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Test: Electrochemistry - 1 - NEET MCQ


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20 Questions MCQ Test Chemistry Class 12 - Test: Electrochemistry - 1

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Test: Electrochemistry - 1 - Question 1

The standard reduction potential at 298 K for the following half cells are given:

Which is the strongest reducing agent:

Detailed Solution for Test: Electrochemistry - 1 - Question 1

Explanation of the Strongest Reducing Agent

  • The strength of a reducing agent is determined by its standard reduction potential. The lower the standard reduction potential, the stronger the reducing agent.
  • Reducing agents donate electrons and are oxidized in the process. The strongest reducing agent is the one that is most easily oxidized, which corresponds to the half-reaction with the lowest standard reduction potential.
  • In the given options,  Zn(s), Cr(s), H2(g), and Fe2+ (aq), Zn(s) is known to have the lowest standard reduction potential.
  • Therefore, Zn(s) is the strongest reducing agent among the given options, making option A the correct answer.
Test: Electrochemistry - 1 - Question 2

Which of the following conditions are satisfied when the cell reaction in the electrochemical cell is spontaneous?

Detailed Solution for Test: Electrochemistry - 1 - Question 2
  • For all spontaneous chemical reactions, the change in Gibbs free energy (ΔG°) is always negative.
  • For a spontaneous reaction in an electrolytic cell, the cell potential (E°cell) should be positive.
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Test: Electrochemistry - 1 - Question 3

What is the observation when the opposing external applied potential to an electrochemical cell is greater than the cell’s potential?

Detailed Solution for Test: Electrochemistry - 1 - Question 3

In an electrochemical cell, when an opposing externally potential is applied and increased slowly, the reaction continues to take place.

  • When the external potential is equal to the potential of the cell, the reaction stops.
  • Once the externally applied potential is greater than the potential of the cell, the reaction goes in the opposite direction and the cell behaves like an electrolytic cell.
Test: Electrochemistry - 1 - Question 4

The standard reduction potentials E°, for the half reactions are as:

The emf for the cell reaction,

Detailed Solution for Test: Electrochemistry - 1 - Question 4

Test: Electrochemistry - 1 - Question 5

The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is:

Detailed Solution for Test: Electrochemistry - 1 - Question 5

The correct order of equivalent conductance at infinite dilution is KCl > NaCl > LiCl.

Anion is same (chloride ion) for all the species. Larger is the size of the cation, greater is the equivalent conductance at infinite dilution and vice versa.

Test: Electrochemistry - 1 - Question 6

Saturated solution of KNO3 is used to make ‘salt-bridge’ because:

Detailed Solution for Test: Electrochemistry - 1 - Question 6

Explanation for the Use of Saturated KNO3 Solution for Salt-Bridge

  • The primary purpose of a salt bridge is to maintain electrical neutrality within the internal circuit of an electrochemical cell. It allows the migration of ions to balance the charge during the redox reactions.
  • A saturated solution of KNO3 is often used to make a salt bridge because of the similar velocities of K+ (potassium ions) and NO3- (nitrate ions) which helps maintain electrical neutrality in the cell, and prevents cell polarization.
Test: Electrochemistry - 1 - Question 7

The specific conductances of four electrolytes in ohm−1cm−1 are given below. Which one offers higher resistance to passage of electric current?

Detailed Solution for Test: Electrochemistry - 1 - Question 7

Specific conductance = Cell constant/Resistance
∴ Specific conductance ∝ 1/Resistance 
∴ The one with the lowest resistance (order 10-9) will have the highest specific conductance.
Hence, option A is correct.

Test: Electrochemistry - 1 - Question 8

In the electrolytic cell, flow of electrons is from:

Detailed Solution for Test: Electrochemistry - 1 - Question 8

Flow of Electrons in an Electrolytic Cell
In an electrolytic cell, the flow of electrons is from: Anode to Cathode through Internal Supply
This means that option B is correct. Here's why:


  • Anode: This is where oxidation takes place in an electrolytic cell. During oxidation, a substance loses electrons. This means that the anode is the source of electrons.
  • Cathode: This is where reduction takes place in an electrolytic cell. During reduction, a substance gains electrons, meaning that the cathode is where electrons are received.
  • Flow of electrons: Since electrons are produced at the anode (through oxidation) and consumed at the cathode (through reduction), the flow of electrons is from the anode to the cathode.
  • Internal supply: In an electrolytic cell, the power supply is connected to the anode and cathode, creating an electric current within the cell. This current forces the electrons to move from the anode to the cathode, hence the term "through internal supply".

