Electrochemistry - 1 - Free MCQ Practice Test with solutions, NEET Chemistry

Answer: Option A, Zn(s).

The relevant standard reduction half-reactions and their standard potentials at 298 K are:

Zn²⁺(aq) + 2e^- ⇌ Zn(s), E° = -0.762 V

Cr³⁺(aq) + 3e^- ⇌ Cr(s), E° = -0.740 V

2H⁺(aq) + 2e^- ⇌ H₂(g), E° = 0.000 V

Fe³⁺(aq) + e^- ⇌ Fe²⁺(aq), E° = +0.770 V

A species is a stronger reducing agent if it is more easily oxidised; when comparing standard reduction potentials, the species with the most negative reduction potential is the strongest reducing agent in its given form.

Comparing the values: -0.762 V < -0.740 V < 0.000 V < +0.770 V. Therefore Zn(s) (E° = -0.762 V) is the most easily oxidised among the options and so is the strongest reducing agent.

Hence, option A is correct.

• 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.

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.

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.

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 NO^3- (nitrate ions) which helps maintain electrical neutrality in the cell, and prevents cell polarization.

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.

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.

• Background:
  • The reaction involves the reduction of iron ions, Fe²⁺ and Fe³⁺, each gaining electrons.
  • The Gibbs free energy change (∆G) for a reaction is related to the electromotive force (E) of the reaction and the number of moles of electrons transferred (n) by the formula:
    ∆G = -nFE
    where:
    F = Faraday's constant (the charge per mole of electrons)
• Calculations for Individual Reactions:
  • For the reduction of Fe²⁺ to Fe:
    Fe²⁺ + 2e^- → Fe
    E = -0.440 V (Given)
    ∆G = -2 x F x (-0.440 V) = 0.880F
  • For the reduction of Fe³⁺ to Fe:
    Fe³⁺ + 3e^- → Fe
    E = -0.036 V (Given)
    ∆G = -3 x F x (-0.036 V) = 0.108F
• Subtracting to Find Reaction of Interest:
  • By subtracting the Gibbs free energy change of reaction (i) from (ii), we find the ∆G for the reaction converting Fe³⁺ to Fe²⁺:
    ∆G for Fe³⁺ + e^- → Fe²⁺:
    ∆G = 0.108F - 0.880F = -0.772F
• Calculating the Standard Electrode Potential (E°):
  • The standard electrode potential for the reaction Fe³⁺ + e^- → Fe²⁺ can be calculated using:
    E° = -∆G / (n x F)
    Since n = 1 (1 mole of electrons),
    E° = -(-0.772F) / (1 x F) = 0.772 volts

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

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.

As given,
MX​_n → M⁺ⁿ + nX⁻

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

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=−nFE_cell​

= −2×96500×0.059

= −11.4 kJmole⁻¹

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.

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

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.

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.

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.