Electrochemistry Test

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Electrochemistry Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

Two half-cells have potentials –0.44 V and 0.799 V, respectively. These two are coupled to make a galvanic cell. Which of the following statements is true?

A
Electrode of half-cell potential –0.44 V acts as anode.

B
Electrode of half-cell potential –0.44 V acts as cathode.

C
Electrode of half-cell potential 0.799 V acts as anode.

D
None of these

Solution

For a spontaneous redox reaction to take place and the electrochemical cell to function, the electrode with a negative reduction potential will constitute negative terminal (i.e. anode) of the galvanic cell.
Question 2
1 / -0

A solution is 0.1 M in each of Pb²⁺ and Fe²⁺ ions. If powders of lead and iron are added in this solution, which of the following facts comes out to be true?

A
More of Fe and Pb²⁺ ions are formed.

B
More of Pb and Fe²⁺ ions are formed.

C
Concentrations of Pb²⁺ and Fe²⁺ ions are increased.

D
Concentrations of Pb²⁺ and Fe²⁺ ions are decreased.

Solution

Since E^o(Fe²⁺ I Fe) is more negative than E^o(Pb²⁺|Pb), the electrode Fe²⁺| Fe will constitute negative terminal of the cell, producing spontaneous reaction.
Cell: Fe(s) | Fe²⁺(aq) || Pb²⁺(aq) | Pb
Cell reaction: Pb²⁺ + Fe → Pb + Fe²⁺

Hence, more of lead and Fe²⁺ ions are formed.
Question 3
1 / -0

Cell constant is defined as

A
l/a

B
a/l

C
la

D
l²/a

Solution

The quantity l/a is called cell constant. It is denoted by the symbol G*. It depends on the distance between the electrodes (l) and their area of cross-section (a).
Question 4
1 / -0

Resistivity is defined as

A
inverse of resistance

B
inverse of conductance

C
inverse of conductivity

D
inverse cell constant

Solution

Electrical resistivity (also known as resistivity, specific electrical resistance) is the resistance offered per unit volume of the material.
It is an intrinsic property that quantifies how strongly a given material opposes the flow of electrical current. It is the reciprocal of conductivity.
Question 5
1 / -0

Given: E°(Ag⁺|Ag) = 0.799 V and E°(Zn²⁺|Zn) = –0.763 V. Which of the following facts is correct?

A
Ag⁺ can be reduced by H₂(g).

B
Ag can oxidise H₂(g) to H⁺.

C
Zn²⁺ can be reduced by H₂(g).

D
Ag can reduce Zn²⁺.

Solution

Ag⁺ ions show more reduction potential than H⁺ ions. The higher the reduction potential, the greater the tendency to be reduced. So, in this process, Ag⁺ is reduced by H₂ (having reduction potential 0 V).
Question 6
1 / -0

Two Faraday of electricity is passed through a solution of CuSO₄. The mass of copper deposited at the cathode is
(Atomic mass of Cu = 63.5 amu)

A
0 g

B
63.5 g

C
2 g

D
127 g

Solution

Cu⁺² + 2e → Cu
So, 2 F charge deposite 1 mol of Cu. Mass deposited = 63.5 g.

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Electrochemistry Test - 3

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Question 1

Electrode of half-cell potential –0.44 V acts as anode.

Electrode of half-cell potential –0.44 V acts as cathode.

Electrode of half-cell potential 0.799 V acts as anode.

None of these

Solution

Question 2

More of Fe and Pb2+ ions are formed.

More of Pb and Fe2+ ions are formed.

Concentrations of Pb2+ and Fe2+ ions are increased.

Concentrations of Pb2+ and Fe2+ ions are decreased.

Solution

Question 3

l/a

a/l

la

l2/a

Solution

Question 4

inverse of resistance

inverse of conductance

inverse of conductivity

inverse cell constant

Solution

Question 5

Ag+ can be reduced by H2(g).

Ag can oxidise H2(g) to H+.

Zn2+ can be reduced by H2(g).

Ag can reduce Zn2+.

Solution

Question 6

0 g

63.5 g

2 g

127 g

Solution

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More of Fe and Pb²⁺ ions are formed.

More of Pb and Fe²⁺ ions are formed.

Concentrations of Pb²⁺ and Fe²⁺ ions are increased.

Concentrations of Pb²⁺ and Fe²⁺ ions are decreased.

l²/a

Ag⁺ can be reduced by H₂(g).

Ag can oxidise H₂(g) to H⁺.

Zn²⁺ can be reduced by H₂(g).

Ag can reduce Zn²⁺.