In summary, in an electrolytic cell, the flow of electrons is from anode to cathode through internal supply as a result of the oxidation and reduction reactions taking place at the anode and cathode, respectively.

Test: Electrochemistry - 1 - Question 9

The emf of the above cell is 0.2905 V. Equilibrium constant for the cell reaction is:

Detailed Solution for Test: Electrochemistry - 1 - Question 9

Test: Electrochemistry - 1 - Question 10

Given, standard electrode potentials, Fe2+ + 2e- →Fe, E° = -0.440V
Fe3+ +3e- → Fe, E° = -0.036V
The standard electrode potential (E°) for Fe3+ + e- → Fe2+ is:

Detailed Solution for Test: Electrochemistry - 1 - Question 10

Fe2+ +2e→Fe, Eo = −0.440 ......(1)
Fe3+ + 3e− → Fe, Eo = −0.036.......(2)
The standard electrode potential for Fe3+ +e →Fe2+ is obtained by substracting the value of the standard electrode potential for first reaction from the standard electrode potential for second reaction. 
−0.036 −(−0.440) = 0.404V

Test: Electrochemistry - 1 - Question 11

Based on the data given blow strongest oxidizing agent will be:

Detailed Solution for Test: Electrochemistry - 1 - Question 11

As MnO4- has highest reduction potential among them so it is strongest oxidizing agent.

Test: Electrochemistry - 1 - Question 12

The measured resistance of a conductance cell was 100 Ω. (MKCl = 74.5 gmol−1 and cell constant =1.25 cm−1). If  the specific conductance in ohm−1 cm−1 is 125 × 10−x, then what is the value of x?

Detailed Solution for Test: Electrochemistry - 1 - Question 12

The expression for specific conductance κ is:

Here R is the resistance and 1/a is the cell constant.
Substitute values in the above expression.

So, the value of x = 4.

Test: Electrochemistry - 1 - Question 13

The solubility of a certain sparingly soluble substance MXn is nearly 1.4 × 10−4 M. If the solubility product is 1.1 × 10−11, what is the value of n?

Detailed Solution for Test: Electrochemistry - 1 - Question 13

As given,
MX​n → M+n + nX

Solubility product = S × (nS)n = nnSn+1 = 1.1 × 10−11 
By comparison, n = 2.

Test: Electrochemistry - 1 - Question 14

The electrochemical cell shown below is a concentration cell.
M|M2+ (saturated solut ion of a sparingly so luble salt, MX2 || M2+ (0.001 mol dm–3)| M.
The emf of the cell depends on the difference in concentration of M2+ ions at the two electrodes. The emf of the cell at 298  is 0.059 V.

Q. 

The value of ΔG (kJ mol-1) for the given cell is (take 1 F = 96500 C mol-1)

Detailed Solution for Test: Electrochemistry - 1 - Question 14

At cathode:M+2(aq)+2e→M(s)
At anode:M(s)+2X(aq)→MX2​(aq)+2e
n - factor of the cell reaction is 2.

ΔG=−nFEcell​

= −2×96500×0.059

= −11.4 kJmole−1

Test: Electrochemistry - 1 - Question 15

For the following cell,

Zn(s) | ZnSO4(aq) || CuSO4(aq) | Cu(s)

When the concentration of Zn2+ is 10 times the concentration of Cu2+, the expression for ΔG (in J mol-1) is [F is Faraday constant; R is gas constant; T is temperature; E° (cell) = 1.1 V]

Detailed Solution for Test: Electrochemistry - 1 - Question 15

Test: Electrochemistry - 1 - Question 16

Galvanization is applying a coating of:

Detailed Solution for Test: Electrochemistry - 1 - Question 16

Galvanization is applying a coating of: Zn

Galvanization is a process used to apply a protective coating of zinc (Zn) onto the surface of another metal. This process is commonly used to prevent corrosion and extend the lifespan of the underlying metal.

Why Zinc (Zn) is used for Galvanization:

1. Corrosion Resistance: Zinc is highly resistant to corrosion, making it an ideal choice for galvanization. When zinc is exposed to the atmosphere, it forms a protective layer of zinc oxide (ZnO) and zinc hydroxide (Zn(OH)2) on its surface. This protective layer acts as a barrier, preventing the underlying metal from coming into contact with corrosive substances such as moisture and oxygen.

