Concept:

Faraday’s second law:

This law states that when the same quantity of electric current is passed through different electrolytes, the masses of the substances deposited are proportional to their respective chemical equivalents or equivalent weights.

i.e.,  \(\frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{E}_{1}}}{{{E}_{2}}}⇒ \frac{{{m}_{1}}}{{{m}_{2}}}=\frac{{{e}_{1}}q}{{{e}_{2}}q}=\frac{{{E}_{1}}}{{{E}_{2}}}...(\because m=eq)\)

Where, 

m₁, m₂ = corresponding masses of two substances liberated

E₁, E₂ = chemical equivalents 

e₁, e₂ = e.c.e or Electrochemical equivalent 

Explanation:

As explained above according to Faraday's law the relation between chemical equivalent and e.c.e can be expressed as

E1/E2 = e1/e2

⇒ e₂ = e1E2/E1

IExplanation:

Given:

 Conductivity, k = 0.146 x 10-3 S cm-1

Resistance, R = 1000

Cell constant = k x R

 = 0.146 X 0-3 X 1000

 = 0.146 cm^-1

 Important Points

EMF of cell:

\(\rm E^0_{cell}=\begin{bmatrix}\rm Standard \ reduction\\\ \rm Potential \ of\ cathode\end{bmatrix}-\begin{bmatrix}\rm Standard \ oxidation\\\ \rm Potential \ of\ anode\end{bmatrix}\)

\(\rm E^0_{cell}=E^0_{cathode}-E^0_{anode}=E^0_{right}-E^0_{left}\)

• Nernst Equation:

​\(\rm M^{n+}_{(aq)}+ne^-\longrightarrow M_{(s)}\)

\(\rm E=E^0-\frac{RT}{nF}ln \frac{[M_{(s)}]}{[M^{n+}_{(aq)}]}\)

\(\rm E=E^0-\frac{2.303RT}{nF}log \frac{[M_{(s)}]}{[M^{n+}_{(aq)}]}\)

• Gibbs Free Energy Change:

​-ΔG = nFE

• Faraday First Law of Electrolysis:

​
m ∝ It or m = ZIt

Where

I = Current in amperes

t = time in seconds

m = mass of the primary product in grams

Z = constant of proportionality

(Electrochemical equivalent). It is the mass of a substance liberated by 1 ampere second of a current (1 coulomb).

The correct answer is option 4 i.e., O2 and H2

Explanation:

• O₂ and H₂ are the gaseous products liberated at anode and cathode respectively when dilute aqueous solution of Sulphuric acid is electrolyzed.
• The cathode rode is the negatively charged electrode and the anode rod is the positively charged electrode.
• The reduction taken place at the cathode which means the electron is gained.
• In the aqueous solution of Sulphuric acid, the hydrogen ions are attracted to the cathode and give Hydrogen gas.
• Bubbles can be observed when hydrogen gas is produced.
• 2H⁺ (aq) + 2e^– → H₂(g)
• Oxygen ions are attracted to the anode rod.
• Oxidation takes place as the electron is lost.
• Please note that the sulphate ion is too stable and does not dissociate readily.
• Instead, either hydroxide ions or water molecules are discharged and oxidized to form oxygen gas.
• 2H₂O (l) – 4e^– → 4H⁺ (aq) + O₂(g) and 4OH^– (aq) – 4e^– → 2H₂O(l) + O₂(g)

The correct answer is The colour of the solution changes from colourless to blue and a silver solid deposits on copper wire.

Concept:

Displacement reaction:

• It is a  chemical reaction wherein a more reactive element displaces the less reactive one from its salt solution.
• For example, CuSO4 + Fe → FeSO4 + Cu, here Fe is more reactive than Cu.

Explanation:

Reactivity series: It refers to the arrangement of metals in the descending order of their reactivities.

Copper is above silver in the reactivity series indicating that it is more reactive than silver. 

Thus, when a copper wire is dipped in silver nitrate solution, it will displace silver and form a blue-coloured copper sulphate solution. 

Concept:

The change in free energy (ΔG) is also a measure of the maximum amount of work that can be performed during a chemical process (ΔG = w_max). ∆G is the change of Gibbs (free) energy for a system and ∆G° is the Gibbs energy change for a system under standard conditions (1 atm, 298K).

