Electrochemistry Test

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Electrochemistry Test - 24

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Weekly Quiz Competition

Question 1
4 / -1

The electrochemical cell stops working after sometime because

A
electrode potential of both the electrodes becomes zero

B
electrode potential of both the electrodes becomes equal

C
one of the electrodes is eaten away

D
the cell reaction gets reversed
Question 2
4 / -1

What flows in the internal circuit of a galvanic cell?

A
Ions

B
Electrons

C
Electricity

D
Atoms
Question 3
4 / -1

Galvanic cell is a device in which

A
chemical energy is converted into electrical energy

B
electrical energy is converted into chemical energy

C
chemical energy is seen in the form of heat

D
thermal energy from an outside source is used to drive the cell reaction
Question 4
4 / -1

In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to

A
generate heat

B
create potential difference between the two electrodes

C
produce high purity water

D
remove adsorbed oxygen from electrode surfaces
Question 5
4 / -1

Which of the following statements (or equation) is correct ?

A
The units of cell emf are V. cm^-1

B
∆G = nF/E_cell

C
In galvanic cell, chemical energy is transformed into electrical energy

D
Oxidation state of Mn in potassium permanganate is +6
Question 6
4 / -1

The best way to prevent rusting of iron is

A
making it cathode

B
putting in saline water

C
Both (1) and (2)

D
None of these
Question 7
4 / -1

Several blocks of magnesium are fixed to the bottom of a ship to

A
keep away the sharks

B
make the ship lighter

C
prevent action of water and salt

D
prevent puncturing by under-sea rocks
Question 8
4 / -1

The process of zinc-plating on iron sheet is known as

A
annealing

B
roasting

C
galvanisation

D
smelting
Question 9
4 / -1

When electric current is passed through a cell having an electrolyte, the positive ions move towards the cathode and the anode.If the cathode is pulled out of the solution

A
the positive and negative ions will move towards the anode

B
the positive ions will start moving towards the anode; the negative ions will stop moving

C
the negative ions will continue to move towards the anode; the positive ions will stop moving

D
the positive and negative ions will start moving randomly
Question 10
4 / -1

What is the standard cell potential for the cell Zn; Zn²⁺(1M); Cu²⁺ (1M); Cu? (E^o for Zn²⁺/Zn = –0.76, E^o for Cu²⁺/Cu = +0.34)

A
‒ 0.76 + ( ‒ 0.34 ) = ‒ 0.42 V

B
‒ 0.34 + 0.76 = + 0.42 V

C
0.34 – (‒ 0.76) = + 1.10 V

D
‒ 0.76 ‒ ( + 0.34) = ‒1.10 V

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Electrochemistry Test - 24

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Electrochemistry Test - 24

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Question 1

electrode potential of both the electrodes becomes zero

electrode potential of both the electrodes becomes equal

one of the electrodes is eaten away

the cell reaction gets reversed

Question 2

Ions

Electrons

Electricity

Atoms

Question 3

chemical energy is converted into electrical energy

electrical energy is converted into chemical energy

chemical energy is seen in the form of heat

thermal energy from an outside source is used to drive the cell reaction

Question 4

generate heat

create potential difference between the two electrodes

produce high purity water

remove adsorbed oxygen from electrode surfaces

Question 5

The units of cell emf are V. cm-1

∆G = nF/Ecell

In galvanic cell, chemical energy is transformed into electrical energy

Oxidation state of Mn in potassium permanganate is +6

Question 6

making it cathode

putting in saline water

Both (1) and (2)

None of these

Question 7

keep away the sharks

make the ship lighter

prevent action of water and salt

prevent puncturing by under-sea rocks

Question 8

annealing

roasting

galvanisation

smelting

Question 9

the positive and negative ions will move towards the anode

the positive ions will start moving towards the anode; the negative ions will stop moving

the negative ions will continue to move towards the anode; the positive ions will stop moving

the positive and negative ions will start moving randomly

Question 10

‒ 0.76 + ( ‒ 0.34 ) = ‒ 0.42 V

‒ 0.34 + 0.76 = + 0.42 V

0.34 – (‒ 0.76) = + 1.10 V

‒ 0.76 ‒ ( + 0.34) = ‒1.10 V

Solution

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The units of cell emf are V. cm^-1

∆G = nF/E_cell