2. Self-Healing: In the event that the zinc coating is scratched or damaged, zinc has the unique ability to self-heal. This means that if the underlying metal is exposed due to a small scratch or abrasion, the zinc coating will react with the surrounding atmosphere and form a new layer of protective zinc oxide, effectively sealing off the damaged area.

3. Sacrificial Protection: Zinc also provides sacrificial protection to the underlying metal. In the presence of an electrolyte, such as moisture, zinc will corrode preferentially over the base metal. This sacrificial corrosion prevents the base metal from corroding, as long as the zinc coating remains intact. This is known as cathodic protection.

4. Ease of Application: Zinc is relatively easy to apply as a coating. It can be applied using various methods, including hot-dip galvanizing, electro-galvanizing, and zinc-rich paint. These methods allow for efficient and uniform coating of the metal surface, ensuring maximum protection against corrosion.

Overall, the use of zinc (Zn) as a coating in the galvanization process provides excellent corrosion resistance, self-healing properties, sacrificial protection, and ease of application. This makes it an ideal choice for protecting various metal structures, including steel, from corrosion and extending their lifespan.

Test: Electrochemistry - 1 - Question 17

An electrochemical cell consists of two half-cell reactions.

The mass of copper (in grams) dissolved on passing 0.5 A current for 1h is [Given, atomic mass of Cu is 63.6, F = 96500 C mol–1]

Detailed Solution for Test: Electrochemistry - 1 - Question 17

m/E = Q/F
m/(63.6/2) = (0.5 x 60 x 60)/96500
m = (1800 x 63.6)/(96500 x 2)
m = 0.59

Test: Electrochemistry - 1 - Question 18

Using the following Latimer diagram for bromine, pH = 0; 

the species undergoing disproportionation is:

Detailed Solution for Test: Electrochemistry - 1 - Question 18


Assigning the oxidation state of Br in each molecule using oxidation number of O=−2,H=+1

Since the oxidation potential of HBrO is less than that of the reduction potential i.e left side potential is less than the right side potential of the conversion.
Thus HBrO is the species that will undergo disproportionation.

Test: Electrochemistry - 1 - Question 19

Conductometric titration curve of an equimolar mixture of HCl and HCN with NaOH (aq) can be given as

Detailed Solution for Test: Electrochemistry - 1 - Question 19

The titration is an example of titration of mixture of acids against strong base (NaOH).

When mixture of acid is titrated with NaOH, strong acid will react first followed by weak acid.

Here, HCl will react with NaOH in the start and get neutralised.

The neutralised reaction is,

HCl + NaOH → NaCl + H2O

During this reaction, the conductance of the solution gradually decreases due to the removal of H+ ions of highly ionised HCl to form unionised water.

This will be the first part of the graph.

When HCl is finished HCN starts reacting with NaOH. Here, the neutralisation reaction is
HCN + NaOH → NaCN + H2O

During the neutralisation of weak acid the conductance increases marginally due to the salts (NaCN) produced which will be the second part of graph.

After the equivalence point, the increase in conductance is appreciable due to the added OH ions.

Thus, the conductance will decrease steeply and then increase marginally and again increase steeply.
Thus, option (d) is correct.

Test: Electrochemistry - 1 - Question 20

In a typical Conductometric titration of a strong acid with a weak base, the curve resembles:

Detailed Solution for Test: Electrochemistry - 1 - Question 20

Here consider a strong acid as HCL and weak base as Ammonium hydroxide that is NH4OH. Suppose the acid has high concentration of H positive ions due to which it show high conductance, when a weak base is added to it, the cation of weak base combines with Cl Negative ion to form ammonium chloride precipitate thus decreasing the conductivity by utilizing H positive ion to form water molecule due to this decrease in conductivity the graph shows on negative fall in conductance after all the edge positive and Cl negative ions are combine end point is reached where no free ions are possible for conduction after this when the basis added to form NH4 positive and was OH negative since it is a weak base it has low conductivity and the graph increases very slowly towards conduction minute changes in increasing conduction can be seen when a weak bases are added

plot a) show high conductivity which is not possible in weak bases
plot c) and d) involve 3 mixture components which is not the case
The perfect diagram would look like this and B is the closest.


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