aq)|AgNO_{3}(aq)|Ag$$

B
$$Pt|H_{2}(g)|HCl(aq)|AgNO_{3}(aq)|Ag$$

C
$$Pt|H_{2}(g)|HCl(aq)|AgCl(s)|Ag$$

D
$$Pt|H_{2}(g)|KCl(aq)|AgCl(s)|Ag$$

Solution

It is given ,

$$H_2(g)+2AgCl(s)\longrightarrow2H^+(aq) + 2Cl^-(aq)+2Ag(s)$$

At low concentrations, the activities of solids and liquids are taken as unity .
the activity of a solute is approximated to its molarity.

Therefore the cell reaction will be

$$Pt(s)|H_2(g)|HCl(aq)|AgCl(s)|Ag$$