The connection between cell potential, Gibbs energy and constant equilibrium are directly related in the following multi-part equation:

∆G⁰ = -nFE⁰_cell

Where, n = number of moles of electrons transferred in the reaction

Calculation:

Given, n = 2

Standard cell potential, E° = 2V

Faraday’s constant, F = 96000 C mol^-1

By substituting these values, we get

∆G0 = -nFE0cell

= - 2(96000) × 2V= - 384000 J/mol

= - 384 kJ/mol

Concept:

• The conductivity is depending on the number of ions present in the unit volume of solution. The decreasing order of the electrical conductivity of the aqueous solution is based on the acid strength.
• The more the acid strength more will be the dissociation of acid into ion and more will be the conductivity.
• Acid strength refers to the tendency of an acid, symbolized by the chemical formula HA, to dissociate into a proton H⁺, and an anion, A⁻. The dissociation of a strong acid in solution is effectively complete, except in its most concentrated solutions.
• Electrical conductivity is the measure of the amount of electrical current a material can carry or its ability to carry a current. Electrical conductivity is also known as specific conductance.

The Order of acidic strength is HCOOH (formic acid) > C₆ H₅ COOH (benzoic acid) > CH₃COOH (Acetic acid).

Explanation:-

Chemical reaction: The chemical reaction is said to be the change that brings new products.

Displacement reaction: 

• The reaction in which cation or anion from the reactant side is displaced to the products is called displacement reaction.
• A displacement reaction is also known as a metal displacement reaction.
• Metals of high reactivity displace the metals of low reactivity from their anions.

Double displacement reaction: The reaction in which both cation and anion is displaced to the products is called double displacement reaction.

Properties of Elements:

• Zinc: Zinc is more reactive metal than Iron and copper.
• Iron: Iron is least reactive than zinc but more than copper.
• Copper: Copper is least reactive than zinc and iron.

Metal reactivity series:

The metals are arranged in increasing order of their reactivity which is known as the metal reactivity series. 

The correct answer Molar conductivity increases with a decrease in concentration.

Molar conductivity:

It is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length. 

Molar conductivity has the SI unit S m² mol⁻¹.

Molar conductivity increases with a decrease in concentration.

 ⇒ Λ_m × M = k × 1000

Where,

Λm - Molar conductivity

V - Volume , k - Specific conductance

M is the molarity

→ As seen in the above equation we can arrive at a conclusion that Λ _m⁰ is inversely proportional to Molarity. 
→ And hence if we increase concentration of solution molar conductivity decreases and vice-versa.

 Additional Information

Specific conductance:

The conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of crosssection 1 sq. cm. Specific conductance is the inverse of resistivity.

⇒ k = 1/p 

where, p - Resistivity

Explanation:

The potential (E) of the PH electrode may be written by means of the Nernst equation as:

\(\rm E = E_0 - \frac{2.3036RT}{F} (\Delta pH)\)

We have to find change in open-circuit voltage across the electrode (i.e.) (E0 - E)

\(\rm E_0- E= \frac{2.3036RT}{F} (\Delta pH)\)

Given,

pH₁ = 6, pH₂ = 8

R = gas constant = 8.31 \(\rm \frac{volt - columb}{mol - k}\)

F = 96500 coulombs mol-1

\(\rm E_0- E= \frac{2.3036 \times 8.31 \times 298 }{96500} (8-6) = 118 \ mV\)

The correct answer is Statement (i) and (ii) are true.

Fuel Cell:

• A device that converts the energy of combustion of fields like hydrogen and methane directly into electrical energy.
• It is an electrochemical reaction.
• It requires a continuous input of fuel and an oxidizing agent which is oxygen.
• Cathode Reaction of fuel cell 
  • O2 + 2H2O + 4e– → 4OH–
• Anode Reaction of fuel cell
  • 2H2 + 4OH– → 4H2O + 4e–
• Hence Statement (i) and (ii) are true while statement (iii) is false.

Applications of fuel cell:

• They are used to power many space expeditions including the Appolo space program.
• Electrical vehicles.
• Some military applications.